{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-2024"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","type_check":"","ques":"Cho h\u00e0m s\u1ed1 $f\\left( x \\right)=a{{x}^{5}}+b{{x}^{3}}+cx-5$ ($a, b, c$ l\u00e0 nh\u1eefng h\u1eb1ng s\u1ed1). <br\/> Bi\u1ebft $f\\left( -3 \\right)=2014$ .T\u00ednh $f\\left( 3 \\right)$. <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $f(3)= \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$.","hint":"T\u00ednh $f(3) +f(-3)$.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span> <b> B\u01b0\u1edbc 1: <\/b> T\u00ednh $f(3);f(-3)$. <br\/> <b> B\u01b0\u1edbc 2: <\/b> T\u00ednh $f(3) +f(-3)$. <br\/> <b> B\u01b0\u1edbc 3:<\/b> T\u00ednh $f(3)$. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/> $f\\left( -3 \\right)\\,$$=a.{{\\left( -3 \\right)}^{5}}+b.{{\\left( -3 \\right)}^{3}}+c.\\left( -3 \\right)-5 =-243a-27b-3c-5$<br\/>$ f\\left( 3 \\right) =a{{.3}^{5}}+b{{.3}^{3}}+c.3-5=243a +27b+3c-5 $ <br\/>$\\Rightarrow f\\left( 3 \\right)+f\\left( -3 \\right)=-10$ <br\/>M\u00e0 $f\\left( -3 \\right)=2014\\Rightarrow f\\left( 3 \\right)=-10-f\\left( -3 \\right)=-10-2014=-2024$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $-2024$.<\/span><\/span>"}]}],"id_ques":41},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $m$ v\u00e0 $k$ \u0111\u1ec3 h\u00e0m s\u1ed1 sau l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t:<br\/>$y=f\\left( x \\right)=k{{x}^{2}}+\\,$$\\left( {{m}^{2}}-mk-6{{k}^{2}} \\right)x\\,$$-9{{x}^{2}}+5$. ","select":["A. $k=0; m\\ne 27$ v\u00e0 $m \\ne - 18$","B. $k= 0; m\\ne -18$","C. $k= 9; m\\ne 27$ v\u00e0 $m \\ne - 18$ ","D. $k= 9; m\\ne 27$ "],"hint":"H\u00e0m s\u1ed1 y=$ax^2+bx +c$ ($a,b,c$ l\u00e0 h\u1eb1ng s\u1ed1) l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t khi $a=0$ v\u00e0 $b \\ne 0$.","explain":"<span class='basic_left'> Ta c\u00f3: <br\/>$y=f\\left( x \\right)=k{{x}^{2}}+\\,$$\\left( {{m}^{2}}-mk-6{{k}^{2}} \\right)x\\,$$-9{{x}^{2}}+5$<br\/>$ \\Leftrightarrow y=f\\left( x \\right)=\\left( k-9 \\right){{x}^{2}}+\\,$$\\left( {{m}^{2}}-mk-6{{k}^{2}} \\right)x+5 $<br\/>H\u00e0m s\u1ed1 \u0111\u00e3 cho l\u00e0 h\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t khi v\u00e0 ch\u1ec9 khi: <br\/>$\\begin{aligned} & \\left\\{ \\begin{aligned} & k-9=0 \\\\ & {{m}^{2}}-mk-6{{k}^{2}}\\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & k=9 \\\\ & m^2-3mk+2mk-6k^2 \\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & k=9 \\\\ & m(m-3.9) +2.9.(m-3.9) \\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & k=9 \\\\ & \\left( m-27 \\right)\\left( m+18 \\right)\\ne 0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & k=9 \\\\ & m\\ne 27;\\,\\,m\\ne -18 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>V\u1eady $k= 9 ; m\\ne 27$ v\u00e0 $m\\ne -18$ th\u00ec h\u00e0m s\u1ed1 \u0111\u00e3 cho l\u00e0 h\u00e0m b\u1eadc nh\u1ea5t. <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 C.<\/span>","column":2}]}],"id_ques":42},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["AB= 5; BC = $\\sqrt{10}$","AB= 5; BC = $2\\sqrt{10}$","AB= 5; BC = $3\\sqrt{10}$"],"ques":"<span class='basic_left'>Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$, cho c\u00e1c \u0111i\u1ec3m $A\\left( -3;2 \\right);\\,\\,B\\left( 1;5 \\right)$ v\u00e0 $C\\left( 2;2 \\right)$. <br\/><b> C\u00e2u 1: <\/b> \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n $AB$ =?; $BC$ = ?<\/span>","hint":"\u00c1p d\u1ee5ng: $AB=\\sqrt{(x_{B}-x_{A})^2+(y_{B}-y_{A})^2} $.","explain":"<span class='basic_left'> \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng $AB$ l\u00e0:<br\/>$\\begin{align}AB & =\\sqrt{{{\\left( {{x}_{B}}-{{x}_{A}} \\right)}^{2}}+{{\\left( {{y}_{B}}-{{y}_{A}} \\right)}^{2}}} \\\\ & =\\sqrt{{{\\left[ 1-(-3) \\right]}^{2}}+{{\\left( 5-2\\right)}^{2}}} \\\\ & =\\sqrt{{{4}^{2}}+{{3}^{2}}} \\\\ & =5 \\\\ \\end{align}$ <br\/>\u0110\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng $BC$ l\u00e0:<br\/>$\\begin{align}BC & =\\sqrt{{{\\left( {{x}_{C}}-{{x}_{B}} \\right)}^{2}}+{{\\left( {{y}_{C}}-{{y}_{B}} \\right)}^{2}}} \\\\ & =\\sqrt{{{\\left( 2-1 \\right)}^{2}}+{{\\left( 2-5\\right)}^{2}}} \\\\ & =\\sqrt{{{1}^{2}}+{{3}^{2}}} \\\\ & =\\sqrt{10} \\\\ \\end{align}$ <span class='basic_pink'><br\/>\u0110\u00e1p \u00e1n \u0111\u00fang l\u00e0 AB= 5; BC = $\\sqrt{10}$<\/span><br\/><span class='basic_green'>Ghi nh\u1edb: <\/span>Ngo\u00e0i c\u00e1ch d\u1ef1ng h\u00ecnh v\u00e0 s\u1eed d\u1ee5ng c\u00e1c h\u1eb1ng \u0111\u1eb3ng th\u1ee9c, h\u1ec7 th\u1ee9c l\u01b0\u1ee3ng trong tam gi\u00e1c \u0111\u1ec3 t\u00ednh \u0111\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng trong h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 ta c\u00f3 th\u1ec3 s\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c sau:<br\/> $AB=\\sqrt{(x_{B}-x_{A})^2+(y_{B}-y_{A})^2} $<\/span>"}]}],"id_ques":43},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<span class='basic_left'>Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$, cho c\u00e1c \u0111i\u1ec3m $A\\left( -3;2 \\right);\\,\\,B\\left( 1;5 \\right)$ v\u00e0 $C\\left( 2;2 \\right)$ .<br\/><b> C\u00e2u 2: <\/b> Tam gi\u00e1c $ABC$ l\u00e0 tam gi\u00e1c g\u00ec?<\/span> ","select":["A. Tam gi\u00e1c c\u00e2n","B. Tam gi\u00e1c vu\u00f4ng","C. Tam gi\u00e1c \u0111\u1ec1u"],"hint":"T\u00ednh \u0111\u1ed9 d\u00e0i $AC$ v\u00e0 so s\u00e1nh v\u1edbi $AB, BC$","explain":"<span class='basic_left'> \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng $AC$ l\u00e0:<br\/>$\\begin{align}AC & =\\sqrt{{{\\left( {{x}_{C}}-{{x}_{A}} \\right)}^{2}}+{{\\left( {{y}_{C}}-{{y}_{A}} \\right)}^{2}}} \\\\ & =\\sqrt{{{\\left( 2-(-3) \\right)}^{2}}+{{\\left( 2-2\\right)}^{2}}} \\\\ & =\\sqrt{{{5}^{2}}+{{0}^{2}}} \\\\ & =5 \\\\ \\end{align}$ <br\/>Theo c\u00e2u 1 tr\u00ean, ta c\u00f3: $AB= 5; BC=\\sqrt{10}$<br\/>$\\Rightarrow AC=AB\\,(=5)$ <br\/> $\\Rightarrow \\Delta ABC$ c\u00e2n t\u1ea1i $A$. <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span>","column":3}]}],"id_ques":44},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["7,5"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'>Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$, cho c\u00e1c \u0111i\u1ec3m $A\\left( -3;2 \\right);\\,\\,B\\left( 1;5 \\right)$ v\u00e0 $C\\left( 2;2 \\right)$.<br\/> <b> C\u00e2u 3: <\/b> T\u00ednh di\u1ec7n t\u00edch tam gi\u00e1c $ABC$. <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> Di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ l\u00e0 _input_ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch)<br\/>(K\u1ebft qu\u1ea3 vi\u1ebft d\u01b0\u1edbi d\u1ea1ng s\u1ed1 th\u1eadp ph\u00e2n)<\/span>","hint":" D\u1ef1ng $BH$ vu\u00f4ng g\u00f3c v\u1edbi $AC$. Khi \u0111\u00f3 di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ b\u1eb1ng n\u1eeda t\u00edch $BH$ v\u1edbi $AC$.","explain":"<span class='basic_left'>Ta c\u00f3 h\u00ecnh v\u1ebd<br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv3/img\/D921_K5.png' \/><\/center>Theo c\u00e2u 2 tr\u00ean, ta c\u00f3: $AC= 5$<br\/>K\u1ebb $BH \\bot AC$ t\u1ea1i $H$. <br\/> T\u1eeb m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9, ta c\u00f3 $BH=3$.<br\/>Di\u1ec7n t\u00edch tam gi\u00e1c $ABC$ l\u00e0:<br\/> $\\dfrac{1}{2}BH.AC=\\dfrac{1}{2}.3.5=7,5$ (\u0111\u01a1n v\u1ecb di\u1ec7n t\u00edch). <span class='basic_pink'><br\/>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $7,5$.<\/span> "}]}],"id_ques":45},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"X\u00e1c \u0111\u1ecbnh h\u00e0m s\u1ed1 $y=f(x)$, bi\u1ebft $f\\left( a-1 \\right)=\\dfrac{1}{2}{{a}^{2}}-3a+\\dfrac{11}{2}$. ","select":["A. $y=f\\left( x \\right)=\\dfrac{1}{2}{{x}^{2}}-x+\\dfrac{9}{2}$","B. $y=f\\left( x \\right)=\\dfrac{1}{2}{{x}^{2}}-2x+2$","C. $y=f\\left( x \\right)=\\dfrac{1}{2}{{x}^{2}}+2x+3$","D. $y=f\\left( x \\right)=\\dfrac{1}{2}{{x}^{2}}-2x+3$ "],"hint":"\u0110\u1eb7t $t=a - 1$. T\u00ednh $f(t)$.","explain":"<span class='basic_left'> \u0110\u1eb7t $t= a- 1\\Rightarrow t + 1= a$.<br\/>$\\begin{align}& f\\left( a-1 \\right) =f\\left( t \\right)\\\\& =\\dfrac{1}{2}{{\\left( t+1 \\right)}^{2}}-3\\left( t+1 \\right)+\\dfrac{11}{2} \\\\ & =\\dfrac{1}{2}{{t}^{2}}+t+\\dfrac{1}{2}-3t-3+\\dfrac{11}{2} \\\\ & =\\dfrac{1}{2}{{t}^{2}}-2t+3\\\\ \\end{align}$ <br\/>H\u00e0m s\u1ed1 c\u1ea7n x\u00e1c \u0111\u1ecbnh l\u00e0 $y=f\\left( x \\right)=\\dfrac{1}{2}{{x}^{2}}-2x+3$.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span>","column":2}]}],"id_ques":46},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"H\u00e0m s\u1ed1 $y=f\\left( x \\right)=\\sqrt{2-x+2\\sqrt{1-x}}$ ngh\u1ecbch bi\u1ebfn tr\u00ean t\u1eadp x\u00e1c \u0111\u1ecbnh c\u1ee7a n\u00f3.","select":["A. \u0110\u00fang","B. Sai"],"hint":"\u0110\u01b0a bi\u1ec3u th\u1ee9c trong c\u0103n c\u1ee7a $f(x)$ v\u1ec1 d\u1ea1ng $(A+B)^2$ r\u1ed3i \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>- T\u00ecm t\u1eadp x\u00e1c \u0111\u1ecbnh (TX\u0110) $D$ c\u1ee7a h\u00e0m s\u1ed1: Gi\u1ea3 s\u1eed $D=(a;b)$ <br\/>- Gi\u1ea3 s\u1eed $x_{1} < x_2\\, (x_{1}; x_2 \\in D)$. X\u00e9t hi\u1ec7u $f(x_{1}) -f(x_{2})$ <br\/>+ N\u1ebfu $f(x_{1}) -f(x_{2}) < 0$ th\u00ec $f(x_{1}) < f(x_{2})$: H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn trong kho\u1ea3ng $(a;b)$ <br\/>+ N\u1ebfu $f(x_{1}) -f(x_{2}) > 0$ th\u00ec $f(x_{1}) > f(x_{2})$: H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn trong kho\u1ea3ng $(a;b)$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/> $y=f\\left( x \\right)=\\sqrt{2-x+2\\sqrt{1-x}}\\,$$=\\sqrt{1-x+2\\sqrt{1-x}+1}\\,$$=\\sqrt{{{\\left( \\sqrt{1-x}+1 \\right)}^{2}}}\\,$$=\\sqrt{1-x}+1$ <br\/>TX\u0110: $x\\le 1$ <br\/>L\u1ea5y ${{x}_{1}}<{{x}_{2}}\\le 1$ , x\u00e9t<br\/>$\\begin{align} &f\\left( {{x}_{1}} \\right)-f\\left( {{x}_{2}} \\right) \\\\& =\\sqrt{1-{{x}_{1}}}+1-\\left( \\sqrt{1-{{x}_{2}}}+1 \\right) \\\\ & =\\sqrt{1-{{x}_{1}}}-\\sqrt{1-{{x}_{2}}} \\\\ & =\\dfrac{1-{{x}_{1}}-1+{{x}_{2}}}{\\sqrt{1-{{x}_{1}}}+\\sqrt{1-{{x}_{2}}}} \\\\ & =\\dfrac{{{x}_{2}}-{{x}_{1}}}{\\sqrt{1-{{x}_{1}}}+\\sqrt{1-{{x}_{2}}}}>0\\,\\,\\,\\,\\,\\,\\,\\left(\\text{V\u00ec}\\,{{x}_{2}}>{{x}_{1}};\\, \\sqrt{1-{{x}_{1}}}+\\sqrt{1-{{x}_{2}}}>0 \\right) \\\\ \\end{align}$ <br\/>Suy ra $f\\left( {{x}_{1}} \\right) > f\\left( {{x}_{2}} \\right)$ <br\/> $\\Rightarrow $ H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean TX\u0110: $x\\le 1$ <br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}]}],"id_ques":47},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"]]],"list":[{"point":10,"width":60,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv3/img\/4.jpg' \/><\/center>Tr\u00ean m\u1eb7t ph\u1eb3ng t\u1ecda \u0111\u1ed9 $Oxy$ cho \u0111i\u1ec3m $A (1; 1)$ v\u00e0 $B (-2;5)$. \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng $AB$ l\u00e0 _input_","hint":"$AB=\\sqrt{(x_{B}-x_{A})^2+(y_{B}-y_{A})^2}$","explain":"<span class='basic_left'>Ta c\u00f3 h\u00ecnh v\u1ebd: <br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv3/img\/D921_K3.png' \/><\/center> <br\/>H\u1ea1 $Am \\bot Oy; Bn \\bot Ox ; Am \\cap Bn=H$ <br\/>Theo h\u00ecnh v\u1ebd ta c\u00f3 $AH= 3; BH = 4$.<br\/>X\u00e9t $\\Delta AHB$ vu\u00f4ng t\u1ea1i $H$ c\u00f3: <br\/>$AB^2=AH^2+HB^2$ (\u0111\u1ecbnh l\u00ed Py - ta - go) <br\/> $\\Rightarrow AB^2=3^2+4^2=25$ <br\/> $\\Rightarrow AB= 5$ <span class='basic_pink'><br\/>Do \u0111\u00f3 s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $5$.<\/span><br\/>Ngo\u00e0i ra, ta c\u00f3 th\u1ec3 s\u1eed d\u1ee5ng c\u00f4ng th\u1ee9c $AB=\\sqrt{(x_{B}-x_{A})^2+(y_{B}-y_{A})^2}$ \u0111\u1ec3 t\u00ednh tr\u1ef1c ti\u1ebfp \u0111\u1ed9 d\u00e0i $AB$.<br\/><b>C\u00e1ch 2:<\/b> \u0110\u1ed9 d\u00e0i \u0111o\u1ea1n th\u1eb3ng $AB$ l\u00e0:<br\/>$\\begin{align}AB & =\\sqrt{{{\\left( {{x}_{B}}-{{x}_{A}} \\right)}^{2}}+{{\\left( {{y}_{B}}-{{y}_{A}} \\right)}^{2}}} \\\\ & =\\sqrt{{{\\left( -2-1 \\right)}^{2}}+{{\\left( 5-1\\right)}^{2}}} \\\\ & =\\sqrt{{{3}^{2}}+{{4}^{2}}} \\\\ & =5 \\\\ \\end{align}$ <\/span> "}]}],"id_ques":48},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a h\u00e0m s\u1ed1 $y=\\sqrt{x-2\\sqrt{x-1}}$ l\u00e0:","select":["A. $x \\ge 2$","B. $x \\ge 1$","C. $1 \\le x \\le 2$","D. C\u1ea3 ba \u0111\u00e1p \u00e1n tr\u00ean \u0111\u1ec1u sai"],"hint":"\u0110\u01b0a bi\u1ec3u th\u1ee9c trong c\u0103n c\u1ee7a $f(x)$ v\u1ec1 d\u1ea1ng $(A+B)^2$ r\u1ed3i \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$y=\\sqrt{x-2\\sqrt{x-1}}\\,$$=\\sqrt{x-1-2\\sqrt{x-1}+1}\\,$$=\\sqrt{{{\\left( \\sqrt{x-1}-1 \\right)}^{2}}}$ <br\/>H\u00e0m s\u1ed1 \u0111\u00e3 cho x\u00e1c \u0111\u1ecbnh khi: $\\sqrt{x-1}$ x\u00e1c \u0111\u1ecbnh. <br\/>$\\Leftrightarrow x -1\\ge 0\\Leftrightarrow x\\ge 1$ <br\/><span class='basic_pink'>\u0110\u00e1p \u00e1n l\u00e0 B.<\/span>","column":2}]}],"id_ques":49},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"select":["A. $y_{min}= -\\dfrac{1}{5}$ khi $x = -2$","B. $y_{min}= -\\dfrac{2}{5}$ khi $x = -2$","C .$y_{min}= -\\dfrac{3}{5}$ khi $x = -2$"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai9/lv3/img\/2.png' \/><\/center>T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 $y=\\dfrac{{{x}^{2}}+4x+2}{{{x}^{2}}+4x+9}$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $y_{min}$=?khi $x=$?","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c th\u00e0nh $m+\\dfrac {n}{f(x)}$ v\u1edbi $m,n \\mathbb Z$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c v\u1ec1 $1-\\dfrac{7}{{{\\left( x+2 \\right)}^{2}}+5}$<br\/>B\u01b0\u1edbc 2: \u0110\u01b0a ph\u00e2n th\u1ee9c t\u00ecm \u0111\u01b0\u1ee3c v\u1ec1 d\u1ea1ng $\\dfrac{a}{{f(x)}^2+b}\\le \\dfrac{a}{b}$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: $x^2+4x+9=(x+2)^2+5$ kh\u00e1c $0$ v\u1edbi m\u1ecdi $x$ n\u00ean h\u00e0m s\u1ed1 x\u00e1c \u0111\u1ecbnh tr\u00ean $ \\mathbb R$.<br\/>$\\begin{align}y & =\\dfrac{{{x}^{2}}+4x+2}{{{x}^{2}}+4x+9} \\\\ & =\\dfrac{\\left( {{x}^{2}}+4x+9 \\right)-7}{{{x}^{2}}+4x+9} \\\\ & =1-\\dfrac{7}{{{x}^{2}}+4x+9} \\\\ & =1-\\dfrac{7}{{{\\left( x+2 \\right)}^{2}}+5} \\\\ \\end{align}$<br\/>Ta c\u00f3 $\\dfrac{7}{{{\\left( x+2 \\right)}^{2}}+5}\\le \\dfrac{7}{5}$<br\/>$\\Rightarrow y= 1-\\dfrac{7}{{{\\left( x+2 \\right)}^{2}}+5}\\ge 1-\\dfrac{7}{5}\\,$$=\\dfrac{-2}{5}$ <br\/>$\\Rightarrow {{y}_{\\min }}=\\dfrac{-2}{5}\\Leftrightarrow {{\\left( x+2 \\right)}^{2}}=0\\,$$\\Leftrightarrow x=-2$<br\/> <span class='basic_pink'>\u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 $y_{min}= -\\dfrac{2}{5}$ khi $x = -2$<\/span> <br\/> <span class='basic_green'><b>L\u01b0u \u00fd:<\/b> <\/span> \u0110\u1ed1i v\u1edbi b\u00e0i t\u00ecm GTLN v\u00e0 GTNN c\u1ee7a ph\u00e2n th\u1ee9c , khi c\u00f3 b\u1eadc c\u1ee7a t\u1eed v\u00e0 b\u1eadc c\u1ee7a m\u1eabu b\u1eb1ng nhau th\u00ec ta c\u00f3 th\u1ec3 th\u1ef1c hi\u1ec7n ph\u00e9p chia \u0111a th\u1ee9c (t\u1ee9c l\u00e0 bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c \u0111\u00f3 th\u00e0nh m\u1ed9t s\u1ed1 th\u1ef1c c\u1ed9ng v\u1edbi m\u1ed9t ph\u00e2n th\u1ee9c c\u00f3 t\u1eed l\u00e0 s\u1ed1 nguy\u00ean).<\/span><\/span> "}]}],"id_ques":50}],"lesson":{"save":0,"level":3}}