Dề sai ko bạn

Chỉ cần ý b thôi 

mình xin lỗi sửa x-1 thành x+1

\(\frac{1-2x-2x^2}{1-x}\)

\(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4x^2-4}{5}\)
\(=\left(\frac{x^2+2x+1+6-x^2+x-3x+3}{2\left(x-1\right)\left(x+1\right)}\right)\frac{4\left(x^2-1\right)}{5}\)
\(=\frac{10}{2\left(x-1\right) \left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=4\)
Vậy giá trị của biểu thức là 4

Ta có: \(H=\left(\frac{x+x^3}{1-x^2}-\frac{x-x^3}{1+x^2}\right):\left(\frac{1+x}{1-x}-\frac{1-x}{1+x}\right)\)

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

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

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

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

a) Ta có: \(C=\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)\right]\)

\(=\dfrac{x\left(x^2-1\right)^2}{x^2+1}:\left[\left(\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}+x\right)\left(\dfrac{\left(1+x\right)\left(1-x+x^2\right)}{\left(1+x\right)}-x\right)\right]\)

\(=\dfrac{x\left(x^2-1\right)^2}{x^2+1}:\left[\left(x^2+2x+1\right)\left(x^2-2x+1\right)\right]\)

\(=\dfrac{x\left(x-1\right)^2\cdot\left(x+1\right)^2}{\left(x^2+1\right)}\cdot\dfrac{1}{\left(x+1\right)^2\cdot\left(x-1\right)^2}\)

\(=\dfrac{x}{x^2+1}\)

b) Thay \(x=-\dfrac{3}{2}\) vào C, ta được:

\(C=\dfrac{-3}{2}:\left(\dfrac{9}{4}+1\right)=\dfrac{-3}{2}:\dfrac{13}{4}=\dfrac{-3}{2}\cdot\dfrac{4}{13}=\dfrac{-6}{13}\)

c) Ta có: \(C=\dfrac{1}{2}\)

nên \(\dfrac{x}{x^2+1}=\dfrac{1}{2}\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x=1\)(Loại)

Bài 1:

\(A=\sqrt{8}-2\sqrt{2}+\sqrt{20}-2\sqrt{5}-2=2\sqrt{2}-2\sqrt{2}+2\sqrt{5}-2\sqrt{5}-2=-2\)\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)

Bài 2:

\(a,P=\left(\frac{1}{x+\sqrt{x}}-\frac{1}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+2\sqrt{x}+1}\left(x>0\right)\)

\(=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\times\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

\(=\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\times\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

\(=\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{x}\cdot\sqrt{x}}\)

\(=\frac{1-x}{x}\)

\(b,\forall x>0\Leftrightarrow\frac{1-x}{x}>\frac{1}{2}\)

\(\Leftrightarrow2\left(1-x\right)>x\)

\(\Leftrightarrow2-2x>x\)

\(\Leftrightarrow-3x>-2\)

\(\Leftrightarrow x< \frac{2}{3}\)

\(\Rightarrow P>\frac{1}{2}\Leftrightarrow\forall0< x< \frac{2}{3}\)