{"segment":[{"time":24,"part":[{"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'> T\u00ecm \u0111\u01a1n th\u1ee9c $P$ bi\u1ebft: <br\/>$P.2{{x}^{k}}{{y}^{k-1}}=-\\dfrac{1}{2}{{x}^{k+2}}{{y}^{k+1}}\\,\\,\\,\\,\\left( k\\in \\mathbb{N};k\\ge 1;x\\ne 0;y\\ne 0 \\right)$ <\/span>","select":["A. $P=-\\dfrac{1}{4}{{x}^{2}}{{y}^{2}}$","B. $P=\\dfrac{1}{4}{{x}^{2}}{{y}^{2}}$","C. $P=-\\dfrac{1}{2}{{x}^{3}}{{y}^{2}}$ ","D. $P=\\dfrac{1}{2}{{x}^{3}}{{y}^{2}}$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Ph\u00e2n t\u00edch $-\\dfrac{1}{2}{{x}^{k+2}}{{y}^{k+1}}$ th\u00e0nh t\u00edch c\u1ee7a \u0111\u01a1n th\u1ee9c $2{{x}^{k}}{{y}^{k-1}}$ v\u1edbi m\u1ed9t \u0111\u01a1n th\u1ee9c kh\u00e1c.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/>$\\begin{align} & P.2{{x}^{k}}{{y}^{k-1}}=-\\dfrac{1}{2}{{x}^{k+2}}{{y}^{k+1}} \\\\ & P.2{{x}^{k}}{{y}^{k-1}}=2.(-\\dfrac{1}{4})x^kx^2y^{k-1}y^2 \\\\ & P.2{{x}^{k}}{{y}^{k-1}}=2{{x}^{k}}{{y}^{k-1}}.\\left( -\\dfrac{1}{4}{{x}^{2}}{{y}^{2}} \\right) \\\\ \\end{align}$ <br\/> Suy ra $P=-\\dfrac{1}{4}{{x}^{2}}{{y}^{2}}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span> <\/span>","column":2}]}],"id_ques":1141},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> Thu g\u1ecdn \u0111\u01a1n th\u1ee9c ${{\\left( {{a}^{n}}{{b}^{n+1}}{{c}^{n}} \\right)}^{k}}.{{\\left( {{a}^{k}}{{b}^{k}}{{c}^{k+1}} \\right)}^{n}}\\,\\,\\,\\,\\,\\left( k,n\\,\\,\\in \\mathbb{N} \\right)$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: <\/span>","select":["A. ${{a}^{n^2k^2}}{{b}^{\\left( 2n+1 \\right)k}}{{c}^{\\left( 2k+1 \\right)n}}$","B. ${{a}^{2nk}}{{b}^{\\left( 2k+1 \\right)n}}{{c}^{\\left( 2n+1 \\right)k}}$","C. ${{a}^{2nk}}{{b}^{\\left( 2n+1 \\right)k}}{{c}^{\\left( 2k+1 \\right)n}}$ ","D. ${{a}^{n^2k^2}}{{b}^{\\left( 2k+1 \\right)n}}{{c}^{\\left( 2n+1 \\right)k}}$"],"explain":"<span class='basic_left'>Ta c\u00f3: <br\/> $\\begin{align} & {{\\left( {{a}^{n}}{{b}^{n+1}}{{c}^{n}} \\right)}^{k}}.{{\\left( {{a}^{k}}{{b}^{k}}{{c}^{k+1}} \\right)}^{n}} \\\\ & ={{a}^{nk}}{{b}^{\\left( n+1 \\right)k}}{{c}^{nk}}.{{a}^{nk}}{{b}^{nk}}{{c}^{\\left( k+1 \\right)n}} \\\\ & =\\left( {{a}^{nk}}{{a}^{nk}} \\right)\\left[ {{b}^{\\left( n+1 \\right)k}}{{b}^{nk}} \\right]\\left[ {{c}^{nk}}{{c}^{\\left( k+1 \\right)n}} \\right] \\\\ & ={{a}^{2nk}}{{b}^{2nk+k}}{{c}^{2nk+n}} \\\\ & ={{a}^{2nk}}{{b}^{\\left( 2n+1 \\right)k}}{{c}^{\\left( 2k+1 \\right)n}} \\\\ \\end{align}$ <br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span> <\/span>","column":2}]}],"id_ques":1142},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'> T\u00ecm b\u1eadc c\u1ee7a t\u00edch c\u00e1c \u0111\u01a1n th\u1ee9c sau: <br\/> $\\left( \\dfrac{1}{2}x{{y}^{2}} \\right)\\left( \\dfrac{2}{3}{{x}^{2}}{{y}^{3}} \\right)\\left( \\dfrac{3}{4}{{x}^{3}}{{y}^{4}} \\right)\\cdots \\left( \\dfrac{99}{100}{{x}^{99}}{{y}^{100}} \\right)$ <\/span>","select":["A. $9997$","B. $9998$","C. $9999$ ","D. $10000$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u00ednh t\u00edch c\u00e1c \u0111\u01a1n th\u1ee9c \u0111\u00e3 cho: H\u1ec7 s\u1ed1 nh\u00e2n h\u1ec7 s\u1ed1, bi\u1ebfn nh\u00e2n bi\u1ebfn.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/>$\\begin{align} & \\left( \\dfrac{1}{2}x{{y}^{2}} \\right)\\left( \\dfrac{2}{3}{{x}^{2}}{{y}^{3}} \\right)\\left( \\dfrac{3}{4}{{x}^{3}}{{y}^{4}} \\right)\\cdots \\left( \\dfrac{99}{100}{{x}^{99}}{{y}^{100}} \\right) \\\\ & =\\left( \\dfrac{1}{2}\\cdot \\dfrac{2}{3}\\cdot \\dfrac{3}{4}\\cdots \\dfrac{99}{100} \\right)\\left( x{{x}^{2}}{{x}^{3}}...{{x}^{99}} \\right)\\left( {{y}^{2}}{{y}^{3}}{{y}^{4}}...{{y}^{100}} \\right) \\\\ & =\\dfrac{1}{100}{{x}^{1+2+3+\\cdots +99}}{{y}^{2+3+4+\\cdots 100}} \\\\ & =\\dfrac{1}{100}{{x}^{\\dfrac{\\left( 1+99 \\right).99}{2}}}{{y}^{\\dfrac{\\left( 2+100 \\right).99}{2}}} \\\\ & =\\dfrac{1}{100}{{x}^{4950}}{{y}^{5049}} \\\\ \\end{align}$ <br\/> B\u1eadc c\u1ee7a t\u00edch c\u00e1c \u0111a th\u1ee9c tr\u00ean l\u00e0: $4950+5049=9999$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span> <\/span>","column":4}]}],"id_ques":1143},{"time":24,"part":[{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 \u0111\u00fang nh\u1ea5t v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["19"],["0"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho $M=3{{a}^{2}}{{x}^{2}}+4{{b}^{2}}{{x}^{2}}-2{{a}^{2}}{{x}^{2}}-3{{b}^{2}}{{x}^{2}}+19\\,\\,\\,\\,\\left( a,b\\,\\,\\ne 0 \\right)$ <br\/> T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t (GTNN) c\u1ee7a $M.$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> GTNN c\u1ee7a $M$ l\u00e0 $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ khi $x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span>","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Bi\u1ebfn \u0111\u1ed5i $M$ v\u1ec1 d\u1ea1ng $A(x)+a$ v\u1edbi $a$ l\u00e0 h\u1eb1ng s\u1ed1 v\u00e0 $A(x)$ l\u00e0 bi\u1ec3u th\u1ee9c kh\u00f4ng \u00e2m v\u1edbi m\u1ecdi $x$ <br\/> Khi \u0111\u00f3 GTNN c\u1ee7a $M$ l\u00e0 $a$ khi $A(x)=0$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & M=3{{a}^{2}}{{x}^{2}}+4{{b}^{2}}{{x}^{2}}-2{{a}^{2}}{{x}^{2}}-3{{b}^{2}}{{x}^{2}}+19 \\\\ & \\,\\,\\,\\,\\,\\,\\,=\\left( 3{{a}^{2}}+4{{b}^{2}}-2{{a}^{2}}-3{{b}^{2}} \\right){{x}^{2}}+19 \\\\ & \\,\\,\\,\\,\\,\\,=\\left( {{a}^{2}}+{{b}^{2}} \\right){{x}^{2}}+19 \\\\ & Do\\,\\,a,b\\,\\,\\ne 0\\,\\,\\Rightarrow {{a}^{2}}+{{b}^{2}}\\,\\, > 0 \\\\ \\end{align}$ <br\/>V\u00e0 ${{x}^{2}}\\ge 0\\,\\,\\forall \\,x$ <br\/> $\\begin{align} & \\Rightarrow \\left( {{a}^{2}}+{{b}^{2}} \\right){{x}^{2}}\\,\\,\\ge 0\\, \\\\ & \\Rightarrow M=\\left( {{a}^{2}}+{{b}^{2}} \\right){{x}^{2}}+19\\,\\ge 19 \\\\ \\end{align}$ <br\/> V\u1eady GTNN c\u1ee7a $M$ l\u00e0 $19$ khi $x=0$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u00e0 $19;0$ <\/span><\/span>"}]}],"id_ques":1144},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho hai \u0111\u01a1n th\u1ee9c $-\\dfrac{1}{3}xy^2z$ v\u00e0 $-\\dfrac{3}{5}x^3y^6z$<br\/> Khi $x;y;z$ l\u1ea5y c\u00e1c gi\u00e1 tr\u1ecb b\u1ea5t k\u00ec kh\u00e1c $0$ th\u00ec hai \u0111\u01a1n th\u1ee9c tr\u00ean c\u00f3 gi\u00e1 tr\u1ecb l\u00e0 hai s\u1ed1 c\u00f9ng d\u1ea5u. <\/span>","select":["\u0110\u00fang","Sai "],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>X\u00e9t t\u00edch c\u1ee7a \u0111\u01a1n th\u1ee9c tr\u00ean: N\u1ebfu t\u00edch c\u1ee7a ch\u00fang l\u00e0 s\u1ed1 d\u01b0\u01a1ng th\u00ec hai \u0111\u01a1n th\u1ee9c \u0111\u00f3 c\u00f9ng d\u1ea5u v\u00e0 ng\u01b0\u1ee3c l\u1ea1i.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> X\u00e9t t\u00edch c\u1ee7a hai \u0111\u01a1n th\u1ee9c \u0111\u00e3 cho: <br\/> $\\begin{align} & -\\dfrac{1}{3}x{{y}^{2}}z.\\left( -\\dfrac{3}{5}{{x}^{3}}{{y}^{6}}z \\right) \\\\ & =\\left[ -\\dfrac{1}{3}.\\left( -\\dfrac{3}{5} \\right) \\right]\\left( x{{x}^{3}} \\right)\\left( {{y}^{2}}{{y}^{6}} \\right)\\left( zz \\right) \\\\ & =\\dfrac{1}{5}{{x}^{4}}{{y}^{8}}{{z}^{2}} \\\\ \\end{align}$ <br\/> V\u1edbi $x;y;z$ b\u1ea5t k\u00ec kh\u00e1c $0$ th\u00ec $\\dfrac{1}{5}{{x}^{4}}{{y}^{8}}{{z}^{2}} > 0$ <br\/> $\\Rightarrow$ Hai \u0111\u01a1n th\u1ee9c \u0111\u00e3 cho c\u00f9ng d\u1ea5u v\u1edbi nhau. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}]}],"id_ques":1145},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho $A=3\\left( {{a}^{2}}+\\dfrac{1}{{{a}^{2}}} \\right){{x}^{2}}{{y}^{4}}{{z}^{6}}$ v\u1edbi $a$ l\u00e0 h\u1eb1ng s\u1ed1 kh\u00e1c $0.$ <br\/> V\u1edbi m\u1ecdi $x;y;z$ th\u00ec $A$ lu\u00f4n kh\u00f4ng \u00e2m. <\/span>","select":["\u0110\u00fang","Sai "],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> \u0110\u00e1nh gi\u00e1 \u0111\u01a1n th\u1ee9c $A$ nh\u1eadn gi\u00e1 tr\u1ecb nh\u01b0 th\u1ebf n\u00e0o v\u1edbi m\u1ecdi $x;y;z$<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: $3\\left( {{a}^{2}}+\\dfrac{1}{{{a}^{2}}} \\right) > 0$ v\u1edbi m\u1ecdi $a$ kh\u00e1c $0$ <br\/> L\u1ea1i c\u00f3: $x^2y^4z^6 \\ge 0$ v\u1edbi m\u1ecdi $x;y;z \\in \\mathbb{R}$ <br\/> V\u1eady $A \\ge 0$ v\u1edbi m\u1ecdi $x;y;z \\in \\mathbb{R}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}]}],"id_ques":1146},{"time":24,"part":[{"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'> Cho ba \u0111\u01a1n th\u1ee9c $-\\dfrac{1}{2}xy^2;\\,-\\dfrac{3}{4}x^3y;\\,2y$ <br\/> T\u00edch c\u1ee7a ba \u0111\u01a1n th\u1ee9c tr\u00ean lu\u00f4n: <\/span>","select":["A. $\\ge 0$","B. $\\le 0$ "],"explain":"<span class='basic_left'> X\u00e9t t\u00edch c\u1ee7a ba \u0111\u01a1n th\u1ee9c \u0111\u00e3 cho: <br\/> $\\begin{align} & -\\dfrac{1}{2}x{{y}^{2}}.\\left( -\\dfrac{3}{4}{{x}^{3}}y \\right).2y \\\\ & =\\left[ -\\dfrac{1}{2}.\\left( -\\dfrac{3}{4} \\right).2 \\right]\\left( x{{x}^{3}} \\right)\\left( {{y}^{2}}yy \\right) \\\\ & =\\dfrac{3}{4}{{x}^{4}}{{y}^{4}}\\,\\,\\ge 0\\,\\,\\forall x;y \\\\ \\end{align}$ <br\/> $\\Rightarrow$ T\u00edch c\u1ee7a ba \u0111\u01a1n th\u1ee9c tr\u00ean lu\u00f4n lu\u00f4n kh\u00f4ng \u00e2m. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":1147},{"time":24,"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'> \u0110\u01a1n th\u1ee9c $3x^{n+3}y^{m-2}$ \u0111\u01b0\u1ee3c vi\u1ebft d\u01b0\u1edbi d\u1ea1ng t\u00edch c\u1ee7a hai \u0111\u01a1n th\u1ee9c trong \u0111\u00f3 m\u1ed9t \u0111\u01a1n th\u1ee9c b\u1eb1ng $\\dfrac{2}{5}x^ny^2$ $(m,n\\in \\mathbb{N}, m> 4)$ <br\/> \u0110\u01a1n th\u1ee9c c\u00f2n l\u1ea1i l\u00e0: <\/span>","select":["A. $\\dfrac{1}{15}x^3y^{m+4}$","B. $\\dfrac{3}{10}x^2y^{m+2}$ ","C. $\\dfrac{2}{15}x^2y^{m}$","D. $\\dfrac{15}{2}x^3y^{m-4}$"],"explain":"<span class='basic_left'> <span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>G\u1ecdi \u0111\u01a1n th\u1ee9c c\u1ea7n t\u00ecm c\u00f3 d\u1ea1ng $ax^py^q \\,(p,q \\in \\mathbb{N})$ <br\/> T\u00ednh t\u00edch hai \u0111\u01a1n th\u1ee9c $-\\dfrac{2}{5}x^ny^2$ $(m,n\\in \\mathbb{N}, m> 4)$ v\u00e0 $ax^py^q$ <br\/> Cho t\u00edch \u0111\u00f3 b\u1eb1ng $3x^{n+3}y^{m-2}$ v\u00e0 t\u00ecm $a;p;q$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> G\u1ecdi \u0111\u01a1n th\u1ee9c c\u1ea7n t\u00ecm c\u00f3 d\u1ea1ng $ax^py^q \\,(p,q \\in \\mathbb{N})$, $a$ l\u00e0 h\u1eb1ng s\u1ed1 v\u00e0 $a\\in \\mathbb{R}$ <br\/> Ta c\u00f3: <br\/> $\\begin{aligned} & 3{{x}^{n+3}}{{y}^{m-2}}=\\dfrac{2}{5}{{x}^{n}}{{y}^{2}}.a\\,{{x}^{p}}{{y}^{q}} \\\\ & 3{{x}^{n+3}}{{y}^{m-2}}=\\dfrac{2}{5}a\\,{{x}^{n+p}}{{y}^{2+q}} \\\\ & \\Rightarrow \\left\\{ \\begin{aligned} & 3=\\dfrac{2}{5}a \\\\ & n+3=n+p \\\\ & m-2=2+q \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & a=\\dfrac{15}{2} \\\\ & p=3 \\\\ & q=m-2-2=m-4 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/> V\u1eady \u0111\u01a1n th\u1ee9c ph\u1ea3i t\u00ecm l\u00e0 $\\dfrac{15}{2}x^3y^{m-4}$<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span>","column":2}]}],"id_ques":1148},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho $M$ l\u00e0 t\u1ed5ng c\u00e1c \u0111\u01a1n th\u1ee9c \u0111\u1ed3ng d\u1ea1ng: <br\/> $M=5a\\,{{x}^{2}}{{y}^{2}}+\\left( -\\dfrac{1}{2}a\\,{{x}^{2}}{{y}^{2}} \\right)+7a{{x}^{2}}{{y}^{2}}+\\left( -a{{x}^{2}}{{y}^{2}} \\right)$ <br\/> <b> C\u00e2u 1. <\/b> \u0110\u1ec3 $M\\ge 0$ th\u00ec $a$ th\u1ecfa m\u00e3n: <br\/> <\/span>","select":["A. $a \\ge -1$","B. $a \\ge 0$ ","C. $a \\ge 1$","D. $a \\ge 2$"],"explain":"<span class='basic_left'> <span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>T\u00ednh t\u1ed5ng c\u00e1c \u0111\u01a1n th\u1ee9c \u0111\u00e3 cho. <br\/> \u0110\u00e1nh gi\u00e1 v\u00e0 t\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $a$ \u0111\u1ec3 $M\\ge 0.$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & M=5a\\,{{x}^{2}}{{y}^{2}}+\\left( -\\dfrac{1}{2}a\\,{{x}^{2}}{{y}^{2}} \\right)+7a{{x}^{2}}{{y}^{2}}+\\left( -a{{x}^{2}}{{y}^{2}} \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,=\\left( 5-\\dfrac{1}{2}+7-1 \\right)a{{x}^{2}}{{y}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,\\,= \\dfrac{21}{2}a{{x}^{2}}{{y}^{2}} \\\\ \\end{align}$ <br\/> Do $x^2y^2\\ge 0$ v\u1edbi m\u1ecdi $x,y$ n\u00ean \u0111\u1ec3 $M \\ge 0$ v\u1edbi m\u1ecdi $x;y$ th\u00ec: <br\/> $\\begin{align} & \\dfrac{21}{2}a\\ge 0 \\\\ & \\Rightarrow a\\ge 0 \\\\ \\end{align}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span>","column":2}]}],"id_ques":1149},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":" <span class='basic_left'> Cho $M$ l\u00e0 t\u1ed5ng c\u00e1c \u0111\u01a1n th\u1ee9c \u0111\u1ed3ng d\u1ea1ng: <br\/> $M=5a\\,{{x}^{2}}{{y}^{2}}+\\left( -\\dfrac{1}{2}a\\,{{x}^{2}}{{y}^{2}} \\right)+7a{{x}^{2}}{{y}^{2}}+\\left( -a{{x}^{2}}{{y}^{2}} \\right)$ <br\/> <b> C\u00e2u 2. <\/b> V\u1edbi $a=2$ th\u00ec s\u1ed1 c\u1eb7p s\u1ed1 nguy\u00ean $(x;y)$ th\u1ecfa m\u00e3n $M=84$ l\u00e0: <br\/> <\/span>","select":["A. $8$","B. $7$ ","C. $6$","D. $5$"],"explain":"<span class='basic_left'> <span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>T\u00ednh t\u1ed5ng c\u00e1c \u0111\u01a1n th\u1ee9c \u0111\u00e3 cho. <br\/> \u0110\u00e1nh gi\u00e1 v\u00e0 t\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $a$ \u0111\u1ec3 $M\\ge 0.$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Theo c\u00e2u 1, ta c\u00f3: $M=5a\\,{{x}^{2}}{{y}^{2}}+\\left( -\\dfrac{1}{2}a\\,{{x}^{2}}{{y}^{2}} \\right)+7a{{x}^{2}}{{y}^{2}}+\\left( -a{{x}^{2}}{{y}^{2}} \\right)=\\dfrac{21}{2}ax^2y^2$ <br\/> V\u1edbi $a=2;M=84$ ta c\u00f3: <br\/> $84=\\dfrac{21}{2}.2x^2y^2$ <br\/> $\\Rightarrow x^2y^2=4$ <br\/> $\\Rightarrow xy=2$ ho\u1eb7c $xy=-2$ <br\/> - N\u1ebfu $xy=2$ th\u00ec $\\left[ \\begin{align} & x=1;y=2 \\\\ & x=2;y=1 \\\\ & x=-1;y=-2 \\\\ & x=-2;y=-1 \\\\ \\end{align} \\right.$ <br\/> - N\u1ebfu $xy=-2$ th\u00ec $\\left[ \\begin{align} & x=-1;y=2 \\\\ & x=2;y=-1 \\\\ & x=1;y=-2 \\\\ & x=-2;y=1 \\\\ \\end{align} \\right.$ <br\/> V\u1eady c\u00f3 $8$ c\u1eb7p s\u1ed1 nguy\u00ean $(x;y)$ th\u1ecfa m\u00e3n \u0111\u1ec1 b\u00e0i. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":4}]}],"id_ques":1150}],"lesson":{"save":0,"level":3}}