a, ĐKXĐ: x≠±3

A=\(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{3-x}{x-3}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x^2-3x}{x^2-9}\right):\dfrac{3x^2}{x+3}\)

A=\(\left(\dfrac{-3}{x+3}\right):\dfrac{3x^2}{x+3}\)

A=\(\dfrac{-1}{x^2}\)

b, Thay x=\(-\dfrac{1}{2}\) (TMĐKXĐ) vào A ta có:

\(\dfrac{-1}{\left(-\dfrac{1}{2}\right)^2}\)=-4

c, A<0 ⇔ \(\dfrac{-1}{x^2}< 0\) ⇔ x²>0 (Đúng với mọi x)

Vậy để A<0 thì x đúng với mọi giá trị (trừ ±3)

\(E=\left[\left(\dfrac{3}{x+1}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)

\(=\left[\left(\dfrac{3}{x+1}-\dfrac{x}{\left(x+1\right)^2}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)

\(=\left[\left(\dfrac{2x+3}{\left(x+1\right)^2}\right):\dfrac{2x^2+3x}{x^2+7x}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)

\(=\left[\dfrac{x^2+7x}{x\left(x+1\right)^2}+\dfrac{3}{x+1}\right].\dfrac{x^2+x}{3x+1}\)

\(=\dfrac{2x\left(2x+5\right)}{x\left(x+1\right)^2}.\dfrac{x^2+x}{3x+1}\)

\(=\dfrac{2x\left(2x+5\right)}{x\left(x+1\right)^2}.\dfrac{x^2+x}{3x+1}=\dfrac{2x\left(2x+5\right)}{\left(x+1\right)\left(3x+1\right)}\)

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

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

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

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

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, Với x khác 1 

\(A=\dfrac{x^2+x+1-3x^2+2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{1}{x^2+x+1}\)

b, Ta có \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\dfrac{-1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)

Vậy với x khác 1 thì bth A luôn nhận gtri âm 

câu d

\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)

\(a,=\dfrac{x^2-20+x^2-7x+10+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\\ b,=\dfrac{10x+15-4x+6+2x-9}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{4}{2x-3}\\ c,=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\\ =\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{x+4-x}{x\left(x+4\right)}=\dfrac{4}{x\left(x+4\right)}\)