{"segment":[{"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'> \u0110a th\u1ee9c $4x^{2004}+6x^{2006}+2007$ kh\u00f4ng c\u00f3 nghi\u1ec7m trong $\\mathbb{R}.$<\/span>","select":["\u0110\u00fang","Sai "],"hint":"Nh\u1eadn \u0111\u1ecbnh v\u1ec1 gi\u00e1 tr\u1ecb c\u1ee7a $x^{2004}$ v\u00e0 $x^{2006}$","explain":"<span class='basic_left'> Ta c\u00f3 $x^{2004} \\ge 0; x^{2006}\\ge 0$ v\u1edbi m\u1ecdi $x$ <br\/> $\\Rightarrow 4x^{2004}+6x^{2006}+2007 \\ge 2007 > 0$ v\u1edbi $\\forall x \\in \\mathbb{R}$ <br\/> Do \u0111\u00f3, \u0111a th\u1ee9c $4x^{2004}+6x^{2006}+2007$ kh\u00f4ng c\u00f3 nghi\u1ec7m trong $\\mathbb{R}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}]}],"id_ques":1241},{"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":[[["-0,2"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","input_hint":["frac"],"ques":" <span class='basic_left'> Cho \u0111a th\u1ee9c sau: <br\/> $Q\\left( x \\right)=4,3+\\left[ -3,4x-\\left( 8x+6 \\right)-2,3 \\right]-\\left( 2,6x-1,2 \\right)$ <br\/> \u0110\u1ec3 $Q(x)=0$ th\u00ec gi\u00e1 tr\u1ecb c\u1ee7a $x$ l\u00e0 _input_ <\/span> ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>R\u00fat g\u1ecdn \u0111a th\u1ee9c $Q(x)$ <br\/> Cho $Q(x)=0$ r\u1ed3i t\u00ecm $x$<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{align} & Q\\left( x \\right)=4,3+\\left[ -3,4x-\\left( 8x+6 \\right)-2,3 \\right]-\\left( 2,6x-1,2 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=4,3+\\left(-3,4x-8x-6-2,3 \\right)-2,6x+1,2 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=4,3+\\left(-11,4x-8,3 \\right)-2,6x+1,2 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=4,3-11,4x-8,3-2,6x+1,2 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=-14x-2,8 \\\\ \\end{align}$ <br\/> \u0110\u1ec3 $Q(x)=0$ ta c\u00f3: <br\/> $\\begin{align} -14x-2,8 &=0 \\\\ 14x &=-2,8 \\\\ x &=-0,2 \\\\ \\end{align}$<\/span>"}]}],"id_ques":1242},{"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":[[["8"],["3"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho hai \u0111a th\u1ee9c sau: <br\/> $A={{x}^{6}}-3{{x}^{3}}+3{{x}^{2}}-6x+2{{x}^{3}}+10-3{{x}^{2}}-{{x}^{6}}$ <br\/> $B=3{{x}^{2}}-5x+2-6{{x}^{2}}+4{{x}^{3}}+5-8{{x}^{5}}-{{x}^{3}}$ <br\/> H\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a \u0111a th\u1ee9c $A-B$ l\u00e0 _input_ <br\/> H\u1ec7 s\u1ed1 t\u1ef1 do c\u1ee7a \u0111a th\u1ee9c $A-B$ l\u00e0 _input_ <\/span> ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> R\u00fat g\u1ecdn hai \u0111a th\u1ee9c $A$ v\u00e0 $B$ \u0111\u00e3 cho v\u00e0 s\u1eafp x\u1ebfp theo l\u0169y th\u1eeba gi\u1ea3m d\u1ea7n. <br\/> T\u00ednh $A-B$ v\u00e0 t\u00ecm h\u1ec7 s\u1ed1 cao nh\u1ea5t v\u00e0 h\u1ec7 s\u1ed1 t\u1ef1 do c\u1ee7a \u0111a th\u1ee9c $A-B$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & A={{x}^{6}}-3{{x}^{3}}+3{{x}^{2}}-6x+2{{x}^{3}}+10-3{{x}^{2}}-{{x}^{6}} \\\\ & \\,\\,\\,\\,\\,\\,=\\left( {{x}^{6}}-{{x}^{6}} \\right)+\\left( -3{{x}^{3}}+2{{x}^{3}} \\right)+\\left( 3{{x}^{2}}-3{{x}^{2}} \\right)-6x+10 \\\\ & \\,\\,\\,\\,\\,\\,=-{{x}^{3}}-6x+10 \\\\ & B=3{{x}^{2}}-5x+2-6{{x}^{2}}+4{{x}^{3}}+5-8{{x}^{5}}-{{x}^{3}} \\\\ & \\,\\,\\,\\,\\,=-8{{x}^{5}}+\\left( 4{{x}^{3}}-{{x}^{3}} \\right)+\\left( 3{{x}^{2}}-6{{x}^{2}} \\right)-5x+\\left( 2+5 \\right) \\\\ & \\,\\,\\,\\,\\,=-8{{x}^{5}}+3{{x}^{3}}-3{{x}^{2}}-5x+7 \\\\ & \\Rightarrow A-B=\\left( -{{x}^{3}}-6x+10 \\right)-\\left( -8{{x}^{5}}+3{{x}^{3}}-3{{x}^{2}}-5x+7 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=-{{x}^{3}}-6x+10+8{{x}^{5}}-3{{x}^{3}}+3{{x}^{2}}+5x-7 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=8{{x}^{5}}+\\left( -{{x}^{3}}-3{{x}^{3}} \\right)+3{{x}^{2}}+\\left( -6x+5x \\right)+\\left( 10-7 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=8{{x}^{5}}-4{{x}^{3}}+3{{x}^{2}}-x+3 \\\\ \\end{align}$ <br\/> L\u0169y th\u1eeba cao nh\u1ea5t b\u1eadc $5$ n\u00ean h\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a \u0111a th\u1ee9c $A-B$ l\u00e0 $8$ <br\/> H\u1ec7 s\u1ed1 t\u1ef1 do l\u00e0 $3$ <br\/> <span class='basic_pink'> V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $8$ v\u00e0 $3$<\/span><\/span>"}]}],"id_ques":1243},{"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'> X\u00e9t \u0111a th\u1ee9c $P(x)=ax^3+bx^2+cx+d$ <br\/> N\u1ebfu $-a+b-c+d=0$ th\u00ec $P(x)$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=-1$<\/span>","select":["\u0110\u00fang","Sai "],"hint":"Ki\u1ec3m tra t\u1ea1i $x=-1$ gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c $P(x)$ c\u00f3 b\u1eb1ng $0$ kh\u00f4ng.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> V\u1edbi $-a+b-c+d=0$ th\u00ec $P(-1)=0$ hay kh\u00f4ng? <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Thay $x=-1$ v\u00e0o \u0111a th\u1ee9c $P(x)$ ta \u0111\u01b0\u1ee3c: <br\/> $a.(-1)^3+b.(-1)^2+c.(-1)+d$ <br\/> $=-a+b-c+d$ <br\/> $=0$ (theo gi\u1ea3 thi\u1ebft) <br\/> Do \u0111\u00f3, $P(-1)=0$ n\u00ean $x=-1$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $P(x)$ v\u1edbi $-a+b-c+d=0$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}]}],"id_ques":1244},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $a= -2; b= \\frac{1}{3}$","B.$a= -2; b = 1$","C. $a= -2; b= -1$"],"ques":"X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 $a,b$ c\u1ee7a \u0111a th\u1ee9c $P(x)=ax+b$ bi\u1ebft r\u1eb1ng $P(1)=-\\dfrac{5}{3}$ v\u00e0 $P\\left( -\\dfrac{1}{2} \\right)=\\dfrac{4}{3}$ <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> $a=$ ?; $b=$ ?","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u1eeb $P(1)=-\\dfrac{5}{3}$ v\u00e0 $P\\left( -\\dfrac{1}{2} \\right)=\\dfrac{4}{3}$ t\u00ecm hai bi\u1ec3u th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa $a$ v\u00e0 $b$ <br\/> T\u1eeb hai bi\u1ec3u th\u1ee9c \u0111\u00f3 bi\u1ebfn \u0111\u1ed5i \u0111\u1ec3 t\u00ecm $a$ v\u00e0 $b$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3 $P(1)=-\\dfrac{5}{3}$ (gi\u1ea3 thi\u1ebft) n\u00ean: <br\/> $\\begin{align} & a.1+b=-\\dfrac{5}{3} \\\\ & \\Rightarrow a+b=-\\dfrac{5}{3} \\\\ & \\Rightarrow b=-\\dfrac{5}{3}-a\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ \\end{align}$ <br\/> V\u00e0 $P\\left( -\\dfrac{1}{2} \\right)=\\dfrac{4}{3}$ n\u00ean: <br\/> $\\begin{align} & a.\\left( -\\dfrac{1}{2} \\right)+b=\\dfrac{4}{3} \\\\ & \\Rightarrow -\\dfrac{1}{2}a+b=\\dfrac{4}{3}\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align}$ <br\/> Thay (1) v\u00e0o (2) ta \u0111\u01b0\u1ee3c: <br\/> $\\begin{align} & -\\dfrac{1}{2}a-\\dfrac{5}{3}-a=\\dfrac{4}{3} \\\\ & \\Rightarrow -\\dfrac{3a}{2}=\\dfrac{4}{3}+\\dfrac{5}{3} \\\\ & \\Rightarrow -\\dfrac{3a}{2}=3 \\\\ & \\Rightarrow a=-2 \\\\ \\end{align}$ <br\/> Thay $a=-2$ v\u00e0o (1) ta c\u00f3: <br\/> $b=-\\dfrac{5}{3}-a=-\\dfrac{5}{3}+2=\\dfrac{1}{3}$<\/span>"}]}],"id_ques":1245},{"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":[[["4"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","ques":" <span class='basic_left'> Cho \u0111a th\u1ee9c $a\\,{{x}^{4}}-6{{x}^{3}}+7-2x+3{{x}^{2}}-4{{x}^{4}}$ <br\/> T\u00ecm $a,$ bi\u1ebft r\u1eb1ng \u0111a th\u1ee9c n\u00e0y c\u00f3 b\u1eadc l\u00e0 $3.$ <br\/> <b> \u0110\u00e1p \u00e1n:<\/b> $a=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","hint":"L\u0169y th\u1eeba c\u00f3 b\u1eadc cao nh\u1ea5t trong \u0111a th\u1ee9c l\u00e0 $3$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> R\u00fat g\u1ecdn \u0111a th\u1ee9c \u0111\u00e3 cho <br\/> B\u1eadc c\u1ee7a \u0111a th\u1ee9c l\u00e0 $3$ n\u00ean l\u0169y th\u1eeba c\u00f3 b\u1eadc cao nh\u1ea5t l\u00e0 $3$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & a\\,{{x}^{4}}-6{{x}^{3}}+7-2x+3{{x}^{2}}-4{{x}^{4}} \\\\ & =\\left( a\\,{{x}^{4}}-4{{x}^{4}} \\right)-6{{x}^{3}}+3{{x}^{2}}-2x+7 \\\\ & =\\left( a-4 \\right){{x}^{4}}-6{{x}^{3}}+3{{x}^{2}}-2x+7 \\\\ \\end{align}$ <br\/> \u0110\u1ec3 \u0111a th\u1ee9c \u0111\u00e3 cho c\u00f3 b\u1eadc l\u00e0 $3$ th\u00ec l\u0169y th\u1eeba c\u00f3 b\u1eadc cao nh\u1ea5t l\u00e0 $3$ <br\/> Do \u0111\u00f3, h\u1ec7 s\u1ed1 c\u1ee7a l\u0169y th\u1eeba b\u1eadc $4$ l\u00e0 $0$ <br\/> $\\Rightarrow a-4=0 \\Rightarrow a=4$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4$<\/span><\/span>"}]}],"id_ques":1246},{"time":24,"part":[{"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 \u0111a th\u1ee9c $-2{{x}^{3}}-7{{x}^{5}}+6{{x}^{2}}-4x+b+a\\,{{x}^{5}}$ <br\/> T\u00ecm $a,b$ bi\u1ebft r\u1eb1ng \u0111a th\u1ee9c n\u00e0y c\u00f3 h\u1ec7 s\u1ed1 cao nh\u1ea5t l\u00e0 $-2$ v\u00e0 h\u1ec7 s\u1ed1 t\u1ef1 do l\u00e0 $3.$ <\/span>","select":["A. $a=-2;b=3$ ho\u1eb7c $a=7;b=3$","B. $a=5;b=3$ ho\u1eb7c $a=7;b=3$","C. $a=5;b=3$ ","D. $a=-2;b=3$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Thu g\u1ecdn \u0111a th\u1ee9c \u0111\u00e3 cho, s\u1eafp x\u1ebfp \u0111a th\u1ee9c theo l\u0169y th\u1eeba gi\u1ea3m d\u1ea7n c\u1ee7a bi\u1ebfn. <br\/> X\u00e9t hai tr\u01b0\u1eddng h\u1ee3p \u0111a th\u1ee9c c\u00f3 b\u1eadc l\u00e0 $5$ v\u00e0 \u0111a th\u1ee9c c\u00f3 b\u1eadc l\u00e0 $3$ th\u1ecfa m\u00e3n h\u1ec7 s\u1ed1 cao nh\u1ea5t l\u00e0 $-2$ v\u00e0 h\u1ec7 s\u1ed1 t\u1ef1 do l\u00e0 $3$<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{align} & -2{{x}^{3}}-7{{x}^{5}}+6{{x}^{2}}-4x+b+a\\,{{x}^{5}} \\\\ & =\\left( -7{{x}^{5}}+a\\,{{x}^{5}} \\right)-2{{x}^{3}}+6{{x}^{2}}-4x+b \\\\ & =\\left( a-7 \\right){{x}^{5}}-2{{x}^{3}}+6{{x}^{2}}-4x+b \\\\ \\end{align}$ <br\/> Do \u0111a th\u1ee9c \u0111\u00e3 cho c\u00f3 h\u1ec7 s\u1ed1 cao nh\u1ea5t l\u00e0 $-2$ v\u00e0 h\u1ec7 s\u1ed1 t\u1ef1 do l\u00e0 $3$ n\u00ean ta c\u00f3 hai tr\u01b0\u1eddng h\u1ee3p: <br\/> + Tr\u01b0\u1eddng h\u1ee3p $1:$ \u0110a th\u1ee9c c\u00f3 b\u1eadc l\u00e0 $5$, h\u1ec7 s\u1ed1 cao nh\u1ea5t l\u00e0 $-2$, h\u1ec7 s\u1ed1 t\u1ef1 do l\u00e0 $3$ <br\/> $\\Rightarrow \\left\\{ \\begin{aligned} & a-7=-2 \\\\ & b=3 \\\\ \\end{aligned} \\right. \\Rightarrow \\left\\{ \\begin{aligned} & a=5 \\\\ & b=3 \\\\ \\end{aligned} \\right. $ <br\/> + Tr\u01b0\u1eddng h\u1ee3p $2:$ \u0110a th\u1ee9c c\u00f3 b\u1eadc l\u00e0 $3,$ h\u1ec7 s\u1ed1 cao nh\u1ea5t l\u00e0 $-2,$ h\u1ec7 s\u1ed1 t\u1ef1 do l\u00e0 $3$ <br\/> $\\Rightarrow \\left\\{ \\begin{aligned} & a-7=0 \\\\ & b=3 \\\\ \\end{aligned} \\right. \\Rightarrow \\left\\{ \\begin{aligned} & a=7 \\\\ & b=3 \\\\ \\end{aligned} \\right. $ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span> <br\/> <b> Nh\u1eadn x\u00e9t:<\/b> \u0110a th\u1ee9c trong b\u00e0i sau khi thu g\u1ecdn c\u00f3 h\u1ec7 s\u1ed1 c\u1ee7a l\u0169y th\u1eeba b\u1eadc $3$ l\u00e0 $-2$ (tr\u00f9ng v\u1edbi h\u1ec7 s\u1ed1 cao nh\u1ea5t) n\u00ean ta ph\u1ea3i x\u00e9t hai tr\u01b0\u1eddng h\u1ee3p nh\u01b0 trong b\u00e0i. <\/span>","column":2}]}],"id_ques":1247},{"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'> Cho \u0111a th\u1ee9c $P\\left( x \\right)={{x}^{99}}-100{{x}^{98}}+100{{x}^{97}}-100{{x}^{96}}+\\cdots +100x-1$ <br\/> $P(99)$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng: <\/span>","select":["A. $1$","B. $100$","C. $99$ ","D. $98$"],"hint":"T\u00e1ch $100x^n=99x^n+x^n$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> Ta th\u1ea5y $x-99=99-99=0$ <br\/> Do \u0111\u00f3, ta t\u00e1ch c\u00e1c h\u1ea1ng t\u1eed c\u00f3 d\u1ea1ng $100x^n=99x^n+x^n$ <br\/> Nh\u00f3m c\u00e1c h\u1ea1ng t\u1eed v\u00e0 ph\u00e2n t\u00edch c\u00e1c nh\u00f3m \u0111\u00f3 b\u1eb1ng c\u00e1ch \u0111\u1eb7t nh\u00e2n t\u1eed chung \u0111\u1ec3 l\u00e0m xu\u1ea5t hi\u1ec7n $x-99$ <br\/> Thay $x=99$ v\u00e0o r\u1ed3i t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3: <br\/> $\\begin{align} & P\\left( x \\right)={{x}^{99}}-100{{x}^{98}}+100{{x}^{97}}-100{{x}^{96}}+\\cdots +100x-1 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{x}^{99}}-99{{x}^{98}}-{{x}^{98}}+99{{x}^{97}}+{{x}^{97}}-99{{x}^{96}}-{{x}^{96}}+\\cdots +99x+x-1 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{x}^{98}}\\left( x-99 \\right)-{{x}^{97}}\\left( x-99 \\right)+{{x}^{96}}\\left( x-99 \\right)-\\cdots -x\\left( x-99 \\right)+\\left( x-1 \\right) \\\\ & \\Rightarrow P\\left( 99 \\right)={{99}^{98}}\\left( 99-99 \\right)-{{99}^{97}}\\left( 99-99 \\right)+{{99}^{96}}\\left( 99-99 \\right)-\\cdots -99\\left( 99-99 \\right)+\\left( 99-1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=0-0+0-\\cdots -0+99-1 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=98 \\\\ \\end{align}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span> <\/span>","column":4}]}],"id_ques":1248},{"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'> X\u00e1c \u0111\u1ecbnh \u0111a th\u1ee9c b\u1eadc hai $P(x)=ax^2+bx+c$ bi\u1ebft r\u1eb1ng $P(1)=0;P(-1)=6$ v\u00e0 $P(-2)=3.$ <\/span>","select":["A. $P(x)=-2x^2-3x+5 $","B. $P(x)=2x^2-3x+1$","C. $P(x)=-2x^2-3x+1$","D. $P(x)=2x^2+5x-3$"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u00ecm ba bi\u1ec3u th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa $a,b$ v\u00e0 $c$ t\u1eeb $P(1)=0;P(-1)=6$ v\u00e0 $P(-2)=3.$ <br\/> T\u1eeb ba bi\u1ec3u th\u1ee9c li\u00ean h\u1ec7 \u0111\u00f3 bi\u1ebfn \u0111\u1ed5i \u0111\u1ec3 t\u00ecm ra $a,b$ v\u00e0 $c$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/>$\\begin{align} & P(1)=0 \\\\ & \\Rightarrow a{{.1}^{2}}+b.1+c=0 \\\\ & \\Rightarrow a+b+c=0\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ \\end{align}$ <br\/> V\u00e0 $P(-1)=6$ n\u00ean: <br\/> $\\begin{align} & a.{{\\left( -1 \\right)}^{2}}+b.\\left( -1 \\right)+c=6 \\\\ & \\Rightarrow a-b+c=6 \\\\ & \\Rightarrow a+c=b+6\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align}$ <br\/> $\\begin{align} & P(-2)=3 \\\\ & \\Rightarrow a.{{\\left( -2 \\right)}^{2}}+b.\\left( -2 \\right)+c=3 \\\\ & \\Rightarrow 4a-2b+c=3\\,\\,\\,\\,\\,\\,\\left( 3 \\right) \\\\ \\end{align}$ <br\/> Thay (2) v\u00e0o (1) ta \u0111\u01b0\u1ee3c: <br\/> $\\begin{align} & b+b+6=0 \\\\ & \\Rightarrow 2b+6=0 \\\\ & \\Rightarrow b=-3 \\\\ \\end{align}$ <br\/> Thay $b=-3$ v\u00e0o (2) v\u00e0 (3) ta \u0111\u01b0\u1ee3c: <br\/> $\\left\\{ \\begin{aligned} & a+c=3 \\\\ & 4a-2.\\left( -3 \\right)+c=3 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & a+c=3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 4 \\right) \\\\ & 4a+c=-3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 5 \\right) \\\\ \\end{aligned} \\right.$ <br\/> T\u1eeb (4) suy ra $c=3-a,$ thay v\u00e0o (5) ta \u0111\u01b0\u1ee3c: <br\/> $\\begin{align} & 4a+3-a=-3 \\\\ & \\Rightarrow 3a=-6 \\\\ & \\Rightarrow a=-2 \\\\ \\end{align}$ <br\/> Suy ra $c=3-a=3-(-2)=5$ <br\/> V\u1eady \u0111a th\u1ee9c c\u1ea7n t\u00ecm l\u00e0 $P(x)=-2x^2-3x+5$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":1249},{"time":24,"part":[{"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 \u0111a th\u1ee9c $f(x)=ax^2+bx+c,$ bi\u1ebft $13a+b+2c=0.$ <br\/> Khi \u0111\u00f3: <\/span>","select":["A. $f(-2).f(3) \\ge 0 $","B. $f(-2).f(3) \\le 0 $"],"explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> T\u00ednh $f(-2)$ v\u00e0 $f(3)$ <br\/> T\u00ednh $f(-2)+f(3)$, k\u1ebft h\u1ee3p v\u1edbi $13a+b+2c=0$ \u0111\u1ec3 t\u00ecm ra gi\u00e1 tr\u1ecb c\u1ee7a $f(-2)+f(3)$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/>$\\begin{align} & f\\left( -2 \\right)=a.{{\\left( -2 \\right)}^{2}}+b.\\left( -2 \\right)+c=4a-2b+c \\\\ & f\\left( 3 \\right)=a{{.3}^{2}}+b.3+c=9a+3b+c \\\\ & \\Rightarrow f\\left( -2 \\right)+f\\left( 3 \\right)=\\left( 4a-2b+c \\right)+\\left( 9a+3b+c \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=4a-2b+c+9a+3b+c \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=13a+b+2c \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=0 \\\\ \\end{align}$ <br\/> Do \u0111\u00f3 $f(-2)$ v\u00e0 $f(3)$ \u0111\u1ed1i nhau. <br\/> $\\Rightarrow f(-2).f(3)\\le 0$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span> <br\/> <b> L\u01b0u \u00fd:<\/b> Hai s\u1ed1 $a$ v\u00e0 $b$ \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 \u0111\u1ed1i nhau khi v\u00e0 ch\u1ec9 khi $a+b=0$ <\/span>","column":2}]}],"id_ques":1250}],"lesson":{"save":0,"level":3}}