\(=x^2+16x+64-2x^2-12x+32+x^2-4x+4\)

\(=100\)

100 nha em, nhân phá ngoặc ra là tính được nhé

a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)

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

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

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

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

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

b) Thay \(x=\dfrac{1}{2}\) vào P, ta được:

\(P=\dfrac{1}{2}+1=\dfrac{3}{2}\)

(-3).8/8.6 rút gọn

Rút gọn biểu thức
\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)+10\)
\(\Leftrightarrow8x+16-5x^2+10x+4\left(x^2+x-2x+2\right)+2\left(x^2+2x-2x+4\right)+10\)
\(\Leftrightarrow18x+26-5x^2+4\left(x^2-x+2\right)+2\left(x^2+4\right)\)
\(\Leftrightarrow18x-5x^2+26+4x^2-4x+8+2x^2+8\)
\(\Leftrightarrow18x-4x-5x^2+4x^2+2x^2+8+26+8\)
\(\Leftrightarrow14x+3x^2+42\)

4x+x^2

a: Đặt \(A=\left(\frac{x^2-2x}{2x^2+8}-\frac{2x^2}{8-4x+2x^2-x^3}\right)\cdot\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\)

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

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

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

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

b: Thay x=1/2 vào A, ta được:

\(A=\frac{\left\lbrack\left(\frac12\right)^2+4\cdot\frac12-4\right\rbrack\left(\frac12+1\right)}{2\left\lbrack\left(\frac12\right)^2+4\right\rbrack}=\frac{\left(\frac14+2-4\right)\cdot\frac32}{2\left(\frac14+4\right)}=\frac{\frac32\cdot\frac{-7}{4}}{2\cdot\frac{17}{4}}=\frac{-21}{8}:\frac{17}{2}=-\frac{21}{8}\cdot\frac{2}{17}=\frac{-21}{68}\)

\(\left(x+2\right)^3-2\left(x^2+6x-5\right)-8\)

\(=x^3+6x^2+12x+8-2x^2-12x+10-8\)

\(=x^3+4x^2+10\)

\(\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}\)

Để biểu thức trên nhận giá trị âm khi \(\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}< 0\)

\(\Rightarrow x^3-2x^2-4x+8< 0\)do \(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-2x\left(x+2\right)< 0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2< 0\Leftrightarrow x< -2\)

Có: \(\left(x+2\right)^2-2\left(x+2\right)\left(x-8\right)+\left(x-8\right)^2=\left[x+2-\left(x-8\right)\right]^2=\left(2+8\right)^2=10^2=100\)

P/s: mình hỏi thật nhá: Bạn ko biết làm à? Mình học lớp 7 mà còn biết làm.