Question

[giúp]

Cho \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

a,Rút gọn

b,Tính giá trị của biểu thức A khi \(|x|=\frac{1}{2}\)

c,Tìm x để A=2

d,tìm x để A<0

e,Tìm x thuộc Z để A thuộc Z

Answer 1

[ST]

ĐKXĐ: \(x\ne0;x\ne\pm2\)

a, \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=\left[\frac{3x^2}{3x\left(x-2\right)\left(x+2\right)}-\frac{6x\left(x+2\right)}{3x\left(x-2\right)\left(x+2\right)}+\frac{3x\left(x-2\right)}{3x\left(x-2\right)\left(x+2\right)}\right]:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(=\frac{3x^2-6x^2-12x+3x^2-6x}{3x\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{-18x}{3x\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{6}\)

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

b, Ta có: \(\left|x\right|=\frac{1}{2}\Rightarrow x=\pm\frac{1}{2}\)

Với \(x=\frac{1}{2}\) thì \(A=\frac{-1}{\frac{1}{2}-2}=\frac{-1}{\frac{-3}{2}}=\frac{2}{3}\)

Với \(x=\frac{-1}{2}\)thì \(A=\frac{-1}{\frac{-1}{2}-2}=\frac{-1}{\frac{-5}{2}}=\frac{2}{5}\)

c, Để A=2 <=> \(\frac{-1}{x-2}=2\Leftrightarrow-1=2x-4\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Vậy x=3/2 thì A=2

d, Để A<0 <=> \(\frac{-1}{x-2}< 0\Leftrightarrow x-2>0\Leftrightarrow x>2\)

Vậy với x>2 thì A<0

e, Để A thuộc Z <=> x-2 thuộc Ư(-1)={1;-1}

Ta có: x-2=1 => x=3 (t/m)

          x-2=-1 => x=1 (t/m)

Vậy x thuộc {3;1} thì A thuộc Z

Answer 2

[Nguyễn Tất Đạt]

a)  \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)(ĐKXĐ: x khác 0; + 2)

\(A=\left(\frac{x^2}{x\left(x^2-4\right)}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right)\)

\(A=\left(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\right):\frac{6}{x+2}\)

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

Vậy \(A=\frac{1}{2-x}.\)

b) \(\left|x\right|=\frac{1}{2}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\). Nếu \(x=\frac{1}{2}\)thì \(A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}.\)

Nếu \(x=-\frac{1}{2}\)thì \(A=\frac{1}{2+\frac{1}{2}}=\frac{2}{5}.\)Vậy ...

c) Để A=2 thì \(\frac{1}{2-x}=2\Rightarrow4-2x=1\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}.\)Vậy ...

d) Để A<0 thì \(\frac{1}{2-x}< 0\Rightarrow2-x< 0\Leftrightarrow x>2.\)Vậy ...

e) Để A thuộc Z thì \(\frac{1}{2-x}\in Z\Rightarrow1⋮2-x\). Mà 2-x thuộc Z (Do x thuộc Z)

Nên \(2-x\in\left\{1;-1\right\}\Rightarrow x\in\left\{1;3\right\}.\)(t/m ĐKXĐ)

Vậy x=1 hay x=3 thì A nguyên.