{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh n\u00e0o sau \u0111\u00e2y l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh tr\u00f9ng ph\u01b0\u01a1ng?","select":["A. $3{{x}^{4}}+{{x}^{3}}+x=0$ ","B. ${{x}^{4}}+2{{x}^{2}}-3=0$","C. ${{x}^{4}}-{{x}^{3}}-x+1=0$ ","D. ${{x}^{3}}+{{x}^{2}}-x+1=0$ "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh tr\u00f9ng ph\u01b0\u01a1ng l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 d\u1ea1ng $a{{x}^{4}}+b{{x}^{2}}+c=0\\,\\,\\,\\left( a\\ne 0 \\right)$<br\/>Suy ra, trong c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean, ch\u1ec9 c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{4}}+2{{x}^{2}}-3=0$ l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh tr\u00f9ng ph\u01b0\u01a1ng.<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh tr\u00f9ng ph\u01b0\u01a1ng kh\u00f4ng ph\u1ea3i l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai, song ta c\u00f3 th\u1ec3 \u0111\u01b0a n\u00f3 v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai b\u1eb1ng c\u00e1ch \u0111\u1eb7t $t={{x}^{2}}.$ Khi \u0111\u00f3, ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh $a{{t}^{2}}+bt+c=0$.<\/span>","column":2}]}],"id_ques":921},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-2"],["2"],["-3"],["3"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{4}}-13{{x}^{2}}+36=0$ (1)<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{$_input_;_input_;_input_;_input_$\\}$","hint":"","explain":"<span class='basic_left'>\u0110\u1eb7t ${{x}^{2}}=t, t\\ge 0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh: ${{t}^{2}}-13t+36=0$ (2)<br\/>$\\Delta ={{13}^{2}}-4.36=25,\\sqrt{\\Delta }=5$ <br\/>Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 ${{t}_{1}}=\\dfrac{13-5}{2}=4$ v\u00e0 ${{t}_{2}}=\\dfrac{13+5}{2}=9$ <br\/>C\u1ea3 hai gi\u00e1 tr\u1ecb $4$ v\u00e0 $9$ \u0111\u1ec1u th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n $t\\ge 0$<br\/>+ V\u1edbi $t={{t}_{1}}=4$ , ta c\u00f3 ${{x}^{2}}=4\\Leftrightarrow x=\\pm 2$ <br\/>+ V\u1edbi $t={{t}_{2}}=9$ , ta c\u00f3 ${{x}^{2}}=9\\Leftrightarrow x=\\pm 3$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1) l\u00e0 $S=\\left\\{ -2;2;3;-3 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2;$ $2;$ $3;$ $-3.$<\/span><\/span>"}]}],"id_ques":922},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh: $3{{x}^{4}}+4{{x}^{2}}+1=0$ . Nh\u1eadn x\u00e9t v\u1ec1 s\u1ed1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh:","select":["A. Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m","B. Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m","C. Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u","D. Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 4 nghi\u1ec7m ph\u00e2n bi\u1ec7t"],"hint":"","explain":"<span class='basic_left'>\u0110\u1eb7t ${{x}^{2}}=t.$ \u0110i\u1ec1u ki\u1ec7n: $t\\ge 0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh: $3{{t}^{2}}+4t+1=0$ (*)<br\/>V\u00ec $a-b+c=3-4+1=0$ n\u00ean (*) c\u00f3 hai nghi\u1ec7m l\u00e0 ${{t}_{1}}=-1;{{t}_{2}}=-\\dfrac{1}{3}$ <br\/>C\u1ea3 hai gi\u00e1 tr\u1ecb $-1$ v\u00e0 $-\\dfrac{1}{3}$ \u0111\u1ec1u kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n $t\\ge 0$<br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho v\u00f4 nghi\u1ec7m.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":923},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh: $5{{x}^{4}}+2{{x}^{2}}-16=10-{{x}^{2}}$ (1) . T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{ -\\sqrt{2};\\sqrt{2};-\\dfrac{13}{5};\\dfrac{13}{5} \\right\\}$","B. $S=\\left\\{ -2;2 \\right\\}$","C. $S=\\left\\{ -\\sqrt{2};\\sqrt{2} \\right\\}$","D. $S=\\varnothing $"],"hint":"Chuy\u1ec3n v\u1ebf \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh tr\u00f9ng ph\u01b0\u01a1ng","explain":"<span class='basic_left'>Ta c\u00f3 $\\left( 1 \\right)\\Leftrightarrow 5{{x}^{4}}+2{{x}^{2}}-16-10+{{x}^{2}}=0\\Leftrightarrow 5{{x}^{4}}+3{{x}^{2}}-26=0$ <br\/>\u0110\u1eb7t ${{x}^{2}}=t.$ \u0110i\u1ec1u ki\u1ec7n: $t\\ge 0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh: $5{{t}^{2}}+3t-26=0$ (2)<br\/>$\\Delta ={{3}^{2}}+4.5.26=529,\\sqrt{\\Delta }=23$ <br\/>Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 <br\/>${{t}_{1}}=\\dfrac{-3-23}{10}=-\\dfrac{13}{5}$ (lo\u1ea1i) v\u00e0 ${{t}_{2}}=\\dfrac{-3+23}{10}=2$ (th\u1ecfa m\u00e3n)<br\/>V\u1edbi $t={{t}_{2}}=2,$ ta c\u00f3 ${{x}^{2}}=2\\Leftrightarrow x=\\pm \\sqrt{2}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1) l\u00e0 $S=\\left\\{ -\\sqrt{2};\\sqrt{2} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":924},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["25"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x-3\\sqrt{x}-10=0$ (1)<br\/>Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x =$ _input_ ","hint":"\u0110\u1eb7t $t=\\sqrt{x}$ ","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ge 0$ <br\/>\u0110\u1eb7t $t=\\sqrt{x}$ ($t\\ge 0$)<br\/>Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: ${{t}^{2}}-3t-10=0$ (2)<br\/>$\\Delta ={{3}^{2}}+4.10=49,\\sqrt{\\Delta }=7$ <br\/>Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0<br\/> ${{t}_{1}}=\\dfrac{3-7}{2}=-2$ (lo\u1ea1i) v\u00e0 ${{t}_{2}}=\\dfrac{3+7}{2}=5$ (th\u1ecfa m\u00e3n)<br\/>$t={{t}_{2}}=5$ $\\Leftrightarrow \\sqrt{x}=5\\Leftrightarrow x=25$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 nghi\u1ec7m l\u00e0 $x=25$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $25.$ <\/span><\/span>"}]}],"id_ques":925},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $x_{1} = \\frac{9}{4}; x_{2}= \\frac{7}{4}$","B. $x_{1} = \\frac{5}{4}; x_{2}= \\frac{7}{4}$","C. $x_{1} = \\frac{7}{4}; x_{2}= \\frac{5}{4}$"],"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{\\left( 4x-5 \\right)}^{2}}-6\\left( 4x-5 \\right)+8=0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=$?; ${{x}_{2}}=$?, bi\u1ebft $x_1 > x_2$","hint":"\u0110\u1eb7t \u1ea9n ph\u1ee5 $t=4x-5$ ","explain":"<span class='basic_left'>\u0110\u1eb7t $t=4x-5.$ <br\/>Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: ${{t}^{2}}-6t+8=0$ <br\/>$\\Delta '=9-8=1 >0$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 ${{t}_{1}}=3+1=4;{{t}_{2}}=3-1=2$ <br\/>+ $t={{t}_{1}}=4$$\\Leftrightarrow 4x-5=4 \\Leftrightarrow x=\\dfrac{9}{4}$<br\/>+ $t={{t}_{2}}=2$ $\\Leftrightarrow 4x-5=2 \\Leftrightarrow x=\\dfrac{7}{4}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=\\dfrac{9}{4};{{x}_{2}}=\\dfrac{7}{4}$ <\/span>"}]}],"id_ques":926},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["7"],["-3"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{12}{x-1}-\\dfrac{8}{x+1}=1$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{$_input_;_input_$\\}$","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/>B\u01b0\u1edbc 2: Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c hai v\u1ebf r\u1ed3i kh\u1eed m\u1eabu th\u1ee9c<br\/>B\u01b0\u1edbc 3: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u1eeba nh\u1eadn \u0111\u01b0\u1ee3c<br\/>B\u01b0\u1edbc 4: Trong c\u00e1c gi\u00e1 tr\u1ecb t\u00ecm \u0111\u01b0\u1ee3c c\u1ee7a \u1ea9n, lo\u1ea1i c\u00e1c gi\u00e1 tr\u1ecb kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh. C\u00e1c gi\u00e1 tr\u1ecb th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\pm 1$ <br\/>Quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho, ta \u0111\u01b0\u1ee3c:<br\/> $\\begin{align} & 12\\left( x+1 \\right)-8\\left( x-1 \\right)=\\left( x+1 \\right)\\left( x-1 \\right) \\\\ & \\Leftrightarrow 12x+12-8x+8={{x}^{2}}-1 \\\\ & \\Leftrightarrow {{x}^{2}}-4x-21=0 \\\\ \\end{align}$<br\/>Ta c\u00f3: $\\Delta '=4+21=25$ <br\/>Suy ra $x_1=2+5=7$ (th\u1ecfa m\u00e3n); $x_2=2-5=-3$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 $S=\\left\\{ 7;-3 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $7;$ $-3.$<\/span><\/span>"}]}],"id_ques":927},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x+2}{x-5}+3=\\dfrac{6}{2-x}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 ","select":["A. $S=\\left\\{ -4;\\dfrac{1}{4} \\right\\}$ ","B. $S=\\left\\{ -4;-\\dfrac{1}{4} \\right\\}$ ","C. $S=\\left\\{ 4;-\\dfrac{1}{4} \\right\\}$"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\left\\{ 2;5 \\right\\}$ <br\/>Quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho, ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{align} & \\left( x+2 \\right)\\left( x-2 \\right)+3\\left( x-5 \\right)\\left( x-2 \\right)=-6\\left( x-5 \\right) \\\\ & \\Leftrightarrow {{x}^{2}}-4+3\\left( {{x}^{2}}-7x+10 \\right)+6\\left( x-5 \\right)=0 \\\\ & \\Leftrightarrow {{x}^{2}}-4+3{{x}^{2}}-21x+30+6x-30=0 \\\\ & \\Leftrightarrow 4{{x}^{2}}-15x-4=0 \\\\ \\end{align}$ <br\/>$\\Delta ={{15}^{2}}+4.4.4=289\\Rightarrow \\sqrt{\\Delta }=17$ <br\/>Suy ra ${{x}_{1}}=\\dfrac{15+17}{8}=4$ (th\u1ecfa m\u00e3n); ${{x}_{2}}=\\dfrac{15-17}{8}=-\\dfrac{1}{4}$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 $S=\\left\\{ 4;-\\dfrac{1}{4} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":3}]}],"id_ques":928},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $\\dfrac{1}{2}$","B. $\\dfrac{3}{2}$","C. $\\dfrac{5}{2}$"],"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{2x}{{{x}^{2}}-1}-\\dfrac{1}{x+1}=2$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $x=$?","hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\left\\{ 1;-1 \\right\\}$<br\/>$\\begin{align} & \\dfrac{2x}{{{x}^{2}}-1}-\\dfrac{1}{x-1}=2 \\\\ & \\Leftrightarrow \\dfrac{2x}{\\left( x-1 \\right)\\left( x+1 \\right)}-\\dfrac{1}{x-1}=2 \\\\ \\end{align}$ <br\/>Quy \u0111\u1ed3ng v\u00e0 kh\u1eed m\u1eabu, ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{align} & 2x-\\left( x-1 \\right)=2\\left( {{x}^{2}}-1 \\right) \\\\ & \\Leftrightarrow 2x-x+1=2{{x}^{2}}-2 \\\\ & \\Leftrightarrow 2{{x}^{2}}-x-3=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( * \\right) \\\\ \\end{align}$ <br\/>Do $a-b+c=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh (*) c\u00f3 nghi\u1ec7m l\u00e0:<br\/>${{x}_{1}}=-1$ (kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n); ${{x}_{2}}=\\dfrac{3}{2}$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m l\u00e0 $x=\\dfrac{3}{2}$<\/span>"}]}],"id_ques":929},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["4"],["-5"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{14}{{{x}^{2}}-9}=1-\\dfrac{1}{3-x}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\{$_input_;_input_$\\}$","hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\pm 3$ <br\/>$\\begin{align} & \\,\\,\\,\\,\\dfrac{14}{{{x}^{2}}-9}=1-\\dfrac{1}{3-x} \\\\ & \\Leftrightarrow \\,\\dfrac{14}{\\left( x-3 \\right)\\left( x+3 \\right)}=1+\\dfrac{1}{x-3} \\\\ \\end{align}$<br\/>$ \\begin{align}& \\Rightarrow14={{x}^{2}}-9+x+3 \\\\ & \\Leftrightarrow {{x}^{2}}+x-20=0 \\\\ \\end{align}$<br\/>$\\Delta =1+4.20=81$ <br\/>Suy ra ${{x}_{1}}=\\dfrac{-1+9}{2}=4$ (th\u1ecfa m\u00e3n); ${{x}_{2}}=\\dfrac{-1-9}{2}=-5$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 $S=\\left\\{ 4;-5 \\right\\}$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $4;$ $-5.$ <\/span><\/span>"}]}],"id_ques":930},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x-1}{4{{x}^{2}}-9}=\\dfrac{2}{2x+3}-\\dfrac{x+1}{3-2x}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 ","select":["A. $S=\\left\\{-1+\\sqrt{5};-1-\\sqrt{5} \\right\\}$ ","B. $S=\\left\\{2+\\sqrt{5};2-\\sqrt{5} \\right\\}$","C. $S=\\left\\{-2+\\sqrt{5};-2-\\sqrt{5} \\right\\}$","D. $S=\\left\\{-4+\\sqrt{5};-4-\\sqrt{5} \\right\\}$"],"hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\left\\{ \\pm \\dfrac{3}{2} \\right\\}$ <br\/>$\\begin{align} & \\,\\dfrac{x-1}{4{{x}^{2}}-9}=\\dfrac{2}{2x+3}-\\dfrac{x+1}{3-2x} \\\\ & \\Leftrightarrow \\dfrac{x-1}{\\left( 2x-3 \\right)\\left( 2x+3 \\right)}=\\dfrac{2}{2x+3}+\\dfrac{x+1}{2x-3} \\\\ \\end{align}$<br\/>$ \\begin{align}& \\Rightarrow x-1=2\\left( 2x-3 \\right)+\\left( x+1 \\right)\\left( 2x+3 \\right) \\\\ & \\Leftrightarrow x-1=4x-6+2{{x}^{2}}+5x+3 \\\\ & \\Leftrightarrow 2{{x}^{2}}+8x-2=0 \\\\ & \\Leftrightarrow {{x}^{2}}+4x-1=0 \\\\ & \\Delta '=4+1=5 \\\\ \\end{align}$<br\/>Suy ra ${{x}_{1}}=-2+\\sqrt{5}$ (th\u1ecfa m\u00e3n); ${{x}_{2}}=-2-\\sqrt{5}$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 $S=\\left\\{-2+\\sqrt{5};-2-\\sqrt{5} \\right\\}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":931},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["0"],["-1"],["-2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{3}}+3{{x}^{2}}+2x=0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\{$_input_;_input_;_input_$\\}$","hint":"Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i ph\u01b0\u01a1ng tr\u00ecnh th\u00e0nh nh\u00e2n t\u1eed \u0111\u1ec3 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch.","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>${{x}^{3}}+3{{x}^{2}}+2x=0$<br\/>$\\begin{aligned} & \\Leftrightarrow x\\left( {{x}^{2}}+3x+2 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & {{x}^{2}}+3x+2=0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh $x^2+3x+2=0$ c\u00f3 $a-b+c=1-3+2=0$ n\u00ean c\u00f3 nghi\u1ec7m l\u00e0 $-1$ v\u00e0 $-2$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 0;-1;-2 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $0;$ $-1;$ $-2.$ <\/span><\/span>"}]}],"id_ques":932},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["0"],["1"],["2"],["-5"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{\\left( {{x}^{2}}+x+1 \\right)}^{2}}={{\\left( 4x-1 \\right)}^{2}}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\{$_input_;_input_;_input_;_input_$\\}$","hint":"Chuy\u1ec3n v\u1ebf v\u00e0 \u00e1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2-b^2$ \u0111\u1ec3 \u0111\u01b0a v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>${{\\left( {{x}^{2}}+x+1 \\right)}^{2}}={{\\left( 4x-1 \\right)}^{2}}$<br\/>$\\begin{aligned} & \\Leftrightarrow {{\\left( {{x}^{2}}+x+1 \\right)}^{2}}-{{\\left( 4x-1 \\right)}^{2}}=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+x+1-4x+1 \\right)\\left( {{x}^{2}}+x+1+4x-1 \\right)=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}-3x+2 \\right)\\left( {{x}^{2}}+5x \\right)=0 \\\\ & \\Leftrightarrow x\\left( x+5 \\right)\\left( {{x}^{2}}-3x+2 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x=-5 \\\\ & {{x}^{2}}-3x+2=0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh $x^2-3x+2=0$ c\u00f3 $a+b+c=1-3+2=0$ n\u00ean c\u00f3 hai nghi\u1ec7m l\u00e0 $1$ v\u00e0 $2$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 0;1;2;-5 \\right\\}$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $0;$ $1;$ $2;$ $-5.$ <\/span><\/span>"}]}],"id_ques":933},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["1"],["-1"],["-2"],["-4"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{\\left( {{x}^{2}}+3x+2 \\right)}^{2}}=6\\left( {{x}^{2}}+3x+2 \\right)$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\{$_input_;_input_;_input_;_input_$\\}$","hint":"Chuy\u1ec3n v\u1ebf \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch ho\u1eb7c \u0111\u1eb7t \u1ea9n ph\u1ee5 $t={{x}^{2}}+3x+2$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & {{\\left( {{x}^{2}}+3x+2 \\right)}^{2}}=6\\left( {{x}^{2}}+3x+2 \\right) \\\\ & \\Leftrightarrow {{\\left( {{x}^{2}}+3x+2 \\right)}^{2}}-6\\left( {{x}^{2}}+3x+2 \\right)=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+3x+2 \\right)\\left[ \\left( {{x}^{2}}+3x+2 \\right)-6 \\right]=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+3x+2 \\right)\\left( {{x}^{2}}+3x-4 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}+3x+2=0 \\\\ & {{x}^{2}}+3x-4=0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+3x+2=0$ c\u00f3 $a-b+c=0$ n\u00ean c\u00f3 nghi\u1ec7m l\u00e0 $-1$ v\u00e0 $-2$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+3x-4=0$ c\u00f3 $a+b+c=0$ n\u00ean c\u00f3 nghi\u1ec7m l\u00e0 $1$ v\u00e0 $-4$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ -1;1;-2;-4 \\right\\}$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1;$ $-1;$ $-2;$ $-4.$ <\/span><br\/><span class='basic_green'>C\u00e1ch kh\u00e1c:<\/span><br\/>\u0110\u1eb7t $t={{x}^{2}}+3x+2$ , ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/> ${{t}^{2}}=6t\\Leftrightarrow {{t}^{2}}-6t=0\\Leftrightarrow t\\left( t-6 \\right)=0\\Leftrightarrow \\left[ \\begin{aligned} & t=0 \\\\ & t=6 \\\\ \\end{aligned} \\right.$ <br\/>+ V\u1edbi $t=0$ $\\Leftrightarrow{{x}^{2}}+3x+2=0$ <br\/> Ph\u01b0\u01a1ng tr\u00ecnh $x^2+3x+2=0$ c\u00f3 $a-b+c=1-3+2=0 $ n\u00ean c\u00f3 nghi\u1ec7m l\u00e0 $-1$ v\u00e0 $-2$<br\/>+V\u1edbi $t=6$ $\\Leftrightarrow{{x}^{2}}+3x+2=6\\Leftrightarrow {{x}^{2}}+3x-4=0$ <br\/> Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+3x-4=0$ c\u00f3 $a+b+c=1+3-4=0$ n\u00ean c\u00f3 nghi\u1ec7m l\u00e0 $1$ v\u00e0 $-4$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ -1;1;-2;-4 \\right\\}$ <\/span>"}]}],"id_ques":934},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{3}}-{{x}^{2}}-3x+3=0$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0","select":["A. $S=\\left\\{1 \\right\\}$ ","B. $S=\\left\\{ -1 \\right\\}$ ","C. $S=\\left\\{ -1;\\sqrt{3};-\\sqrt{3} \\right\\}$ ","D. $S=\\left\\{ 1;\\sqrt{3};-\\sqrt{3} \\right\\}$ "],"hint":"Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i ph\u01b0\u01a1ng tr\u00ecnh th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$\\begin{aligned} & {{x}^{3}}-{{x}^{2}}-3x+3=0 \\\\ & \\Leftrightarrow {{x}^{3}}-{{x}^{2}}-\\left( 3x-3 \\right)=0 \\\\ & \\Leftrightarrow {{x}^{2}}\\left( x-1 \\right)-3\\left( x-1 \\right)=0 \\\\ & \\Leftrightarrow \\left( x-1 \\right)\\left( {{x}^{2}}-3 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x-1=0 \\\\ & {{x}^{2}}-3=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=1 \\\\ & x=\\pm \\sqrt{3} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\left\\{ 1;\\sqrt{3};-\\sqrt{3} \\right\\}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":935},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>S\u1ed1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{4}}+5{{x}^{3}}+15x-9=0$ l\u00e0:","select":["A. 0 nghi\u1ec7m","B. 2 nghi\u1ec7m","C. 3 nghi\u1ec7m ","D. 4 nghi\u1ec7m"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho b\u1eb1ng c\u00e1ch ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$\\begin{aligned} & {{x}^{4}}+5{{x}^{3}}+15x-9=0 \\\\ & \\Leftrightarrow \\left( {{x}^{4}}-9 \\right)+\\left( 5{{x}^{3}}+15x \\right)=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}-3 \\right)\\left( {{x}^{2}}+3 \\right)+5x\\left( {{x}^{2}}+3 \\right)=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}+3 \\right)\\left( {{x}^{2}}-3+5x \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}+3=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & {{x}^{2}}+5x-3=0\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/> Ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00f4 nghi\u1ec7m v\u00ec $x^2+3>0$ v\u1edbi m\u1ecdi $x$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 $\\Delta =25+12=37$ $>0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $x=\\dfrac{-5\\pm \\sqrt{37}}{2}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":936},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{6}}-{{x}^{3}}-6=0$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<\/span>","select":["A. $S=\\left\\{-\\sqrt[3]{3};\\sqrt[3]{2}\\right\\}.$ ","B. $S=\\left\\{\\sqrt[3]{3};-\\sqrt[3]{2}\\right\\}.$","C. $S=\\left\\{\\sqrt{3};-\\sqrt{2}\\right\\}.$ ","D. $S=\\left\\{-\\sqrt{3};-\\sqrt{2}\\right\\}.$ "],"hint":"\u0110\u1eb7t \u1ea9n ph\u1ee5 $t={{x}^{3}}$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/> B\u01b0\u1edbc 1: \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai b\u1eb1ng c\u00e1ch \u0111\u1eb7t \u1ea9n ph\u1ee5 $t={{x}^{3}}$<br\/>B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh m\u1edbi v\u1edbi \u1ea9n $t$.<br\/>B\u01b0\u1edbc 3: T\u00ecm $x$: \u00c1p d\u1ee5ng $x^3=t \\Leftrightarrow x=\\sqrt[3]{t}$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110\u1eb7t $t={{x}^{3}},$ ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: ${{t}^{2}}-t-6=0$ <br\/>$\\Delta =1+4.6=25\\Rightarrow \\sqrt{\\Delta }=5$ <br\/>Suy ra ${{t}_{1}}=\\dfrac{1+5}{2}=3;{{t}_{2}}=\\dfrac{1-5}{2}=-2$ <br\/>+V\u1edbi $t={{t}_{1}}=3$ $\\Leftrightarrow{{x}^{3}}=3\\Leftrightarrow x=\\sqrt[3]{3}$ <br\/>+V\u1edbi $t={{t}_{1}}=-2$ $\\Leftrightarrow {{x}^{3}}=-2\\Leftrightarrow x=\\sqrt[3]{-2}=-\\sqrt[3]{2}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\left\\{\\sqrt[3]{3};-\\sqrt[3]{2}\\right\\}.$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":937},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["1"],["-2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{\\left( {{x}^{2}}+x \\right)}^{2}}+4{{x}^{2}}+4x=12$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\{$_input_;_input_$\\}$","hint":"\u0110\u1eb7t \u1ea9n ph\u1ee5 ${{x}^{2}}+x=t$","explain":"<span class='basic_left'>\u0110\u1eb7t ${{x}^{2}}+x=t,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>${{t}^{2}}+4t=12\\Leftrightarrow {{t}^{2}}+4t-12=0$ <br\/>$\\Delta '=4+12=16\\Rightarrow \\sqrt{\\Delta '}=4$<br\/>Suy ra ${{t}_{1}}=-2+4=2;{{t}_{2}}=-2-4=-6$ <br\/>+ $t={{t}_{1}}=2$ $\\Leftrightarrow {{x}^{2}}+x=2$<br\/>$\\Leftrightarrow {{x}^{2}}+x-2=0\\Leftrightarrow \\left[ \\begin{align} & x=1 \\\\ & x=-2 \\\\ \\end{align} \\right.\\,\\left(\\text{do}\\, a+b+c=0 \\right)$ <br\/>+ $t={{t}_{1}}=-6$ $\\Leftrightarrow {{x}^{2}}+x=-6$<br\/>$\\Leftrightarrow {{x}^{2}}+x+6=0$ (ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m v\u00ec $\\Delta =-23<0$ )<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\left\\{ 1;-2 \\right\\}$ <br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1;$ $-2.$<\/span><\/span>"}]}],"id_ques":938},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x-\\sqrt{x-1}-3=0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $x=$_input_","hint":"\u0110\u1eb7t \u1ea9n ph\u1ee5 $t=\\sqrt{x-1}$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: \u0110\u1eb7t \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh x\u00e1c \u0111\u1ecbnh<br\/>B\u01b0\u1edbc 2: \u0110\u1eb7t \u1ea9n ph\u1ee5 v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh theo \u1ea9n m\u1edbi.<br\/>B\u01b0\u1edbc 3: Tr\u1edf v\u1ec1 \u1ea9n ban \u0111\u1ea7u v\u00e0 x\u00e1c \u0111\u1ecbnh t\u1eadp nghi\u1ec7m.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n $x\\ge 1$<br\/>Ta c\u00f3:<br\/>$x-\\sqrt{x-1}-3=0\\Leftrightarrow x-1-\\sqrt{x-1}-2=0$<br\/>\u0110\u1eb7t $t=\\sqrt{x-1}$ ($t\\ge 0$ ). <br\/>Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: <br\/>$\\begin{aligned} & {{t}^{2}}-t-2=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned}& t=-1\\, \\\\ & t=2\\,\\, \\\\ \\end{aligned} \\right.\\,\\left(\\text{do}\\, a-b+c=0 \\right) \\\\ & \\Leftrightarrow t=2\\,\\left( \\text{do} \\, t\\ge 0\\right) \\\\ & \\Leftrightarrow \\sqrt{x-1}=2 \\\\ & \\Leftrightarrow x-1=4 \\\\ & \\Leftrightarrow x=5 \\\\ \\end{aligned}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m l\u00e0 $x=5$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $5.$ <\/span><\/span>"}]}],"id_ques":939},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["5"],["-5"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $3{{x}^{2}}-14\\left| x \\right|-5=0$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\{$_input_;_input_$\\}$","hint":"\u0110\u1eb7t \u1ea9n ph\u1ee5 $t=\\left| x \\right|$","explain":"<span class='basic_left'>\u0110\u1eb7t $t=\\left| x \\right|,t\\ge 0$ .<br\/> Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: $3{{t}^{2}}-14t-5=0$ <br\/>$\\Delta '={{7}^{2}}+3.5=64\\Rightarrow \\sqrt{\\Delta '}=8$ <br\/>Suy ra ${{t}_{1}}=\\dfrac{7+8}{3}=5$ (th\u1ecfa m\u00e3n); ${{t}_{2}}=\\dfrac{7-8}{3}=-\\dfrac{1}{3}$ (lo\u1ea1i)<br\/>V\u1edbi $t={{t}_{1}}=5\\Rightarrow \\left| x \\right|=5\\Leftrightarrow x=\\pm 5$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 5;-5 \\right\\}$<br\/> <span class='basic_green'>C\u00e1ch 2:<\/span> D\u00f9ng \u0111\u1ecbnh ngh\u0129a gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i \u0111\u1ec3 chia tr\u01b0\u1eddng h\u1ee3p gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh <br\/>+ V\u1edbi $x\\ge 0$ th\u00ec $\\left| x \\right|=x$ .<br\/> Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: $3{{x}^{2}}-14x-5=0$ <br\/>$\\Delta '={{7}^{2}}+3.5=64$<br\/>$\\Rightarrow$ ${{x}_{1}}=\\dfrac{7+8}{3}=5$ (th\u1ecfa m\u00e3n); ${{x}_{2}}=\\dfrac{7-8}{3}=-\\dfrac{1}{3}$ (lo\u1ea1i)<br\/>+ V\u1edbi $x<0$ th\u00ec $\\left| x \\right|=-x$ <br\/> Ta \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng tr\u00ecnh: $3{{x}^{2}}+14x-5=0$ <br\/>$\\Delta '={{7}^{2}}+3.5=64$<br\/>$\\Rightarrow$ ${{x}_{1}}=\\dfrac{-7+8}{3}=\\dfrac{1}{3}$ (lo\u1ea1i); ${{x}_{2}}=\\dfrac{-7-8}{3}=-5$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ 5;-5 \\right\\}$ <br\/><span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $5;$ $-5.$<\/span><\/span>"}]}],"id_ques":940}],"lesson":{"save":0,"level":1}}