Cho biểu thức
(\(\dfrac{x^2-2x}{2x^2+8}\) - \(\dfrac{2x^2}{8-4x+2x^2-x^3}\)).(1-\(\dfrac{1}{x}\)-\(\dfrac{2}{x^2}\))(x≠0;x≠2)
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24 tháng 7 2021
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:
\(P=\dfrac{1}{2}+1=\dfrac{3}{2}\)
15 tháng 1 2022
a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)
b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)
\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}=\dfrac{x+1}{2x}\)
c: Thay x=2017 vào C, ta được:
\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)
30 tháng 10 2022
\(A=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\left(\dfrac{x\left(x^2-4x+4\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x\left(x^2-4x+4+4x\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x+1}{2x}\)
16 tháng 1 2018
( x2−2x / 2x2+8 − 2x2 / 8−4x+2x2−x3 ).(1− 1/x − 2/x2 )
=[ x2−2x / 2(x2+4) − 2x2 / 2(x2+4)−x(x2+4) ]. x2−x−2 / x2
=[x2−2x / 2(x2+4) − 2x2 / (2−x)(x2+3)] . x2−x−2 / x2
=(x2−2x)(2−x)−4x2 / 2(2−x)(x2+4) . x2+x−2x−2 / x2
= −x(x2+4) / 2(2−x)(x2+4). (x+1)(x−2) / x2
=x+1 / 2x
26 tháng 6 2022
\(A=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-2x^2-2x^2+4x+4x^2}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
HP
2 tháng 2 2021
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
HP
2 tháng 2 2021
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
20 tháng 5 2022
a: \(M=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)^2+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-4x^2+4x+4x^2}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
b: Thay x=1/2 vào M, ta được:
\(M=\left(\dfrac{1}{2}+1\right):\left(2\cdot\dfrac{1}{2}\right)=\dfrac{3}{2}\)
6 tháng 4 2018
1. ĐKXĐ: \(x\ne0;x\ne2\)
Ta có: \(A=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(A=\left[\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right]\left(\dfrac{x^2-x-2}{x^2}\right)\)
\(A=\left[\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(4+x^2\right)\left(2-x\right)}\right]\left(\dfrac{x^2-x-2}{x^2}\right)\)
\(A=\left[\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(4+x^2\right)\left(x-2\right)}\right]\left(\dfrac{x^2-x-2}{x^2}\right)\)
\(A=\dfrac{x\left(x-2\right)^2+2.2x^2}{2\left(x^2+4\right)\left(x-2\right)}.\dfrac{\left(x^2-2x\right)+\left(x-2\right)}{x^2}\)
\(A=\dfrac{x\left(x^2-4x+4\right)+4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\dfrac{\left(x+1\right)\left(x-2\right)}{x^2}\)
\(A=\dfrac{\left(x^3+4x\right)\left(x+1\right)\left(x-2\right)}{2x^2\left(x^2+4\right)\left(x-2\right)}\)
\(A=\dfrac{x\left(x^2+4\right)\left(x+1\right)\left(x-2\right)}{2x^2\left(x^2+4\right)\left(x-2\right)}\)
\(A=\dfrac{x+1}{2x}\)
