{"segment":[{"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'> \u0110\u01a1n th\u1ee9c n\u00e0o sau \u0111\u00e2y \u0111\u1ed3ng d\u1ea1ng v\u1edbi \u0111\u01a1n th\u1ee9c $- 3x^3y^2$? <\/span> ","select":["A. $4x^2y^3$ ","B. $-xy$","C. $2(xy)^2$ ","D. $2(xy)^2.(-3x)$ "],"hint":"","explain":" <span class='basic_left'> Ta c\u00f3: <br\/> $\\circ \\,\\,4x^2y^3$ c\u00f3 ph\u1ea7n bi\u1ebfn l\u00e0 $x^2y^3$ <br\/> $\\circ \\,\\,-xy$ c\u00f3 ph\u1ea7n bi\u1ebfn l\u00e0 $xy$ <br\/> $\\circ \\,\\,2(xy)^2$ c\u00f3 ph\u1ea7n bi\u1ebfn l\u00e0 $(xy)^2=x^2y^2$ <br\/> $\\circ \\,\\, 2(xy)^2.(-3x)=-6x^3y^2$ c\u00f3 ph\u1ea7n bi\u1ebfn l\u00e0 $x^3y^2$ <br\/> Do \u0111\u00f3, \u0111\u01a1n th\u1ee9c $2(xy)^2.(-3x)$ \u0111\u1ed3ng d\u1ea1ng v\u1edbi \u0111\u01a1n th\u1ee9c $-3x^3y^2$ \u0111\u00e3 cho. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span> ","column":2}],"id_ques":1311},{"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'> H\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a \u0111a th\u1ee9c $-4{{x}^{3}}-2{{x}^{2}}+7x+2012$ l\u00e0: <\/span> ","select":["A. $-4$ ","B. $3$","C. $-2$ ","D. $2012$ "],"hint":"","explain":" <span class='basic_left'> \u0110a th\u1ee9c $-4{{x}^{3}}-2{{x}^{2}}+7x+2012$ c\u00f3 l\u0169y th\u1eeba b\u1eadc cao nh\u1ea5t l\u00e0 $3$ <br\/> Do \u0111\u00f3, h\u1ec7 s\u1ed1 cao nh\u1ea5t c\u1ee7a \u0111a th\u1ee9c l\u00e0 $-4$<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":4}],"id_ques":1312},{"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'> \u0110a th\u1ee9c ${{x}^{4}}y+2x{{y}^{2}}-3xyz-6{{x}^{2}}y{{z}^{3}}$ c\u00f3 b\u1eadc l\u00e0: <\/span> ","select":["A. $17$ ","B. $6$","C. $5$","D. $3$ "],"hint":"","explain":" <span class='basic_left'> \u0110a th\u1ee9c ${{x}^{4}}y+2x{{y}^{2}}-3xyz-6{{x}^{2}}y{{z}^{3}}$ g\u1ed3m c\u00f3 $4$ h\u1ea1ng t\u1eed. <br\/> Trong \u0111\u00f3, h\u1ea1ng t\u1eed $-6x^2yz^3$ c\u00f3 b\u1eadc l\u00e0 $6$ v\u00e0 l\u00e0 b\u1eadc cao nh\u1ea5t trong c\u00e1c h\u1ea1ng t\u1eed c\u1ee7a \u0111a th\u1ee9c <br\/> Do \u0111\u00f3, b\u1eadc c\u1ee7a \u0111a th\u1ee9c \u0111\u00e3 cho l\u00e0 $6$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":4}],"id_ques":1313},{"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'> Thu g\u1ecdn \u0111\u01a1n th\u1ee9c $1,25x.(-4{{x}^{3}}{{y}^{2}}).\\left(9\\dfrac{1}{3}x{{y}^{2}}{{z}^{4}}\\right)$ \u0111\u01b0\u1ee3c \u0111\u01a1n th\u1ee9c c\u00f3 ph\u1ea7n bi\u1ebfn l\u00e0: <\/span> ","select":["A. ${{x}^{4}}{{y}^{2}}{{z}^{4}}$ ","B. ${{x}^{4}}{{y}^{4}}{{z}^{4}}$","C. ${{x}^{5}}{{y}^{4}}{{z}^{4}}$","D. ${{x}^{5}}{{y}^{4}}$ "],"hint":"Thu g\u1ecdn \u0111\u01a1n th\u1ee9c, ta l\u1ea5y h\u1ec7 s\u1ed1 nh\u00e2n h\u1ec7 s\u1ed1, bi\u1ebfn nh\u00e2n bi\u1ebfn.","explain":" <span class='basic_left'>Ta c\u00f3: <br\/> $\\begin{align} & 1,25x.(-4{{x}^{3}}{{y}^{2}}).\\left(9\\dfrac{1}{3}x{{y}^{2}}{{z}^{4}}\\right) \\\\ & =\\left[ \\dfrac{5}{4}.\\left( -4 \\right).\\dfrac{28}{3} \\right]\\left( x{{x}^{3}}x \\right)\\left({{y}^{2}}{{y}^{2}} \\right){{z}^{4}} \\\\ & =-\\dfrac{140}{3}{{x}^{5}}{{y}^{4}}{{z}^{4}} \\\\ \\end{align}$ <br\/> Ph\u1ea7n bi\u1ebfn c\u1ee7a \u0111\u01a1n th\u1ee9c sau khi thu g\u1ecdn l\u00e0 $x^5y^4z^4$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}],"id_ques":1314},{"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'> T\u00ecm \u0111a th\u1ee9c $M$ bi\u1ebft: <br\/>$M-(-{{x}^{2}}y+x{{y}^{2}}-1)=(5{{x}^{2}}y-x{{y}^{2}}+5x-3)$ <\/span> ","select":["A. $4{{x}^{2}}y+5x-4$ ","B. $4{{x}^{2}}y-2xy^2+5x-4$","C. $6{{x}^{2}}y-2xy^2+5x-2$","D. $6{{x}^{2}}y+5x-4$ "],"hint":"\u0110a th\u1ee9c b\u1ecb tr\u1eeb b\u1eb1ng \u0111a th\u1ee9c hi\u1ec7u c\u1ed9ng v\u1edbi \u0111a th\u1ee9c tr\u1eeb.","explain":" <span class='basic_left'>Ta c\u00f3: <br\/> $\\begin{align} & M-(-{{x}^{2}}y+x{{y}^{2}}-1)=(5{{x}^{2}}y-x{{y}^{2}}+5x-3) \\\\ & \\Rightarrow M=\\left( 5{{x}^{2}}y-x{{y}^{2}}+5x-3 \\right)+\\left( -{{x}^{2}}y+x{{y}^{2}}-1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=5{{x}^{2}}y-x{{y}^{2}}+5x-3-{{x}^{2}}y+x{{y}^{2}}-1 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=4{{x}^{2}}y+5x-4 \\\\ \\end{align}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":2}],"id_ques":1315},{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["3"],["4"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'> T\u00ecm nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $A(x)=(x-3)(16-4x)$<br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $ \\left[ \\begin{align} & x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x =\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$ <\/span> ","hint":"Cho \u0111a th\u1ee9c b\u1eb1ng $0$ v\u00e0 t\u00ecm $x$","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{aligned} & A\\left( x \\right)=0 \\\\ & \\Rightarrow \\left( x-3 \\right)\\left( 16-4x \\right)=0 \\\\ & \\Rightarrow \\left[ \\begin{aligned} & x-3=0 \\\\ & 16-4x=0 \\\\ \\end{aligned} \\right. \\\\ & \\Rightarrow \\left[ \\begin{aligned} & x=3 \\\\ & x=4 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3;4$<\/span><\/span>"}],"id_ques":1316},{"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":5,"ques":" <span class='basic_left'> Cho c\u00e1c \u0111a th\u1ee9c: <br\/> $f(x) = x^2 \u2013 x + 5 - (4x^2 + x^3 - 4x + 3)$ <br\/> $g(x) = - (2x^2 - 4x + 1) - ( -3x^3 + 5x^2 - 2)$ <br\/> Bi\u1ebft r\u1eb1ng $g(x) - k(x) = f(x).$ <br\/> <b> C\u00e2u a.<\/b> T\u00ednh $k(x).$ <\/span> ","select":["A. $k(x)=4{{x}^{3}}-10{{x}^{2}}+7x-1$ ","B. $k(x)=4{{x}^{3}}-4{{x}^{2}}+x-1$","C. $k(x)=3{{x}^{3}}-10{{x}^{2}}+x-3$","D. $k(x)=3{{x}^{3}}-4{{x}^{2}}+x-1$ "],"hint":"\u0110a th\u1ee9c tr\u1eeb b\u1eb1ng \u0111a th\u1ee9c b\u1ecb tr\u1eeb tr\u1eeb \u0111i \u0111a th\u1ee9c hi\u1ec7u","explain":" <span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>- R\u00fat g\u1ecdn c\u00e1c \u0111a th\u1ee9c $f(x);g(x)$ <br\/> - T\u00ecm $k(x)$ t\u1eeb $g(x) - k(x) = f(x).$ <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3: <br\/> $\\begin{align} & f(x)={{x}^{2}}-x+5-(4{{x}^{2}}+{{x}^{3}}-4x+3) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,={{x}^{2}}-x+5-4{{x}^{2}}-{{x}^{3}}+4x-3 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=-{{x}^{3}}-3{{x}^{2}}+3x+2 \\\\ & g(x)=-(2{{x}^{2}}-4x+1)-(-3{{x}^{3}}+5{{x}^{2}}-2) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=-2{{x}^{2}}+4x-1+3{{x}^{3}}-5{{x}^{2}}+2 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=3{{x}^{3}}-7{{x}^{2}}+4x+1 \\\\ & g\\left( x \\right)-k\\left( x \\right)=f\\left( x \\right) \\\\ & \\Rightarrow k\\left( x \\right)=g\\left( x \\right)-f\\left( x \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( 3{{x}^{3}}-7{{x}^{2}}+4x+1 \\right)-\\left( -{{x}^{3}}-3{{x}^{2}}+3x+2 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=3{{x}^{3}}-7{{x}^{2}}+4x+1+{{x}^{3}}+3{{x}^{2}}-3x-2 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=4{{x}^{3}}-4{{x}^{2}}+x-1 \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\ \\end{align}$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":2}],"id_ques":1317},{"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":[[3]],"list":[{"point":5,"ques":" <span class='basic_left'> Cho c\u00e1c \u0111a th\u1ee9c: <br\/> $f(x) = x^2 \u2013 x + 5 - (4x^2 + x^3 - 4x + 3)$ <br\/> $g(x) = - (2x^2 - 4x + 1) - ( -3x^3 + 5x^2 - 2)$ <br\/> Bi\u1ebft r\u1eb1ng $g(x) - k(x) = f(x).$ <br\/> <b> C\u00e2u b.<\/b> T\u00ecm nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $k(x).$ <\/span> ","select":["A. $x=0$ ","B. $x=1;x=\\pm \\dfrac{1}{2}$","C. $x=1$","D. $x=\\pm 1$ "],"hint":"","explain":" <span class='basic_left'> Theo c\u00e2u a, ta c\u00f3: $k(x)=4{{x}^{3}}-4{{x}^{2}}+x-1$ <br\/> Cho $k(x)=0$ ta \u0111\u01b0\u1ee3c: <br\/> $4x^3-4x^2+x-1=0$ <br\/> $\\Rightarrow 4x^2(x-1)+(x-1)=0$ <br\/> $\\Rightarrow (x-1)(4x^2+1)=0$ <br\/> $\\Rightarrow x-1=0 \\,\\,(do\\, 4x^2+1 \\ge 1 > 0\\,\\,\\forall x)$ <br\/> $\\Rightarrow x=1$ <br\/> Nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c $k(x)$ l\u00e0 $x=1$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span> ","column":2}],"id_ques":1318},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["f","t","f","t"]],"list":[{"point":5,"image":"","col_name":["Kh\u1eb3ng \u0111\u1ecbnh","\u0110\u00fang","Sai"],"arr_ques":["\u0110a th\u1ee9c $6{{x}^{2}}{{y}^{3}}-2{{z}^{2}}+7$ c\u00f3 b\u1eadc l\u00e0 $7$ "," Trong tam gi\u00e1c t\u00f9, c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c t\u00f9 l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t","Tr\u1ecdng t\u00e2m tam gi\u00e1c l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c \u0111\u00f3","Kh\u00f4ng t\u1ed3n t\u1ea1i tam gi\u00e1c m\u00e0 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh l\u1ea7n l\u01b0\u1ee3t l\u00e0 $3 cm, 4 cm, 7 cm.$"],"hint":"","explain":[" <span class='basic_left'>Sai, v\u00ec \u0111a th\u1ee9c $6{{x}^{2}}{{y}^{3}}-2{{z}^{2}}+7$ c\u00f3 b\u1eadc l\u00e0 $5$<\/span>","<span class='basic_left'>\u0110\u00fang, v\u00ec: Trong m\u1ed9t tam gi\u00e1c th\u00ec ch\u1ec9 c\u00f3 nhi\u1ec1u nh\u1ea5t l\u00e0 m\u1ed9t g\u00f3c t\u00f9 (g\u00f3c l\u1edbn nh\u1ea5t c\u00f3 trong m\u1ed9t tam gi\u00e1c). M\u00e0 c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c l\u1edbn h\u01a1n th\u00ec l\u1edbn h\u01a1n. Do \u0111\u00f3, c\u1ea1nh \u0111\u1ed1i di\u1ec7n v\u1edbi g\u00f3c t\u00f9 l\u00e0 c\u1ea1nh l\u1edbn nh\u1ea5t.<\/span>","<span class='basic_left'>Sai, v\u00ec: Tr\u1ecdng t\u00e2m tam gi\u00e1c l\u00e0 giao c\u1ee7a ba \u0111\u01b0\u1eddng trung tuy\u1ebfn. T\u00e2m c\u1ee7a \u0111\u01b0\u1eddng trong ngo\u1ea1i ti\u1ebfp tam gi\u00e1c l\u00e0 giao c\u1ee7a ba \u0111\u01b0\u1eddng trung tr\u1ef1c. <\/span>","<span class='basic_left'>\u0110\u00fang v\u00ec: Ta th\u1ea5y $3+4=7$ kh\u00f4ng th\u1ecfa m\u00e3n b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c.<\/span>"]}],"id_ques":1319},{"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'> Tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c l\u00e0 giao c\u1ee7a: <\/span> ","select":["A. Ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c ","B. Ba \u0111\u01b0\u1eddng trung tuy\u1ebfn c\u1ee7a tam gi\u00e1c","C. Ba \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a tam gi\u00e1c","D. Ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c "],"hint":"","explain":" <span class='basic_left'> Tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c l\u00e0 giao c\u1ee7a ba \u0111\u01b0\u1eddng cao c\u1ee7a tam gi\u00e1c. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span> ","column":1}],"id_ques":1320},{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-9"],["3"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'> T\u00ecm c\u00e1c h\u1ec7 s\u1ed1 $b,c$ bi\u1ebft $g(x) = 5x^2 + bx + c$ v\u00e0 $g(2) = 5; g(1) = -1.$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $ \\left\\{ \\begin{align} & b=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & c =\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$ <\/span> ","hint":"","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & g\\left( 2 \\right)=5 \\\\ & \\Rightarrow {{5.2}^{2}}+b.2+c=5 \\\\ & \\Rightarrow 2b+c=5-20 \\\\ & \\Rightarrow 2b+c=-15\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & g\\left( 1 \\right)=-1 \\\\ & \\Rightarrow {{5.1}^{2}}+b.1+c=-1 \\\\ & \\Rightarrow b+c=-1-5 \\\\ & \\Rightarrow b+c=-6\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align}$ <br\/> T\u1eeb (1) ta c\u00f3: <br\/> $2b+c=-15 \\Rightarrow b+(b+c)=-15 \\Rightarrow b-6=-15 \\Rightarrow b=-9$ <br\/> Thay $b=-9$ v\u00e0o (2) ta \u0111\u01b0\u1ee3c: $-9+c=-6 \\Rightarrow c=3$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-9;3$<\/span><\/span>"}],"id_ques":1321},{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" <span class='basic_left'> \u0110a th\u1ee9c $x^4+3x^2+1$ c\u00f3 b\u1ed1n nghi\u1ec7m ph\u00e2n bi\u1ec7t. <\/span> ","select":["\u0110\u00fang","Sai "],"hint":"","explain":" <span class='basic_left'> Ta c\u00f3: $x^4 \\ge 0;x^2\\ge 0$ v\u1edbi m\u1ecdi $x$ <br\/> $\\Rightarrow 3x^2 \\ge 0$ v\u1edbi m\u1ecdi $x$ <br\/> $\\Rightarrow x^4+3x^2+1 \\ge 1 > 0$ v\u1edbi m\u1ecdi $x$ <br\/> Do \u0111\u00f3, \u0111a th\u1ee9c $x^4+3x^2+1$ kh\u00f4ng c\u00f3 nghi\u1ec7m v\u1edbi m\u1ecdi $x$ <br\/> $\\Rightarrow$ Kh\u1eb3ng \u0111\u1ecbnh \u0111\u1ec1 b\u00e0i \u0111\u00e3 cho l\u00e0 Sai. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span><\/span> ","column":2}],"id_ques":1322},{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'> Bi\u1ebft $x+y=0,$ t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a \u0111a th\u1ee9c sau: <br\/> $C = 2x + 2y + 3xy( x + y ) + 5( x^3y^2 + x^2y^3 ) + 2$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $ C=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ <\/span> ","hint":"Ph\u00e2n t\u00edch \u0111a th\u1ee9c $C$ l\u00e0m xu\u1ea5t hi\u1ec7n $x+y$","explain":"<span class='basic_left'> Ta c\u00f3: <br\/> $\\begin{align} & C=2x+2y+3xy(x+y)+5({{x}^{3}}{{y}^{2}}+{{x}^{2}}{{y}^{3}})+2 \\\\ & \\,\\,\\,\\,\\,=2\\left( x+y \\right)+3xy\\left( x+y \\right)+5{{x}^{2}}{{y}^{2}}\\left( x+y \\right)+2 \\\\ & \\,\\,\\,\\,\\,=2.0+3xy.0+5{{x}^{2}}{{y}^{2}}.0+2\\,\\,\\,\\left( do\\,\\,x+y=0 \\right) \\\\ & \\,\\,\\,\\,\\,=2 \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$<\/span><\/span>"}],"id_ques":1323},{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2018"],["0"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'> T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $A=2025-|x^2+7|$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $ Max\\,A=\\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> ","hint":"S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a, ph\u00e1 d\u1ea5u gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/> - Ph\u00e1 d\u1ea5u gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i <br\/> - \u0110\u00e1nh gi\u00e1 $A \\le a$ ($a$ l\u00e0 h\u1eb1ng s\u1ed1) th\u00ec khi \u0111\u00f3 $Max\\, A=a$ <br\/> - T\u00ecm gi\u00e1 tr\u1ecb $x$ th\u1ecfa m\u00e3n $A=a$<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> V\u00ec $x^2+7 > 0$ n\u00ean $|x^2+7|=x^2+7$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & A=2025-\\left| {{x}^{2}}+7 \\right| \\\\ & \\,\\,\\,\\,\\,=2025-{{x}^{2}}-7 \\\\ & \\,\\,\\,\\,\\,=2018-{{x}^{2}} \\\\ & Do\\,\\,{{x}^{2}}\\,\\ge \\,\\,0\\,\\,\\,\\,\\forall x \\\\ & \\Rightarrow A=2018-{{x}^{2}}\\,\\,\\le 2018\\,\\,\\,\\forall x \\\\ & \\Rightarrow Max\\,A=2018\\,\\,khi\\,\\,x=0 \\\\ & \\,\\,\\,\\, \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2018;0$<\/span><\/span>"}],"id_ques":1324},{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" <span class='basic_left'> Cho \u0111a th\u1ee9c $P(x)=ax^3+bx^2+cx+d$ v\u1edbi $a,b,c,d$ l\u00e0 c\u00e1c s\u1ed1 nguy\u00ean. Bi\u1ebft r\u1eb1ng $P(x)\\, \\vdots\\, 5$ v\u1edbi m\u1ecdi $x$ l\u00e0 s\u1ed1 nguy\u00ean. Khi \u0111\u00f3 c\u00e1c s\u1ed1 nguy\u00ean $a,b,c$ c\u0169ng chia h\u1ebft cho $5.$ <\/span> ","select":["\u0110\u00fang","Sai "],"hint":"","explain":" <span class='basic_left'> Theo b\u00e0i $P(x) \\, \\vdots\\, 5$ v\u1edbi m\u1ecdi $x\\in \\mathbb{Z}$ <br\/> $\\Rightarrow P(0)=d \\, \\vdots\\, 5$ <br\/> Ta c\u00f3: $P(1)=(a+b+c+d) \\, \\vdots\\, 5$ (1) <br\/> $P(-1)=(-a+b-c+d) \\, \\vdots\\, 5$ (2) <br\/> T\u1eeb (1) v\u00e0 (2) suy ra $(b+d) \\, \\vdots\\, 5$ <br\/> M\u00e0 $d \\, \\vdots\\, 5 \\Rightarrow b\\, \\vdots\\, 5$ <br\/> $P(2)=(8a+4b+2c+d)\\, \\vdots\\, 5$ m\u00e0 $b \\, \\vdots\\, 5; d\\, \\vdots\\, 5$ n\u00ean $(8a+2c) \\, \\vdots\\, 5$ (3) <br\/> M\u1eb7t kh\u00e1c, t\u1eeb (1) v\u00e0 (2) ta l\u1ea1i c\u00f3: <br\/> $(a+b+c+d)-(-a+b-c+d)=2(a+c)\\,\\vdots \\,5$ <br\/> $\\Rightarrow (a+c)\\,\\vdots\\,5$ (4) <br\/> T\u1eeb (3) v\u00e0 (4) suy ra $a \\, \\vdots\\, 5; c\\, \\vdots\\, 5$ <br\/> $\\Rightarrow$ Kh\u1eb3ng \u0111\u1ecbnh \u0111\u1ec1 b\u00e0i \u0111\u00e3 cho l\u00e0 \u0110\u00fang. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span><\/span> ","column":2}],"id_ques":1325},{"time":3,"title":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A,$ \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c g\u00f3c $B$ c\u1eaft $AC$ t\u1ea1i $E.$ V\u1ebd $EH$ vu\u00f4ng g\u00f3c v\u1edbi $BC\\, (H \\in BC).$ Ch\u1ee9ng minh r\u1eb1ng: $BE$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $AH.$","title_trans":"S\u1eafp x\u1ebfp l\u1ea1i c\u00e1c b\u01b0\u1edbc \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh ho\u00e0n ch\u1ec9nh","temp":"sequence","correct":[[[2],[4],[3],[1],[5]]],"list":[{"point":5,"image":"","left":["$\\Rightarrow \\Delta ABE = \\Delta HBE$ (C\u1ea1nh huy\u1ec1n \u2013 g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow BA = BH$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng)","V\u00e0 $EA = EH $ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow E$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AH$","$\\Rightarrow B$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AH$","X\u00e9t tam gi\u00e1c vu\u00f4ng $ABE$ v\u00e0 tam gi\u00e1c vu\u00f4ng $HBE$ c\u00f3: <br\/> + $\\widehat {B_1} = \\widehat{B_2}$ (gi\u1ea3 thi\u1ebft) ; $BE $ chung","$\\Rightarrow EB$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111\u1ecdan th\u1eb3ng $AH$"],"top":80,"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai31/lv3/img\/D7B31_K1_13.png' \/><\/center> X\u00e9t tam gi\u00e1c vu\u00f4ng $ABE$ v\u00e0 tam gi\u00e1c vu\u00f4ng $HBE$ c\u00f3: <br\/> + $\\widehat {B_1} = \\widehat{B_2}$ (gi\u1ea3 thi\u1ebft) <br\/> + $BE $ chung <br\/> $\\Rightarrow \\Delta ABE = \\Delta HBE$ (C\u1ea1nh huy\u1ec1n \u2013 g\u00f3c nh\u1ecdn) <br\/> $\\Rightarrow BA = BH$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng)<br\/> $\\Rightarrow B$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AH$ <br\/> V\u00e0 $EA = EH $ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow E$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AH$ <br\/> $\\Rightarrow EB$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a \u0111\u1ecdan th\u1eb3ng $AH$ <\/span>"}],"id_ques":1326},{"title":"\u0110i\u1ec1n c\u00e1c s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["16"],["3"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'> Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $AB = 6cm; BC = 10 cm.$ Tr\u00ean tia \u0111\u1ed1i c\u1ee7a tia $AB$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $A $ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a \u0111o\u1ea1n th\u1eb3ng $BD.$ G\u1ecdi $K$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a c\u1ea1nh $BC,$ \u0111\u01b0\u1eddng th\u1eb3ng $DK$ c\u1eaft c\u1ea1nh $AC$ t\u1ea1i $M.$ T\u00ednh $MC.$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $ MC=\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$ $cm$ <\/span> ","hint":"","explain":"<span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai31/lv3/img\/D7B31_K1_15.png' \/><\/center> X\u00e9t $\\Delta BCD$ c\u00f3: <br\/> + $CA$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn (v\u00ec $BA=AD$ theo gi\u1ea3 thi\u1ebft) <br\/> + $DK$ l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn (do $BK=KC$ theo gi\u1ea3 thi\u1ebft) <br\/> M\u00e0 $CA \\cap DK =M$ <br\/> $\\Rightarrow M$ l\u00e0 tr\u1ecdng t\u00e2m c\u1ee7a $\\Delta BCD$ <br\/> $\\Rightarrow CM=\\dfrac{2}{3}AC$ (1) <br\/> X\u00e9t $\\Delta ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3: <br\/> $BC^2=AB^2+AC^2$ (\u0111\u1ecbnh l\u00ed Pytago) <br\/> $\\Rightarrow 10^2=6^2+AC^2$ <br\/> $\\Rightarrow AC^2=64$ <br\/> $\\Rightarrow AC=8\\,(cm)$ <br\/> T\u1eeb (1) suy ra $CM=\\dfrac{2}{3}.8=\\dfrac{16}{3}\\,(cm)$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $16;3$<\/span><\/span>"}],"id_ques":1327},{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00e2y \u0110\u00fang hay Sai","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" <span class='basic_left'> Cho $\\Delta ABC$ c\u00f3 $\\widehat{B}=90^o,$ \u0111\u01b0\u1eddng cao $BH.$ Tr\u00ean tia $HA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $HD = HC.$ T\u1eeb $A$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi $BD,$ c\u1eaft $BD$ t\u1ea1i $K,$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $BH$ t\u1ea1i $I.$ <br\/> <b> C\u00e2u a. <\/b> Khi \u0111\u00f3 $BA \\bot DI.$ <\/span> ","select":["\u0110\u00fang","Sai "],"hint":"","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai31/lv3/img\/D7B31_K1_16.png' \/><\/center> Theo b\u00e0i $AK \\bot BD \\Rightarrow IK \\bot BD$ <br\/> $DH \\bot BI$ (do $AH \\bot BH$) <br\/> X\u00e9t $\\Delta BDI$ c\u00f3: <br\/> $\\left\\{ \\begin{align} & IK\\,\\,\\bot \\,BD \\\\ & DH\\,\\bot \\,BI \\\\ & IK\\,\\,\\cap \\,DH\\,=A \\\\ \\end{align} \\right.$ <br\/> $\\Rightarrow A$ l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a $\\Delta BDI$ <br\/> $\\Rightarrow BA$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng cao c\u1ee7a $\\Delta BDI$ <br\/> $\\Rightarrow BA\\,\\bot \\,DI$ <br\/> $\\Rightarrow$ Kh\u1eb3ng \u0111\u1ecbnh \u1edf \u0111\u1ec1 b\u00e0i l\u00e0 \u0110\u00fang. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span><\/span> ","column":2}],"id_ques":1328},{"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":5,"ques":" <span class='basic_left'> Cho $\\Delta ABC$ c\u00f3 $\\widehat{B}=90^o; AB < AC,$ \u0111\u01b0\u1eddng cao $BH.$ Tr\u00ean tia $HA$ l\u1ea5y \u0111i\u1ec3m $D$ sao cho $HD = HC.$ T\u1eeb $A$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng vu\u00f4ng g\u00f3c v\u1edbi $BD,$ c\u1eaft $BD$ t\u1ea1i $K,$ c\u1eaft \u0111\u01b0\u1eddng th\u1eb3ng $BH$ t\u1ea1i $I.$ <br\/> <b> C\u00e2u b. <\/b> \u0110\u1ec3 $\\Delta BDI$ \u0111\u1ec1u th\u00ec $\\Delta ABC$ c\u1ea7n c\u00f3 th\u00eam \u0111i\u1ec1u ki\u1ec7n l\u00e0: <\/span> ","select":["A. $\\widehat{A}=45^o$ ","B. $\\widehat{A}=60^o$","C. $\\widehat{A}=30^o$"],"hint":"","explain":" <span class='basic_left'> <center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai31/lv3/img\/D7B31_K1_16.png' \/><\/center> Theo c\u00e2u a, ta c\u00f3 $BA\\bot DI$ <br\/> M\u00e0 $BC \\bot AB$ (gi\u1ea3 thi\u1ebft) $\\Rightarrow DI \/\/BC$ <br\/> $\\Rightarrow \\widehat{IDB}+\\widehat{DBC}=180^o$ (hai g\u00f3c trong c\u00f9ng ph\u00eda) (1) <br\/> $\\Rightarrow \\widehat{IDC}=\\widehat {DCB}$ (hai g\u00f3c so le trong) (2) <br\/> M\u00e0 $\\widehat{DCB}=\\widehat{CDB}$ (v\u00ec ta d\u1ec5 d\u00e0ng ch\u1ec9 ra \u0111\u01b0\u1ee3c $\\Delta BDC$ c\u00e2n t\u1ea1i $B$) (3) <br\/> T\u1eeb (2) v\u00e0 (3) suy ra $\\widehat{IDC}=\\widehat{CDB}$ <br\/> $\\Rightarrow DH$ v\u1eeba l\u00e0 \u0111\u01b0\u1eddng cao v\u1eeba l\u00e0 ph\u00e2n gi\u00e1c trong $\\Delta BDI$ <br\/> $\\Rightarrow \\Delta BDI$ c\u00e2n t\u1ea1i $D$ <br\/> \u0110\u1ec3 $\\Delta BDI$ \u0111\u1ec1u th\u00ec $\\widehat {IDB}=60^o$ <br\/> $\\Rightarrow \\widehat{CDB}=\\dfrac{\\widehat{IDB}}{2}=30^o$ <br\/> Suy ra $\\widehat{BCD} =\\widehat{CDB} = 30^{o}$ (tam gi\u00e1c $BDC$ c\u00e2n t\u1ea1i $B$) <br\/> Tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $B$ c\u00f3 $\\widehat{C} = 30^{o}\\Rightarrow \\widehat{BAC}= 60^{o}$ <br\/> V\u1eady \u0111\u1ec3 $\\Delta BDI$ \u0111\u1ec1u th\u00ec $\\Delta ABC$ c\u1ea7n th\u00eam \u0111i\u1ec1u ki\u1ec7n l\u00e0 $\\widehat{BAC}=60^o$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span> ","column":3}],"id_ques":1329},{"time":3,"title":"Cho $\\Delta ABC$ vu\u00f4ng t\u1ea1i $A,$ \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $B$ c\u1eaft $AC$ t\u1ea1i $D.$ V\u1ebd $DH \\bot BC\\, (H \\in BC).$ Tr\u00ean tia \u0111\u1ed1i c\u1ee7a $AB$ l\u1ea5y \u0111i\u1ec3m $K$ sao cho $AK = HC.$ Ch\u1ee9ng minh ba \u0111i\u1ec3m $K, D, H$ th\u1eb3ng h\u00e0ng.","title_trans":"S\u1eafp x\u1ebfp l\u1ea1i c\u00e1c b\u01b0\u1edbc \u0111\u1ec3 \u0111\u01b0\u1ee3c b\u00e0i ch\u1ee9ng minh ho\u00e0n ch\u1ec9nh","temp":"sequence","correct":[[[4],[1],[5],[6],[2],[3]]],"list":[{"point":5,"image":"","left":["$\\Rightarrow D$ l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c $BKC$","X\u00e9t hai tam gi\u00e1c vu\u00f4ng $ABD$ v\u00e0 $HBD$ c\u00f3: $BD$ l\u00e0 c\u1ea1nh chung; $DA=DH$ ($D$ n\u1eb1m tr\u00ean tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $B$)","M\u1eb7t kh\u00e1c, ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\Delta CAK=\\Delta KHC$ (c \u2013 g \u2013 c) <br\/> $\\Rightarrow KH \\bot BC$","$\\Rightarrow KH$ l\u00e0 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb \u0111\u1ec9nh $K$ c\u1ee7a $\\Delta BKC$ n\u00ean $KH$ ph\u1ea3i \u0111i qua tr\u1ef1c t\u00e2m $H$ <br\/> V\u1eady ba \u0111i\u1ec3m $K,D,H$ th\u1eb3ng h\u00e0ng","$\\Rightarrow \\Delta ABD=\\Delta HBD$ (c\u1ea1nh huy\u1ec1n \u2013 c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow AB=BH$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng)","M\u00e0 $AK=HC$ (gi\u1ea3 thi\u1ebft) $\\Rightarrow \\Delta BKC$ c\u00e2n t\u1ea1i $B$ <br\/> Khi \u0111\u00f3, $BD$ v\u1eeba l\u00e0 ph\u00e2n gi\u00e1c, v\u1eeba l\u00e0 \u0111\u01b0\u1eddng cao xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh $B$ c\u1ee7a $\\Delta BKC$"],"top":80,"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop7/toan/daiso/bai31/lv3/img\/D7B31_K1_18.png' \/><\/center> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $ABD$ v\u00e0 $HBD$ c\u00f3: <br\/> + $BD$ l\u00e0 c\u1ea1nh chung<br\/> + $DA=DH$ ($D$ n\u1eb1m tr\u00ean tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $B$)<br\/>$\\Rightarrow \\Delta ABD=\\Delta HBD$ (c\u1ea1nh huy\u1ec1n \u2013 c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow AB=BH$ (hai c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) <br\/> M\u00e0 $AK=HC$ (gi\u1ea3 thi\u1ebft) $\\Rightarrow BK=BC$ <br\/> $\\Rightarrow \\Delta BKC$ c\u00e2n t\u1ea1i $B$ <br\/> Khi \u0111\u00f3, $BD$ v\u1eeba l\u00e0 ph\u00e2n gi\u00e1c, v\u1eeba l\u00e0 \u0111\u01b0\u1eddng cao xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh $B$ c\u1ee7a $\\Delta BKC$ <br\/> $\\Rightarrow D$ l\u00e0 tr\u1ef1c t\u00e2m c\u1ee7a tam gi\u00e1c $BKC$ <br\/> M\u1eb7t kh\u00e1c, ta ch\u1ee9ng minh \u0111\u01b0\u1ee3c: $\\Delta CAK=\\Delta KHC$ (c \u2013 g \u2013 c) <br\/> $\\Rightarrow \\widehat{KHC}=\\widehat{KAC}=90^o$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow KH \\bot BC$ <br\/> $\\Rightarrow KH$ l\u00e0 \u0111\u01b0\u1eddng cao k\u1ebb t\u1eeb \u0111\u1ec9nh $K$ c\u1ee7a $\\Delta BKC$ n\u00ean $KH$ ph\u1ea3i \u0111i qua tr\u1ef1c t\u00e2m $D$ <br\/> V\u1eady ba \u0111i\u1ec3m $K,D,H$ th\u1eb3ng h\u00e0ng <\/span>"}],"id_ques":1330}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":59}}