J = $\left(1+\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)$ . $\left(1-\frac{2-\sqrt{2}}{1-\sqrt{2}}\right)$

M = $\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\frac{1}{\sqrt{6}}$

N = $\frac{6}{1+\sqrt{7}}+\frac{1}{\sqrt{7}}$

O = $\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{2-\sqrt{3}}$

Q = $\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)$

S = $\frac{2}{\sqrt{5}+1}-\sqrt{\frac{2}{3-\sqrt{5}}}$

Các thầy cô, các bạn giúp em với ạ. Em cảm ơn !

Chứng minh các đẳng thức sau:

a)$326+223−432=66" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:20.32px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; text-align:left; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">$32√6+2√23−4√32=√66$\$\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{6}$

b) x6x+2x3+6x):6x=213" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:20.32px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; text-align:left; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">(x√6x+√2x3+√6x):√6x=213$\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2\mathrm{x}}{3}}+\sqrt{6\mathrm{x}}\right):\sqrt{6\mathrm{x}}=2\frac{1}{3}$ với x > 0.

Bài 1: Rút gọn biểu thức:
$a)\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}$
$b)\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}$
$c)\sqrt{9-2\sqrt{14}}-\sqrt{9-2\sqrt{14}}$
$d)\sqrt{2(4-\sqrt{7})}$
$e)\sqrt{a+1+2\sqrt{a}}$
$f)-\sqrt{2(2-\sqrt{3})}+\sqrt{2(2+\sqrt{3})}$
$g)\sqrt{x+y-2\sqrt{xy}}$ $(x\ge y)$
$h)\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$
$i)\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}$
$j)\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5-2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}-\sqrt{7+2\sqrt{10}}}}$
$k)\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}$
$l)\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}$
$m\left)\right(4+\sqrt{15}\left)\right(\sqrt{10}-\sqrt{6}\left)\right(4-\sqrt{15})$
$n\left)\sqrt{3-\sqrt{5}}\right(\sqrt{10}-\sqrt{2}\left)\right(3+\sqrt{5})$
$o)\sqrt{31+2}.\sqrt{6+\sqrt{5+\sqrt{2}}}.+\sqrt{3+\sqrt{3+\sqrt{5+\sqrt{2}}}}.\sqrt{3-\sqrt{3+\sqrt{5+\sqrt{2}}}}$

Bài 2: Phân tích đa thức thành nhân tử:
$a)x-\sqrt{x}$
$b)x-1$
$c)x-2\sqrt{2}+1$
$d)-x+3\sqrt{x}+4$
$e)2x-\sqrt{x}-1$
$f)x-\sqrt{2}-2$
$g)x\sqrt{x}+1$
$h)2x+\sqrt{x}-3$
$i)x\sqrt{x}+9x+14\sqrt{x}$
$j)2x\sqrt{x}+5x+3\sqrt{x}$

Bài 3: Cho biểu thức
$E=(\frac{3x-2}{x^2-9}-\frac{5x+1}{x^2-3x}-\frac{x+1}{x^2+3x}):\frac{x+2}{x^2+3x}-\frac{4^2-2x}{x^2-x-6}$
a) Tìm ĐKXĐ
b) Rút gọn.
c) Tìm giá trị nguyên cuả x để E nhận giá trị nguyên.

Giúp mk vs (ko hỉu j thì thôi nha )

Giải hệ phương trình:\(\hept{ }\)
{x+10+y−1=11y+10+x−1=11" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">√x+10+√y−1=11√y+10+√x−1=11

J = $\left(1+\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)$ . $\left(1-\frac{2-\sqrt{2}}{1-\sqrt{2}}\right)$

M = $\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\frac{1}{\sqrt{6}}$

N = $\frac{6}{1+\sqrt{7}}+\frac{1}{\sqrt{7}}$

O = $\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{2-\sqrt{3}}$

Q = $\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)$

S = $\frac{2}{\sqrt{5}+1}-\sqrt{\frac{2}{3-\sqrt{5}}}$

Các thầy cô, các bạn giúp em với ạ. Em cảm ơn !

Có bn nào bt lm thì giúp mk nha!!! Thanks

1, A= $\frac{\sqrt{x}+4}{\sqrt{x}-1}$         B= $\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\left(x\ge 0,x\ne 1\right)$

Tìm x để $\frac{A}{B}\ge \frac{x}{4}+5$biết B= $\frac{1}{\sqrt{x}-1}$

2, A= $\frac{4\left(\sqrt{x}+1\right)}{25-x}$ B= $\left(\frac{15-5x}{x-25}+\frac{2}{\sqrt{x}+5}\right):\frac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge 0,x\ne 25\right)$

Tìm giá trị nguyên của x để P= A.B đặt giá trị nguyên lớn nhất

GIÚP MK VỚI! THANKS

x−2+10−x=x2−12x+40" role="presentation" style="background-color:rgb(247, 247, 247); border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:14.04px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap" class="MathJax_CHTML mjx-chtml">√x−2+√10−x=x2−12x+40${\displaystyle \sqrt{x-2}+\sqrt{10-x}=x^2-12x+40}$

giúp  mình với!!

Cho P=xxy+x+3+yyz+y+1+3zxz+3z+3" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-table; float:none; font-size:20.34px; font-style:normal; font-weight:normal; letter-spacing:normal; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">√x√xy+√x+3+√y√yz+√y+1+3√z√xz+3√z+3${\displaystyle \frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+3}}+{\displaystyle \frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}}+{\displaystyle \frac{3\sqrt{z}}{\sqrt{xz}+3\sqrt{z}+3}}$ và xyz =9. Tính $\sqrt{10P-1}$

. Cho P=xxy+x+3+yyz+y+1+3zxz+3z+3" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-table; float:none; font-size:20.34px; font-style:normal; font-weight:normal; letter-spacing:normal; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">√x√xy+√x+3+√y√yz+√y+1+3√z√xz+3√z+3${\displaystyle \frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+3}}+{\displaystyle \frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}}+{\displaystyle \frac{3\sqrt{z}}{\sqrt{xz}+3\sqrt{z}+3}}$ và xyz =9. Tính $\sqrt{10P-1}$