1. Tìm x
|x(x-4)|=x
jup mình nha!!!
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a: \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{5}\\x-\dfrac{3}{4}=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{20}\\x=\dfrac{11}{20}\end{matrix}\right.\)
\(a,\left|x-\dfrac{3}{4}\right|=\dfrac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{5}\\x-\dfrac{3}{4}=\dfrac{-1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{20}\\x=\dfrac{11}{20}\end{matrix}\right.\)
\(b,\dfrac{-1}{3}+\left|x\right|=0,5\)
\(\Leftrightarrow\left|x\right|=\dfrac{5}{6}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{-5}{6}\end{matrix}\right.\)
\(x+\dfrac{3}{4}=\dfrac{28}{6}+\dfrac{1}{2}\\ x+\dfrac{3}{4}=\dfrac{31}{6}\\ x=\dfrac{31}{6}-\dfrac{3}{4}\\ x=\dfrac{53}{12}\)
\(x+\dfrac{3}{4}=\dfrac{28+3}{6}=\dfrac{31}{6}\Leftrightarrow x=\dfrac{31}{6}-\dfrac{3}{4}=\dfrac{62-9}{12}=\dfrac{53}{12}\)
Lời giải:
$(x+2)+(x+4)+(x+6)+...+(x+32)=352$
$(x+x+...+x)+(2+4+6+...+32)=352$
$16x+272=352$
$16x=352-272=80$
$x=80:16=5$
ket qua =0