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TA CÓ \(\left(2X-3\right)^2-4.\left(X+1\right)^2=5\)
=>\(\left(2X-3\right)^2-\left(4X+4\right)^2=5\)
=>\(\left(2X-3\right)^2=5;\left(4X+4\right)^2=5\)
=>\(\orbr{\begin{cases}2X-3=5\\4X+4=5\end{cases}}\) =>\(\orbr{\begin{cases}2X=8\\4X=1\end{cases}}\) =>\(\orbr{\begin{cases}X=4\\X=\frac{1}{4}\end{cases}}\)
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\(\left|x-\dfrac{1}{2}\right|\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\\\left|2x-\dfrac{3}{4}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow2x-\dfrac{3}{4}\ge0\)
\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=1\Rightarrow x=\dfrac{3}{2}\\x-\dfrac{1}{2}=-1\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
\(2x-\dfrac{3}{4}\ge0\Rightarrow2x\ge\dfrac{3}{4}\Rightarrow x\ge\dfrac{3}{2}\)
Vậy xảy ra khi:
\(x=\dfrac{3}{2}\)
a.x=4
giai chi tiet gium minh voi