Question

Bài 8: Tìm x, biết a)|x-5|+2=6,7;c)(x-5)²=x-5;e*)|x-5|+x=1 b)2x²=3x;d*)|x-1|+x=1;f*)|x²+3|-x=5

Answer 1

a) \(\left|x-5\right|+2=6< =>\left[{}\begin{matrix}x-5+2=6\\5-x+2=6\end{matrix}\right.< =>\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
b) \(2x^2=3x< =>2x^2-3x=0< =>x\left(2x-3\right)=0\)
\(< =>\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
c) \(\left(x-5\right)^2=\left(x-5\right)< =>\left(x-5\right)^2-\left(x-5\right)=0\)
\(< =>\left(x-5\right)\left(x-5-1\right)=0\)
\(< =>\left[{}\begin{matrix}x-5=0\\x-6=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
d) \(\left|x-1\right|+x=1< =>\left[{}\begin{matrix}x-1+x=0\\1-x+x=0\left(koTM\right)\end{matrix}\right.\)\(< =>2x=1< =>x=\dfrac{1}{2}\)
e) \(\left|x-5\right|+x=1< =>\left[{}\begin{matrix}x-5+x=1\\5-x+x=1\left(koTM\right)\end{matrix}\right.\)
\(< =>2x=6< =>x=3\)
f) \(\left|x^2+3\right|-2=5< =>x^2+3-2=5< =>x^2=4< =>x=2\)

Answer 2

Bài của bạn Linh Nguyễn làm sai đề tùm lum