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HD
0
LQ
1
21 tháng 1 2017
Ta có: \(\frac{x}{x^2+x+1}=\frac{-2}{3}\)
\(\Leftrightarrow\frac{x^2+x+1}{x}=-1,5\)
\(\Leftrightarrow x+1+\frac{1}{x}=-1,5\)
\(\Leftrightarrow x+\frac{1}{x}=-2,5\)
Ta lại có: \(A=\frac{x^2}{x^4+x^2+1}\)
\(\Leftrightarrow\frac{1}{A}=\frac{x^4+x^2+1}{x^2}=x^2+1+\frac{1}{x^2}\)
\(=\left(x+\frac{1}{x}\right)^2-1=\left(-2,5\right)^2-1=5,25\)
11 tháng 7 2019
\(A=\frac{x^2}{x^2-1}-\frac{2x^2}{x^4-1}-\frac{1}{x^2+1}\)ĐK \(x\ne1\)
\(=\frac{x^2}{x^2-1}-\frac{2x^2}{\left(x^2-1\right)\left(x^2+1\right)}-\frac{1}{x^2+1}\)
\(=\frac{x^2\left(x^2+1\right)-2x^2-1\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)
\(=\frac{x^4+x^2-2x^2-x^2+1}{\left(x^2-1\right)\left(x^2+1\right)}\)
\(=\frac{x^4-2x^2+1}{\left(x^2-1\right)\left(x^2+1\right)}\)
\(=\frac{x^4-x^2-x^2+1}{\left(x^2-1\right)\left(x^2+1\right)}\)
\(=\frac{x^2\left(x^2-1\right)-\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)
\(=\frac{\left(x^2-1\right)\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)
\(=\frac{x^2-1}{x^2+1}\)
Thay \(x=-\frac{2}{3}\)ta có
\(\frac{\left(\frac{-2}{3}\right)^2-1}{\left(-\frac{2}{3}\right)^2+1}=\frac{\frac{4}{9}-1}{\frac{4}{9}+1}=-\frac{5}{9}:\frac{13}{9}=-\frac{5}{13}\)
1 tháng 8 2016
\(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4x^2-4}{5}\)
\(=\left(\frac{x^2+2x+1+6-x^2+x-3x+3}{2\left(x-1\right)\left(x+1\right)}\right)\frac{4\left(x^2-1\right)}{5}\)
\(=\frac{10}{2\left(x-1\right)
\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=4\)
Vậy giá trị của biểu thức là 4
TQ
2
20 tháng 11 2019
Ta có:
\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)
= \(\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
= \(\frac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
= \(\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
= \(\frac{1}{x^2+x+1}\)
20 tháng 11 2019
Bài làm
\(\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}+\frac{1}{1-x}\)
\(=\frac{x^2+2}{x^3-1}+\frac{2}{x^2+x+1}-\frac{1}{x-1}\)
MTC = ( x - 1 )( x2 + x + 1 )
\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{(x^2+x+1)\left(x-1\right)}-\frac{1\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\)
\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-2}{(x^2+x+1)\left(x-1\right)}-\frac{x^2+x+1}{(x-1)\left(x^2+x+1\right)}\)
\(=\frac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{1}{x^2+x+1}\)
# Học tốt #
HM
0
13 tháng 7 2017
\(C=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[6]{\left(7-4\sqrt{3}\right).\left(7+4\sqrt{3}\right)}-x}{\sqrt[4]{\left(9+4\sqrt{5}\right).\left(9-4\sqrt{5}\right)}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1+\sqrt{x}\right).\left(1-\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\sqrt{x}+1-\sqrt{x}=1\)
25 tháng 12 2021
a: \(P=\dfrac{x+x-2-2x-4}{\left(x+2\right)\left(x-2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{2}=\dfrac{-3}{x-2}\)