Question

bài 1: tìm x

a) x+3/4=x+1/2

b) |x-1|=2x-5

c) (x-3)^6=(3-x)^10

Answer 1

a: =>2(x+3)=4(x+1)

=>x+3=2(x+1)

=>2x+2=x+3

=>x=1

b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\\left(2x-5-x+1\right)\left(2x-5+x-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\\left(x-4\right)\left(3x-6\right)=0\end{matrix}\right.\Leftrightarrow x=4\)

c: \(\Leftrightarrow\left(x-3\right)^6\left[\left(x-3\right)^4-1\right]=0\)

=>(x-3)(x-4)(x-2)=0

hay \(x\in\left\{3;4;2\right\}\)