Thực hiện phép tính
a) \(\frac{\text{x + 9}}{x^2 - 9}-\frac{\text{3}}{\text{x^2 + 3x}}\)

b) \(\frac{\text{3x + 5 }}{\text{x^2 - 5x }}+\frac{\text{ 25 - x }}{\text{25 - 5x }}\)

c) \(\frac{\text{3 }}{\text{2x }}+\frac{\text{3x - 3 }}{\text{2x - 1 }}+\frac{ 2x^2 + 1 }{\text{4x^2 - 2x }}\)

d) \(\frac{\text{1}}{\text{3x - 2 }}-\frac{1}{\text{3x + 2 }}- \frac{\text{3x - 6}}{\text{4 - 9x^2}}\)
e) \(\frac{\text{18 }}{\text{(x - 3)(x^2 - 9) }}-\frac{\text{3 }}{\text{x^2 - 6x + 9 }}-\frac{\text{x}}{\text{x^2 - 9}}\)
g) \(\frac{\text{x + 2 }}{\text{x + 3 }}-\frac{\text{5 }}{\text{x^2 + x - 6 }}+\frac{\text{1}}{\text{2 - x}}\)
h) \(\frac{\text{4x }}{\text{x + 2 }}-\frac{\text{3x }}{\text{x - 2 }}+\frac{\text{12x}}{\text{x^2 - 4}}\)
i) \(\frac{\text{ x + 1 }}{\text{ x - 1 }}-\frac{\text{ x - 1 }}{\text{ x + 1 }}-\frac{\text{4}}{\text{1 - x^2}}\)
k) \(\frac{\text{ 3x + 21 }}{\text{ x^2 - 9 }}+\frac{\text{2 }}{\text{x + 3 }}-\frac{\text{3}}{\text{x - 3}}\)

Ai giải giúp mấy bài toán vs

Bài 1:

A=\(\sqrt{\frac{1}{\text{√}2+1}-\frac{\text{√}8-\text{√}10}{2-\text{√}5}}\)

B=\(\frac{5\text{√}5}{\text{√}5+2}+\frac{\text{√}5}{\text{√}5-1}-\frac{3\text{√}5}{3+\text{√}5}\)

Bài 2 rút gọn biểu thức

A=\(\left(\frac{x+\sqrt[]{xy}}{\text{√}x+\text{√}y}-2\right):\frac{1}{\text{√}x+2}\) với x :y >0

B=\(\left(\frac{a}{a-2\text{√}a}+\frac{a}{\text{√}a-2}\right):\frac{\text{√}a+1}{a-4\text{√}a+4}\)

Bài 3 cho biểu thức

P=\(\left(\frac{x-2}{x+2\text{√}x}+\frac{1}{\text{√}x+2}\right)\frac{\text{√}x+1}{\text{√}x-1}\)

a)Rút gọn P

b)tìm x để P=\(\text{√}x+\frac{5}{2}\)

bài 4 rút gọn biểu thức 

A=\(\frac{1}{x+\text{√}x}+\frac{2\text{√}x}{x-1}-\frac{1}{x-\text{√}x}\)

B=\(\left(\frac{x}{x+3\text{√}x}+\frac{1}{\text{√}x+3}\right):\left(1-\frac{2}{\text{√}x}+\frac{6}{x+3\text{√}x}\right)\)

Bài 5

A=\(\left(\frac{2}{\text{√}x-3}-\frac{1}{\text{√}x+3}-\frac{x}{\text{√}x\left(x-9\right)}\right):\text{(√}x+3-\frac{x}{\text{√}x-3}\)

a)rút gọn A

b)tìm gtri x để A= -1/4

AI GIẢI GIÙM MÌNH ĐI MÌNH TẠ ƠN

\(\text{Tìm }x,\text{ }\text{biết:}\)

\(1\text{)}\text{ }2.\left(x-\frac{1}{3}\right)-3\left(x-\frac{1}{2}\right)=\frac{1}{2}x\)

\(2\text{) }-3\left(x-\frac{1}{4}\right)-\frac{1}{3}\left(x+\frac{1}{2}\right)=x\)

\(\text{3) }\frac{3}{2}\left(x-\frac{5}{3}\right)-\frac{4}{5}=x+1\)

\(\text{4) }\frac{1}{6}\left(2.x-3\right)=\frac{1}{2}\left(-x+\frac{1}{4}-\frac{2}{3}\right)\)

\(\text{5) }-\frac{2}{3}\left(x-\frac{1}{4}\right)=\frac{1}{3}\left(2.x-1\right)\)

1.Giải pt sau:(\(\sqrt{2}\) +2)(x\(\sqrt{2}\) -1)=2x\(\sqrt{2}\) -\(\sqrt{2}\)

2.Cho pt: 2(a-1).x-a(x-1)=2a+3

3.Giải pt sau:

a) \(\frac{2}{x+\frac{\text{1}}{\text{1}+\frac{x+\text{1}}{x-2}}}=\frac{6}{3x-\text{1}}\)

b) \(\frac{\frac{x+\text{1}}{x-\text{1}}-\frac{x-\text{1}}{x+\text{1}}}{\text{1}+\frac{x+\text{1}}{x-\text{1}}}=\frac{x-\text{1}}{2\left(x+\text{1}\right)}\)

3/a)\(M\left(x\right)=\frac{1}{2}x^2-3x-x^3+3\)

\(=-x^3+\frac{1}{2}x^2-3x+3\)

\(N\left(x\right)=-4x+x^2+\frac{1}{2}x^3+6\)

\(=\frac{1}{2}x^3+x^2-4x^3+6\)

b)Ta có:\(\text{A}\left(x\right)=M\left(x\right)-N\left(x\right)\)

hay \(\text{A}\left(x\right)=\left(-x^3+\frac{1}{2}x^2-3x+3\right)-\left(\frac{1}{2}x^3+x^2-4x+6\right)\)

\(=-x^3+\frac{1}{2}x^2-3x+3-\frac{1}{2}x^3-x^2+4x-6\)

\(=-\frac{3}{2}x^3-\frac{1}{2}x^2+x-3\)

Đặt\(\text{A}\left(x\right)=0\)

\(\Rightarrow-\frac{3}{2}x^3-\frac{1}{2}x^2+x-3=0\)

\(-\frac{3}{2}x^3-\frac{1}{2}x^2=-x+3\)

\(-2\left(x^3-x^2\right)=-x+3\)

\(x^3-x^2+x=3+2=5\)

\(x^2=5\)

\(\Rightarrow x=\sqrt{5}\)

Vậy \(\text{A}\left(x\right)\) có 1 nghiệm là \(\sqrt{5}\)

1)\(\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx\)

2)\(\int\frac{dx}{\left(e^x+1\right)\left(x^2+1\right)}\)

3)\(\int\frac{1+2x\sqrt{1-x^2}+2x^2}{1+x+\sqrt{1+x^2}}\)dx

4)\(\int\frac{sin^6x+c\text{os}^6x}{1+6^x}dx\)

5)\(\int_0^{\frac{\pi}{2}}\frac{\sqrt{c\text{os}x}}{\sqrt{s\text{inx}}+\sqrt{c\text{os}x}}dx\)

6)\(\int\frac{x^4}{2^x+1}dx\)

7)\(\int_0^{\frac{\pi^2}{4}}sin\sqrt{x}dx\)

8)\(\int\sqrt[6]{1-c\text{os}^3x}.s\text{inx}.c\text{os}^5xdx\)

9)\(\int\sqrt{\frac{1}{4x}+\frac{\sqrt{x}+e^x}{\sqrt{x}.e^x}}dx\)

10)\(\int\frac{c\text{os}x+s\text{inx}}{\left(e^xs\text{inx}+1\right)s\text{inx}}dx\)