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)

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

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

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

b: Khi x=2 thì \(A=\dfrac{\left(4+2\right)\left(2+1\right)}{\left(2\cdot2+1\right)\left(2-1\right)}=\dfrac{18}{5}\)

Bài này đã có tại đây:

Cho biểu thức:  \(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)Với ... - Hoc24

Lời giải:

a.

\(A=\left[\frac{(2+x)^2}{(2-x)(2+x)}+\frac{4x^2}{(2-x)(2+x)}-\frac{(2-x)^2}{(2-x)(2+x)}\right]:\frac{x(x-3)}{x^2(2-x)}\)

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

b.

Khi $x=12$ thì $A=\frac{4.12^2}{12-3}=64$

c. 

$A=1\Leftrightarrow \frac{4x^2}{x-3}=1$

$\Leftrightarrow 4x^2=x-3$

$\Leftrightarrow 4x^2-x+3=0$

$\Leftrightarrow (2x-\frac{1}{4})^2=-\frac{47}{16}< 0$ (vô lý)

Vậy không tồn tại $x$

d. Để $A$ nguyên thì $\frac{4x^2}{x-3}$ nguyên

$\Leftrightarrow 4x^2\vdots x-3$

$\Leftrightarrow 4(x^2-9)+36\vdots x-3$

$\Leftrightarrow 36\vdots x-3$

$\Leftrightarrow x-3\in\left\{\pm 1;\pm 2;\pm 3;\pm 4;\pm 9; \pm 12; \pm 36\right\}$

Đến đây bạn có thể tự tìm $x$ được rồi, chú ý ĐKXĐ để loại ra những giá trị không thỏa mãn.

e.

$A>4\Leftrightarrow \frac{4x^2}{x-3}>4$

$\Leftrightarrow \frac{x^2}{x-3}>1$

$\Leftrightarrow \frac{x^2-x+3}{x-3}>0$

$\Leftrightarrow x-3>0$ (do $x^2-x+3>0$ với mọi $x$ thuộc ĐKXĐ)

$\Leftrightarrow x>3$. Kết hợp với đkxđ suy ra $x>3$

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2-2x}{1-x}\right)\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)}{x-1}\)

\(=\dfrac{-6}{\left(x+2\right)\left(x-1\right)}\)

b: Thay x=-4 vào A, ta được:

\(A=-\dfrac{6}{\left(-4+2\right)\left(-4-1\right)}=\dfrac{-6}{-2\cdot\left(-5\right)}=\dfrac{-6}{10}=\dfrac{-3}{5}\)

a: \(P=\dfrac{x+x-2-2x-4}{\left(x+2\right)\left(x-2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{2}=\dfrac{-3}{x-2}\)

`a)(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4,x ne 9)`

`=(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`

`=(2sqrtx-9+(sqrtx-3)(sqrtx+3)+(2sqrtx+1)(sqrtx-2))/(x-5sqrtx+6)`

`=(2sqrtx-9+x-9+2x-3sqrtx-2)/(x-5sqrtx+6)`

`=(3x-sqrtx-20)/