{"segment":[{"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho c\u00e1c \u0111\u01a1n th\u1ee9c: <br\/> $\\begin{align} & A=\\dfrac{2}{3}{{x}^{2}}y.\\dfrac{3}{4}{{x}^{4}}{{y}^{3}} \\\\ & B=-0,5x{{y}^{2}}.{{x}^{5}}{{y}^{2}} \\\\ \\end{align}$ <br\/> Hi\u1ec7u $A-B$ l\u00e0: <\/span> ","select":["A. $0$ ","B. $\\dfrac{1}{2}{{x}^{6}}{{y}^{4}}$","C. $2{{x}^{6}}{{y}^{4}}$ ","D. ${{x}^{6}}{{y}^{4}}$ "],"hint":"Thu g\u1ecdn c\u00e1c \u0111\u01a1n th\u1ee9c r\u1ed3i t\u00ednh $A-B$","explain":" <span class='basic_left'>Ta c\u00f3: <br\/> $\\begin{align} & A=\\dfrac{2}{3}{{x}^{2}}y.\\dfrac{3}{4}{{x}^{4}}{{y}^{3}}=\\left( \\dfrac{2}{3}.\\dfrac{3}{4} \\right)\\left( {{x}^{2}}{{x}^{4}} \\right)\\left( y{{y}^{3}} \\right)=\\dfrac{1}{2}{{x}^{6}}{{y}^{4}} \\\\ & B=-0,5x{{y}^{2}}.{{x}^{5}}{{y}^{2}}=-0,5\\left( x{{x}^{5}} \\right)\\left( {{y}^{2}}{{y}^{2}} \\right)=-0,5{{x}^{6}}{{y}^{4}} \\\\ & \\Rightarrow A-B=\\dfrac{1}{2}{{x}^{6}}{{y}^{4}}-\\left( -0,5{{x}^{6}}{{y}^{4}} \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\dfrac{1}{2}{{x}^{6}}{{y}^{4}}+\\dfrac{1}{2}{{x}^{6}}{{y}^{4}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( \\dfrac{1}{2}+\\dfrac{1}{2} \\right){{x}^{6}}{{y}^{4}} \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{x}^{6}}{{y}^{4}} \\\\ \\end{align}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":4}],"id_ques":1301},{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["2","3"]],"list":[{"point":10,"img":"","ques":"<span class='basic_left'> H\u00e3y ch\u1ecdn c\u00e1c \u0111\u01a1n th\u1ee9c \u0111\u1ed3ng d\u1ea1ng v\u1edbi \u0111\u01a1n th\u1ee9c $-7xz^2$ <\/span>","hint":"","column":2,"number_true":2,"select":["A. $-7,5x^2z^3$","B. $-8xz^2$","C. $9xz^2$","D. $8x^2z$"],"explain":"<span class='basic_left'> \u0110\u01a1n th\u1ee9c $-7xz^2$ c\u00f3 ph\u1ea7n bi\u1ebfn l\u00e0 $xz^2$ <br\/> C\u00e1c \u0111\u01a1n th\u1ee9c $-8xz^2$ v\u00e0 $9xz^2$ c\u0169ng c\u00f3 ph\u1ea7n bi\u1ebfn l\u00e0 $xz^2$ n\u00ean hai \u0111\u01a1n th\u1ee9c n\u00e0y \u0111\u1ed3ng d\u1ea1ng v\u1edbi \u0111\u01a1n th\u1ee9c $-7xz^2$ <br\/> <span class='basic_pink'>Do \u0111\u00f3 c\u00f3 hai \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B v\u00e0 C.<\/span> <\/span>"}],"id_ques":1302},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c $A=-\\dfrac{3}{4}{{x}^{2}}{{y}^{3}}+8{{y}^{2}}-2{{y}^{2}}-3{{x}^{2}}{{y}^{3}}$ t\u1ea1i $x=-1;y=-2$ l\u00e0: <\/span> ","select":["A. $54$ ","B. $6$","C. $\\dfrac{37}{2}$ ","D. $-\\dfrac{27}{4}$ "],"hint":"Thu g\u1ecdn \u0111a th\u1ee9c $A$ r\u1ed3i thay gi\u00e1 tr\u1ecb c\u1ee7a $x,y$ v\u00e0o \u0111\u1ec3 t\u00ednh.","explain":" <span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & A=-\\dfrac{3}{4}{{x}^{2}}{{y}^{3}}+8{{y}^{2}}-2{{y}^{2}}-3{{x}^{2}}{{y}^{3}} \\\\ & \\,\\,\\,\\,\\,\\,=\\left( -\\dfrac{3}{4}{{x}^{2}}{{y}^{3}}-3{{x}^{2}}{{y}^{3}} \\right)+\\left( 8{{y}^{2}}-2{{y}^{2}} \\right) \\\\ & \\,\\,\\,\\,\\,\\,=-\\dfrac{15}{4}{{x}^{2}}{{y}^{3}}+6{{y}^{2}} \\\\ \\end{align}$ <br\/> Thay $x=-1;y=-2$ v\u00e0o k\u1ebft qu\u1ea3 tr\u00ean ta \u0111\u01b0\u1ee3c: <br\/> $A=-\\dfrac{15}{4}.{(-1)^{2}}.{(-2)^{3}}+6.{(-2)^{2}}=54$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}],"id_ques":1303},{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","3"]],"list":[{"point":10,"img":"","ques":"<span class='basic_left'> Cho \u0111a th\u1ee9c $f(x)=(2x^2-3x+1)-(x^2-7x-2)$ <br\/> C\u00e1c gi\u00e1 tr\u1ecb n\u00e0o sau \u0111\u00e2y l\u00e0 nghi\u1ec7m c\u1ee7a $f(x)$? <\/span>","hint":"","column":2,"number_true":2,"select":["A. $-1$","B. $-2$","C. $-3$","D. $-4$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>- Thu g\u1ecdn \u0111a th\u1ee9c $f(x)$ <br\/> - Th\u1eed c\u00e1c gi\u00e1 tr\u1ecb \u0111\u00e3 cho xem gi\u00e1 tr\u1ecb n\u00e0o th\u1ecfa m\u00e3n $f(x)=0$ th\u00ec \u0111\u00f3 l\u00e0 nghi\u1ec7m c\u1ee7a $f(x)$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $f(x)=(2x^2-3x+1)-(x^2-7x-2)=2x^2-3x+1-x^2+7x+2=x^2+4x+3$ <br\/> - Thay $x=-1$ v\u00e0o $f(x)$ ta \u0111\u01b0\u1ee3c: <br\/> $(-1)^2+4.(-1)+3=0$ <br\/> $\\Rightarrow x=-1$ l\u00e0 nghi\u1ec7m c\u1ee7a $f(x)$ <br\/> - Thay $x=-2$ v\u00e0o $f(x)$ ta \u0111\u01b0\u1ee3c: <br\/> $(-2)^2+4.(-2)+3=-1$ <br\/> $\\Rightarrow x=-2$ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a $f(x)$ <br\/> - Thay $x=-3$ v\u00e0o $f(x)$ ta \u0111\u01b0\u1ee3c: <br\/> $(-3)^2+4.(-3)+3=0$ <br\/> $\\Rightarrow x=-3$ l\u00e0 nghi\u1ec7m c\u1ee7a $f(x)$ <br\/> - Thay $x=-4$ v\u00e0o $f(x)$ ta \u0111\u01b0\u1ee3c: <br\/> $(-4)^2+4.(-4)+3=3$ <br\/> $\\Rightarrow x=-4$ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a $f(x)$ <br\/> <span class='basic_pink'>Do \u0111\u00f3 c\u00f3 hai \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A v\u00e0 C.<\/span> <\/span>"}],"id_ques":1304},{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-7"],["3"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","input_hint":["frac"],"ques":"T\u00ecm $x$ bi\u1ebft $\\left( 2x-1 \\right)-\\left( 5x+1 \\right)=\\left( x+3 \\right)-\\left( x-2 \\right)$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $x=$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"","explain":"<span class='basic_left'>Ta c\u00f3: <br\/> $\\begin{align} & \\left( 2x-1 \\right)-\\left( 5x+1 \\right)=\\left( x+3 \\right)-\\left( x-2 \\right) \\\\ & \\Rightarrow 2x-1-5x-1=x+3-x+2 \\\\ & \\Rightarrow -3x-2=5 \\\\ & \\Rightarrow -3x=7 \\\\ & \\Rightarrow x=\\dfrac{-7}{3} \\\\ \\end{align}$ <\/span>"}],"id_ques":1305},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> X\u00e1c \u0111\u1ecbnh d\u1ea5u c\u1ee7a $c,$ bi\u1ebft r\u1eb1ng $2a^3bc$ tr\u00e1i d\u1ea5u v\u1edbi $-3a^5b^3c^2$ ($a,b,c$ kh\u00e1c $0$) <\/span> ","select":["A. $c > 0$ ","B. $c < 0$ "],"hint":"Hai s\u1ed1 tr\u00e1i d\u1ea5u th\u00ec t\u00edch c\u1ee7a ch\u00fang nh\u1ecf h\u01a1n $0$","explain":" <span class='basic_left'> $2a^3bc$ v\u00e0 $-3a^5b^3c^2$ tr\u00e1i d\u1ea5u n\u00ean $(2a^3bc)(-3a^5b^3c^2) < 0$ <br\/> $\\Rightarrow -6a^8b^4c^3 < 0$ <br\/> Do $a^8b^4 \\ge 0$ v\u00e0 $a,b,c \\ne 0$ n\u00ean $-6a^8b^4 < 0$ <br\/> \u0110\u1ec3 $-6a^8b^4c^3 < 0$ th\u00ec $ c^3 > 0 \\Rightarrow c > 0$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}],"id_ques":1306},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho hai \u0111a th\u1ee9c: <br\/> $P=5{{x}^{4}}-3{{x}^{2}}+9{{x}^{3}}-2{{x}^{4}}+4+5x$ <br\/> $Q=-10x+5+8{{x}^{3}}+3{{x}^{2}}+{{x}^{3}}$ <br\/> T\u00ednh $M=P-Q$ r\u1ed3i t\u00ecm h\u1ec7 s\u1ed1 c\u1ee7a l\u0169y th\u1eeba b\u1eadc $3$ trong \u0111a th\u1ee9c $M.$ <br\/> H\u1ec7 s\u1ed1 c\u1ee7a l\u0169y th\u1eeba b\u1eadc $3$ trong \u0111a th\u1ee9c $M$ l\u00e0: <\/span> ","select":["A. $18$ ","B. $0$","C. $1$ ","D. $9$ "],"hint":"","explain":" <span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & P=5{{x}^{4}}-3{{x}^{2}}+9{{x}^{3}}-2{{x}^{4}}+4+5x \\\\ & \\,\\,\\,\\,\\,=3{{x}^{4}}+9{{x}^{3}}-3{{x}^{2}}+5x+4 \\\\ & Q=-10x+5+8{{x}^{3}}+3{{x}^{2}}+{{x}^{3}} \\\\ & \\,\\,\\,\\,\\,\\,=9{{x}^{3}}+3{{x}^{2}}-10x+5 \\\\ & \\Rightarrow M=P-Q=\\left( 3{{x}^{4}}+9{{x}^{3}}-3{{x}^{2}}+5x+4 \\right)-\\left( 9{{x}^{3}}+3{{x}^{2}}-10x+5 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=3{{x}^{4}}+9{{x}^{3}}-3{{x}^{2}}+5x+4-9{{x}^{3}}-3{{x}^{2}}+10x-5 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=3{{x}^{4}}-6{{x}^{2}}+15x-1 \\\\ \\end{align}$ <br\/> H\u1ec7 s\u1ed1 c\u1ee7a l\u0169y th\u1eeba b\u1eadc $3$ trong \u0111a th\u1ee9c $M$ l\u00e0 $0$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":4}],"id_ques":1307},{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["-5"],["1"],["3"],["9"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'> Cho bi\u1ec3u th\u1ee9c $f\\left( x \\right)=\\dfrac{x+5}{x-2}$ <br\/> T\u00ecm $x$ nguy\u00ean \u0111\u1ec3 $f(x)$ c\u00f3 gi\u00e1 tr\u1ecb nguy\u00ean. <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> $x\\in \\,\\,\\left\\{ { \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} } \\right\\}$ <\/span> ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>\u0110\u01b0a $f(x)$ v\u1ec1 d\u1ea1ng $a+\\dfrac{b}{A(x)}$ v\u1edbi $a;b\\in \\mathbb{Z}$ v\u00e0 $A(x)$ l\u00e0 bi\u1ec3u th\u1ee9c ch\u1ee9a \u1ea9n $x$ <br\/> \u0110\u1ec3 $f(x)\\in \\mathbb{Z}$ th\u00ec $A(x) \\in \u01af(b)$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $f\\left( x \\right)=\\dfrac{x+5}{x-2}=\\dfrac{\\left( x-2 \\right)+7}{x-2}=\\dfrac{x-2}{x-2}+\\dfrac{7}{x-2}=1+\\dfrac{7}{x-2}$ <br\/> \u0110\u1ec3 $f(x)$ nguy\u00ean th\u00ec $\\dfrac{7}{x-2} \\in \\mathbb{Z}$ <br\/> $\\Rightarrow (x-2)\\in \u01af(7)=\\{-7;-1;1;7\\}$ <br\/> Ta c\u00f3 b\u1ea3ng sau: <br\/> <table><tr><th>$x-2$<\/th><th>-7<\/th><th>-1<\/th><th>1<\/th><th>7<\/th><\/tr><tr><td>$x$<\/td><td>-5<\/td><td>1<\/td><td>3<\/td><td>9<\/td><\/tr><\/table> <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-5;1;3;9$<\/span><\/span>"}],"id_ques":1308},{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'> Cho \u0111a th\u1ee9c $Q=a\\,{{x}^{2}}{{y}^{2}}-2xy+3xy-2{{x}^{2}}{{y}^{2}}+5$ <br\/> Bi\u1ebft r\u1eb1ng \u0111a th\u1ee9c $Q$ c\u00f3 b\u1eadc l\u00e0 $4$ v\u00e0 $a$ l\u00e0 s\u1ed1 nguy\u00ean t\u1ed1 nh\u1ecf h\u01a1n $5.$ T\u00ecm gi\u00e1 tr\u1ecb c\u1ee7a $a.$ <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> $a=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> - R\u00fat g\u1ecdn \u0111a th\u1ee9c $Q$ <br\/> - T\u00ecm $a$ t\u1eeb d\u1eef ki\u1ec7n b\u1eadc c\u1ee7a $Q$ v\u00e0 \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $a$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & Q=a\\,{{x}^{2}}{{y}^{2}}-2xy+3xy-2{{x}^{2}}{{y}^{2}}+5 \\\\ & \\,\\,\\,\\,\\,\\,=\\left( a-2 \\right){{x}^{2}}{{y}^{2}}+xy+5 \\\\ \\end{align}$ <br\/> \u0110a th\u1ee9c $Q$ c\u00f3 b\u1eadc l\u00e0 $4$ n\u00ean $a-2 \\ne 0 \\Rightarrow a\\ne 2$ (1) <br\/> M\u00e0 $a$ l\u00e0 s\u1ed1 nguy\u00ean t\u1ed1 nh\u1ecf h\u01a1n $5$ n\u00ean $a=2$ ho\u1eb7c $a=3$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $a=3$ th\u1ecfa m\u00e3n y\u00eau c\u1ea7u b\u00e0i to\u00e1n. <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3$<\/span><\/span>"}],"id_ques":1309},{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $F=\\dfrac{3a-5}{2a+b}-\\dfrac{4b+5}{a+3b}$ v\u1edbi $a-b=5$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $F=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u1eeb $a-b=5$ r\u00fat $a$ theo $b$ <br\/> \u0110\u01b0a $F$ v\u1ec1 bi\u1ec3u th\u1ee9c ch\u1ec9 ch\u1ee9a bi\u1ebfn $b$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> T\u1eeb $a-b=5 \\Rightarrow a=b+5$ <br\/> Khi \u0111\u00f3 ta c\u00f3: <br\/> $\\begin{align} & F=\\dfrac{3\\left( b+5 \\right)-5}{2(b+5)+b}-\\dfrac{4b+5}{(b+5)+3b} \\\\ & \\,\\,\\,\\,=\\dfrac{3b+15-5}{2b+10+b}-\\dfrac{4b+5}{b+5+3b} \\\\ & \\,\\,\\,\\,=\\dfrac{3b+10}{3b+10}-\\dfrac{4b+5}{4b+5} \\\\ & \\,\\,\\,\\,=1-1 \\\\ & \\,\\,\\,\\,=0 \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$ <\/span><\/span>"}],"id_ques":1310}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":44}}