Question

Tìm x,y biết:

a)|x+4/15|-|-3.75|=-|-2.5|

b)|1/2-1/3+x|=-1/4-|y|

c)|x-y|+|y+9/25|=0

d)|x(x²-5/4)|=x

Answer 1

a: \(\left|x+\dfrac{4}{15}\right|-\left|-3.75\right|=-\left|-2.5\right|\)

\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.5+3.75=1.25=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{5}{4}\\x+\dfrac{4}{15}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{59}{60}\\x=-\dfrac{91}{60}\end{matrix}\right.\)

c: \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)

d: Ta có: \(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{5}{4}\right)=x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x\left(x^2-\dfrac{9}{4}\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{3}{2}\right\}\)