{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["2"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c:$\\left( 2-\\sqrt{3} \\right)\\left( \\sqrt{6}+\\sqrt{2} \\right)\\sqrt{2+\\sqrt{3}}=$_input_","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $\\sqrt{6}+\\sqrt{2}=\\sqrt {2} (\\sqrt {3}+1)$. <br\/>Nh\u00f3m $\\sqrt {2}$ v\u00e0 $\\sqrt{2}+\\sqrt{3}$ r\u1ed3i th\u1ef1c hi\u1ec7n ph\u00e9p nh\u00e2n v\u00e0 bi\u1ebfn \u0111\u1ed5i.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Ph\u00e2n t\u00edch bi\u1ec3u th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed, nh\u00f3m v\u00e0 bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 d\u1ea1ng $(A \\pm B)^2$ <br\/><b>B\u01b0\u1edbc 2:<\/b> \u00c1p d\u1ee5ng: V\u1edbi $A$ l\u00e0 m\u1ed9t bi\u1ec3u th\u1ee9c, ta c\u00f3 $\\sqrt{A^2}=|A|=\\left\\{ \\begin{align} & A\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,{A}\\ge {0} \\\\ & -A\\,\\,\\,\\,\\text{n\u1ebfu}\\,\\,\\,\\,\\,\\,\\,\\,{A}<{0} \\\\\\end{align} \\right.$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align*} &\\left( 2-\\sqrt{3} \\right)\\left( \\sqrt{6}+\\sqrt{2} \\right)\\sqrt{2+\\sqrt{3}} \\\\ &=\\left( 2-\\sqrt{3} \\right)\\left( \\sqrt{3}+1 \\right).\\sqrt{2}.\\sqrt{2+\\sqrt{3}} \\\\ &=\\left( 2-\\sqrt{3} \\right)\\left( \\sqrt{3}+1 \\right)\\sqrt{4+2\\sqrt{3}} \\\\ &=\\left( 2-\\sqrt{3} \\right)\\left( \\sqrt{3}+1 \\right)\\sqrt{{{\\left( \\sqrt{3}+1 \\right)}^{2}}} \\\\ &=\\left( 2-\\sqrt{3} \\right)\\left( \\sqrt{3}+1 \\right)\\left( \\sqrt{3}+1 \\right) \\\\ &=\\left( 2-\\sqrt{3} \\right)\\left( 4+2\\sqrt{3} \\right) \\\\ &=8+4\\sqrt{3}-4\\sqrt{3}-6\\\\ &=2 \\\\ \\end{align*}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $2$<\/span><\/span>"}]}],"id_ques":551},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{5}{3}$","B. $\\dfrac{1}{3}$","C. $\\dfrac{7}{3}$"],"ques":"T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c $A=\\sqrt{\\dfrac{\\sqrt{a}-1}{\\sqrt{b}+1}}:\\sqrt{\\dfrac{\\sqrt{b}-1}{\\sqrt{a}+1}}$ v\u1edbi $a=7,25$ v\u00e0 $b= 3,25$<br\/>\u0110\u00e1p \u00e1n: $A=$ ?","hint":"\u00c1p d\u1ee5ng quy t\u1eafc t\u00ednh th\u01b0\u01a1ng c\u1ee7a hai c\u0103n b\u1eadc hai.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> \u00c1p d\u1ee5ng quy t\u1eafc t\u00ednh th\u01b0\u01a1ng c\u1ee7a hai c\u0103n b\u1eadc hai.<br\/><b>B\u01b0\u1edbc 2:<\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $A$<br\/><b>B\u01b0\u1edbc 3:<\/b> Thay $a=7,25$ v\u00e0 $b=3,25$ v\u00e0o bi\u1ec3u th\u1ee9c $A$ \u0111\u00e3 r\u00fat g\u1ecdn. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n: $a,b \\ge 0$ v\u00e0 $b \\le 1$<br\/> Ta c\u00f3:<br\/>$\\begin{aligned} A&=\\sqrt{\\dfrac{\\sqrt{a}-1}{\\sqrt{b}+1}}:\\sqrt{\\dfrac{\\sqrt{b}-1}{\\sqrt{a}+1}} \\\\&=\\sqrt {\\dfrac{\\sqrt{a}-1}{\\sqrt{b}+1}:\\dfrac{\\sqrt{b}-1}{\\sqrt{a}+1}}\\\\ & =\\sqrt{\\dfrac{\\sqrt{a}-1}{\\sqrt{b}+1}.\\dfrac{\\sqrt{a}+1}{\\sqrt{b}-1}} \\\\ & =\\sqrt{\\dfrac{a-1}{b-1}} \\\\ \\end{aligned}$ <br\/>Thay $a= 7,25$ v\u00e0 $b= 3,25$ v\u00e0o bi\u1ec3u th\u1ee9c $A$ \u0111\u00e3 r\u00fat g\u1ecdn, ta \u0111\u01b0\u1ee3c:<br\/> $A=\\sqrt{\\dfrac{7,25-1}{3,25-1}}=\\sqrt{\\dfrac{6,25}{2,25}}=\\sqrt{\\dfrac{625}{225}}=\\dfrac{25}{15}=\\dfrac{5}{3}$<\/span><\/span>"}]}],"id_ques":552},{"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":"K\u1ebft qu\u1ea3 c\u1ee7a ph\u00e9p t\u00ednh $\\sqrt{3-\\sqrt{5}}+\\sqrt{3+\\sqrt{5}}$ l\u00e0: ","select":["A. $2\\sqrt{10}$","B. $2\\sqrt{5}$ ","C. $\\sqrt{5}$","D. $\\sqrt{10}$"],"hint":"\u0110\u1eb7t $A=\\sqrt{3-\\sqrt{5}}+\\sqrt{3+\\sqrt{5}}$. Nh\u00e2n c\u1ea3 hai v\u1ebf v\u1edbi $\\sqrt{2}$ \u0111\u1ec3 \u0111\u01b0a c\u00e1c bi\u1ec3u th\u1ee9c trong c\u0103n \u1edf v\u1ebf ph\u1ea3i v\u1ec1 d\u1ea1ng b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t t\u1ed5ng, m\u1ed9t hi\u1ec7u. Sau \u0111\u00f3 ti\u1ebfp t\u1ee5c khai c\u0103n.","explain":"<span class='basic_left'>\u0110\u1eb7t $A=\\sqrt{3-\\sqrt{5}}+\\sqrt{3+\\sqrt{5}}$ (*)<br\/>Nh\u00e2n hai v\u1ebf c\u1ee7a (*) v\u1edbi $\\sqrt{2},$ ta c\u00f3:<br\/> $\\begin{aligned} &A\\sqrt{2}=\\sqrt{2}\\left( \\sqrt{3-\\sqrt{5}}+\\sqrt{3+\\sqrt{5}} \\right) \\\\ & =\\sqrt{6-2\\sqrt{5}}+\\sqrt{6+2\\sqrt{5}} \\\\& = \\sqrt{5 -2\\sqrt{5} + 1} + \\sqrt{5 + 2\\sqrt{5} + 1} \\\\ & =\\sqrt{{{\\left( \\sqrt{5}-1 \\right)}^{2}}}+\\sqrt{{{\\left( \\sqrt{5}+1 \\right)}^{2}}} \\\\ & =\\sqrt{5}-1+\\sqrt{5}+1 \\,\\, (V\u00ec \\,\\sqrt{5} > 1)\\\\ & =2\\sqrt{5} \\end{aligned}$ <br\/>Do \u0111\u00f3 $A=A:\\sqrt{2}=2\\sqrt{5}:\\sqrt{2}=\\sqrt{10}$<br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":553},{"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":" Ph\u01b0\u01a1ng tr\u00ecnh $\\sqrt{2x-1}=\\sqrt{x-1}$ c\u00f3 nghi\u1ec7m l\u00e0: ","select":["A. $x=\\dfrac{1}{2}$","B. $x=-\\dfrac{1}{2}$ ","C. $x= 0$","D. $x\\in \\varnothing$"],"hint":"B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<br\/><b>B\u01b0\u1edbc 2:<\/b> B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $\\left\\{ \\begin{aligned} & 2x-1\\ge 0 \\\\ & x-1\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ge \\dfrac{1}{2} \\\\ & x\\ge 1 \\\\ \\end{aligned} \\right.\\Leftrightarrow x\\ge 1$ <br\/>$\\begin{align} & \\,\\,\\,\\sqrt{2x-1}=\\sqrt{x-1} \\\\ & \\Leftrightarrow 2x-1=x-1 \\\\ & \\Leftrightarrow x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=0 \\,\\,\\,\\text(lo\u1ea1i)\\\\ \\end{align}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m. <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><br\/><i>L\u01b0u \u00fd:<\/i> Sau khi t\u00ecm \u0111\u01b0\u1ee3c gi\u00e1 tr\u1ecb c\u1ee7a $x$, c\u1ea7n x\u00e9t xem gi\u00e1 tr\u1ecb $x$ c\u00f3 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh hay kh\u00f4ng, r\u1ed3i k\u1ebft lu\u1eadn nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<\/span><\/span>","column":2}]}],"id_ques":554},{"time":24,"part":[{"title":" \u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng:","title_trans":"","temp":"fill_the_blank","correct":[[["290"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $2\\left( \\sqrt{\\dfrac{x-1}{4}}-3 \\right)=\\sqrt{\\dfrac{4x-4}{9}}-\\dfrac{1}{3}$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: $S=${_input_} ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<br\/><b>B\u01b0\u1edbc 2:<\/b> Bi\u1ebfn \u0111\u1ed5i t\u01b0\u01a1ng \u0111\u01b0\u01a1ng hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 \u0111\u01b0a v\u1ec1 d\u1ea1ng: $\\sqrt {f(x)}=a$ v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x-1\\ge 0\\Leftrightarrow x\\ge 1$ <br\/> Ta c\u00f3:<br\/> $\\begin{align} & 2\\left( \\sqrt{\\dfrac{x-1}{4}}-3 \\right)=\\sqrt{\\dfrac{4x-4}{9}}-\\dfrac{1}{3} \\\\ & \\Leftrightarrow \\sqrt{x-1}-6=\\dfrac{2}{3}\\sqrt{x-1}-\\dfrac{1}{3} \\\\ & \\Leftrightarrow \\dfrac{1}{3}\\sqrt{x-1}=\\dfrac{17}{3} \\\\ & \\Leftrightarrow \\sqrt{x-1}=17 \\\\ & \\Leftrightarrow x-1=289 \\\\ & \\Leftrightarrow x=290 \\ (\\text{th\u1ecfa m\u00e3n}) \\end{align}$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{290\\}$ <br\/> <span class='basic_pink'> V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $290$ <\/span> <\/span><\/span> "}]}],"id_ques":555},{"time":24,"part":[{"title":" \u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":10,"width":50,"content":"","type_input":"","ques":"T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $C=x+\\sqrt{2-{{x}^{2}}}$ <br\/><b> \u0110\u00e1p \u00e1n:<\/b> $C_{\\max}=$_input_ ","hint":"B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf, s\u1eed d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cauchy: V\u1edbi $a, b \\ge 0$, ta c\u00f3: $ a + b \\ge 2 \\sqrt{ab}$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 c\u0103n th\u1ee9c c\u00f3 ngh\u0129a<br\/><b>B\u01b0\u1edbc 2:<\/b> B\u00ecnh ph\u01b0\u01a1ng hai v\u1ebf<br\/><b>B\u01b0\u1edbc 3:<\/b> S\u1eed d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cauchy, x\u00e1c \u0111\u1ecbnh gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c. <br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/><span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $2-{{x}^{2}}\\ge 0\\Leftrightarrow {{x}^{2}}\\le 2\\,$$\\,\\Leftrightarrow -\\sqrt{2}\\le x\\le \\sqrt{2}$ <br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & {{C}^{2}}={{\\left( x+\\sqrt{2-{{x}^{2}}} \\right)}^{2}} \\\\ & ={{x}^{2}}+2\\sqrt{{{x}^{2}}\\left( 2-{{x}^{2}} \\right)}+2-{{x}^{2}} \\\\ & =2+2\\sqrt{{{x}^{2}}\\left( 2-{{x}^{2}} \\right)} \\\\ \\end{aligned}$<br\/>\u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cauchy cho hai s\u1ed1 kh\u00f4ng \u00e2m: $x^2$ v\u00e0 $2-x^2$ ta c\u00f3: <br\/>$2\\sqrt{{{x}^{2}}\\left( 2-{{x}^{2}} \\right)}$$\\le {{x}^{2}}+2-{{x}^{2}}=2$ <br\/>Suy ra ${{C}^{2}}\\le 2+2=4\\Rightarrow C\\le 2$<br\/>$\\begin{aligned} \\Rightarrow&{{C}_{\\max }}=2\\Leftrightarrow x =\\sqrt{2-{{x}^{2}}} \\\\ \\Leftrightarrow\\,\\, &{{x}^{2}}=2-{{x}^{2}} \\\\ \\Leftrightarrow\\,\\, &{{x}^{2}}=1 \\\\ \\Leftrightarrow \\,\\,&x=\\pm 1 \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'> V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 ${{C}_{\\max }}=2$ khi $x=\\pm 1$ <\/span><br\/><i>L\u01b0u \u00fd: <\/i>B\u1ea5t \u0111\u1eb3ng th\u1ee9c Cauchy: V\u1edbi c\u00e1c s\u1ed1 $a, b$ kh\u00f4ng \u00e2m, ta lu\u00f4n c\u00f3: $\\dfrac{a+b}{2} \\ge \\sqrt{ab}$<br\/>D\u1ea5u b\u1eb1ng x\u1ea3y ra khi $a=b$<\/span><\/span> "}]}],"id_ques":556},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai3/lv3/img\/2.jpg' \/><\/center>V\u1edbi $a\\ge 0,\\,\\,b\\ge 0$, ta c\u00f3: $\\dfrac{a+b}{2}\\le \\sqrt{ab}$ ","select":["A. \u0110\u00fang","B. Sai "],"hint":"X\u00e9t hi\u1ec7u $\\dfrac{a+b}{2}-\\sqrt{ab}$","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$\\dfrac{a+b}{2}-\\sqrt{ab}\\,\\,$$=\\dfrac{a+b-2\\sqrt{ab}}{2}\\,\\,$$=\\dfrac{{{\\left( \\sqrt{a}-\\sqrt{b} \\right)}^{2}}}{2}\\ge 0$$\\,\\,\\,\\forall a\\ge 0,\\,\\,b\\ge 0$ <br\/>Suy ra $\\dfrac{a+b}{2}\\ge \\sqrt{ab}$<br\/>Do \u0111\u00f3 kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai<br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":557},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai3/lv3/img\/4.jpg' \/><\/center>K\u1ebft qu\u1ea3 ph\u00e9p t\u00ednh $\\left( 3\\sqrt{2}+3 \\right)\\left( \\sqrt{2}-1 \\right)$ l\u00e0 m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean. ","select":["A. \u0110\u00fang","B. Sai "],"hint":"Th\u1ef1c hi\u1ec7n ph\u00e9p nh\u00e2n hai bi\u1ec3u th\u1ee9c.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} &\\left( 3\\sqrt{2}+3 \\right)\\left( \\sqrt{2}-1 \\right) \\\\ & =3\\left( \\sqrt{2}+1 \\right)\\left( \\sqrt{2}-1 \\right) \\\\ & =3\\left( 2-1 \\right)\\\\ &=3 \\\\ \\end{align}$ <br\/>V\u1eady k\u1ebft qu\u1ea3 ph\u00e9p t\u00ednh l\u00e0 m\u1ed9t s\u1ed1 t\u1ef1 nhi\u00ean.<br\/>Do \u0111\u00f3 kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang .<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A <\/span>","column":2}]}],"id_ques":558},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":" $\\sqrt{\\sqrt{2}-1}.\\sqrt{2-\\sqrt{3-\\sqrt{2}}}\\,$$\\,.\\sqrt{2+\\sqrt{3-\\sqrt{2}}}=$ _input_ ","hint":" Nh\u00f3m hai th\u1eeba s\u1ed1 th\u1ee9 2 v\u00e0 3: \u00c1p d\u1ee5ng quy t\u1eafc nh\u00e2n hai c\u0103n th\u1ee9c v\u00e0 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2-b^2$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$ \\sqrt{\\sqrt{2}-1}.\\sqrt{2-\\sqrt{3-\\sqrt{2}}}\\,$$\\,.\\sqrt{2+\\sqrt{3-\\sqrt{2}}}$<br\/>$ =\\sqrt{\\sqrt{2}-1}.\\,$$\\sqrt{\\left( 2-\\sqrt{3-\\sqrt{2}} \\right)\\left( 2+\\sqrt{3-\\sqrt{2}} \\right)} $<br\/>$ =\\sqrt{\\sqrt{2}-1}.\\sqrt{4-3+\\sqrt{2}} $<br\/>$ =\\sqrt{\\sqrt{2}-1}.\\sqrt{\\sqrt{2}+1} $<br\/>$ =\\sqrt{2-1}$<br\/>$=1 $<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n l\u00e0 $1$<\/span>"}]}],"id_ques":559},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u $> , < , =$ th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["<"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai3/lv3/img\/5.jpg' \/><\/center><br\/>$\\sqrt{2008}+\\sqrt{2010}$ _input_$2\\sqrt{2009}$ ","hint":"X\u00e9t hi\u1ec7u ${{\\left( 2\\sqrt{2009} \\right)}^{2}}-{{\\left( \\sqrt{2008}+\\sqrt{2010} \\right)}^{2}}$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> X\u00e9t hi\u1ec7u ${{\\left( 2\\sqrt{2009} \\right)}^{2}}-{{\\left( \\sqrt{2008}+\\sqrt{2010} \\right)}^{2}}$ <br\/><b>B\u01b0\u1edbc 2:<\/b> Bi\u1ebfn \u0111\u1ed5i v\u1ec1 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(A -B)^2$. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> X\u00e9t hi\u1ec7u ${{\\left( 2\\sqrt{2009} \\right)}^{2}}-{{\\left( \\sqrt{2008}+\\sqrt{2010} \\right)}^{2}}$<br\/>$=4.2009-(2008+2\\sqrt{2008}.\\sqrt{2010}+2010) $<br\/>$ =4.2009-2008-2\\sqrt{2008}.\\sqrt{2010}\\,\\,$$-2010 $ <br\/> $= (2.2009 - 2008) + (2.2009 - 2010) - 2\\sqrt{2008}.\\sqrt{2010}$<br\/>$ =2010+2008-2\\sqrt{2008}.\\sqrt{2010} $<br\/>$ ={{\\left( \\sqrt{2010}-\\sqrt{2008} \\right)}^{2}}>0$<br\/> Suy ra $2\\sqrt{2009}>\\sqrt{2008}+\\sqrt{2010}$<br\/><span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n l\u00e0 $<$<\/span><\/span>"}]}],"id_ques":560}],"lesson":{"save":0,"level":3}}