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

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 !

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 )

Cho tam giác ABC có AB=c, BC=a, AC=b và $P=\frac{a+b+c}{2}$

CMR: a) (P-a)(P-b)(P-c) $\le$$\frac{1}{8}abc$

b) $\frac{1}{P-a}+\frac{1}{P-b}+\frac{1}{P-c}\ge 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

c) $\frac{1}{{\left(P-a\right)}^2}+\frac{1}{{\left(P-b\right)}^2}+\frac{1}{{\left(P-c\right)}^2}\ge \frac{P}{\left(P-a\right)\left(P-b\right)\left(P-c\right)}$

Cho tam giác AB có AB=c, BC=a, CA=b

CMR: a) (b+c-a) (c+a-b) (a+b-c) $\le$abc (1)

b) $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ge 3\left(2\right)$

c) $\frac{a^2}{b+c-1}+\frac{b^2}{c+a-b}+\frac{c^2}{a+b-c}\ge a+b+c\left(3\right)$

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.

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 !

x−2x−3+x+1x+3+x−4x−99−x" 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; padding:1px 0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">P=√x−2√x−3+√x+1√x+3+x−4√x−99−x$P={\displaystyle \frac{\sqrt{x}-2}{\sqrt{x}-3}}+{\displaystyle \frac{\sqrt{x}+1}{\sqrt{x}+3}}+{\displaystyle \frac{x-4\sqrt{x}-9}{9-x}}$và x+53−x" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; 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; padding:1px 0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">Q=√x+53−√x$Q={\displaystyle \frac{\sqrt{x}+5}{3-\sqrt{x}}}$(với$x\ge 0;x\ne 9$)

1, Rút gọn P

2, Tìm x để P=3

3, Tìm $M=P:Q$. Tìm x để 12" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; 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; padding:1px 0px; position:relative; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">|M|<12

Cho a,b,c >0 ,thõa mãn : a+b+c =1

CM: $\frac{3}{ab+bc+ac}+\frac{2}{a^2+b^2+c^2}>14$

Áp dụng bunhia nha các bạn

Chứng minh rằng với mọi số dương a,b,c,d&gt;0" role="presentation" style="border:0px; color:rgb(0, 0, 128); direction:ltr; display:inline-block; float:none; font-family:times new roman; font-size:18.18px; 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-wrap:normal" class="MathJax_CHTML mjx-chtml">a,b,c,d>0$a,b,c,d>0$ thì
a&#x2212;db+d+d&#x2212;bb+c+b&#x2212;cc+a+c&#x2212;aa+d&#x2265;0" role="presentation" style="border:0px; color:rgb(0, 0, 128); direction:ltr; display:inline-block; float:none; font-family:times new roman; font-size:18.18px; 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-wrap:normal" class="MathJax_CHTML mjx-chtml">a−db+d+d−bb+c+b−cc+a+c−aa+d≥0$\frac{a-d}{b+d}+\frac{d-b}{b+c}+\frac{b-c}{c+a}+\frac{c-a}{a+d}\ge 0$