tim x biet (2x-3)^2-4.(x+1)^2=5
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\(2\left(|x-1|+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)
\(TH1:x\ge1\Rightarrow|x-1|=x-1\)
\(\Rightarrow2\left(x-1+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)
\(\Rightarrow2\left(2x-\frac{9}{5}\right)=2x-\frac{2}{5}\Rightarrow4x-\frac{18}{5}=2x-\frac{2}{5}\)
\(\Rightarrow4x-2x=\frac{18}{5}-\frac{2}{5}\Rightarrow2x=\frac{16}{5}\Rightarrow x=\frac{16}{5}:2=\frac{16}{10}=\frac{8}{5}\)
\(TH2:x< 1\Rightarrow|x-1|=-x+1\)
\(\Rightarrow2\left(-x+1+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)
\(\Rightarrow2\left(1-\frac{4}{5}\right)=2x-\frac{2}{5}\Rightarrow2\cdot\frac{1}{5}=2x-\frac{2}{5}\)
\(\Rightarrow2x-\frac{2}{5}=\frac{2}{5}\Rightarrow2x=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}\Rightarrow x=\frac{4}{5}:2=\frac{4}{10}=\frac{2}{5}\)
TA CÓ \(\left(2X-3\right)^2-4.\left(X+1\right)^2=5\)
=>\(\left(2X-3\right)^2-\left(4X+4\right)^2=5\)
=>\(\left(2X-3\right)^2=5;\left(4X+4\right)^2=5\)
=>\(\orbr{\begin{cases}2X-3=5\\4X+4=5\end{cases}}\) =>\(\orbr{\begin{cases}2X=8\\4X=1\end{cases}}\) =>\(\orbr{\begin{cases}X=4\\X=\frac{1}{4}\end{cases}}\)