\(\left|x-1\right|< 5-7=-2\)( vô lí ) 

do \(\left|x-1\right|\ge0\forall x;-2< 0\)

\(2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\Leftrightarrow\frac{3}{2}-10x=\frac{4}{5}-3x\)

\(\Leftrightarrow-7x=\frac{4}{5}-\frac{3}{2}=\frac{8-15}{10}=-\frac{7}{10}\Leftrightarrow x=\frac{1}{10}\)

\(\left(x-5\right)^4-2^2.4=53\Leftrightarrow\left(x-5\right)^4-16=53\)

\(\Leftrightarrow\left(x-5\right)^4=69\Leftrightarrow\left[\left(x-5\right)^2\right]^2=\sqrt{69}^2\)

TH1 : \(\left(x-5\right)^2=\sqrt{69}\Rightarrow x-5=\sqrt{\sqrt{69}}\Rightarrow x=\sqrt{\sqrt{69}}+5=\sqrt[4]{69}+5\)

TH2 : \(\left(x-5\right)^2=-\sqrt{69}\)( vô lí ) 

\(2x^2y.x^3y^2x=2x^5y^3x=2x^6y^3\)bậc 9 

-> chọn D 

a) Ta có: 

VT = |x + 1| + |x + 2| + |2x - 3| \(\ge\)|x + 1 + x + 2| + |3 - 2x| =  |2x + 3| + |3 - 2x| \(\ge\)|2x + 3 + 3 - 2x| = 6

VP = 6

Dấu "=" xảy ra<=> \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\)  => \(\orbr{\begin{cases}x\ge-1\\x\le-2\end{cases}}\)và \(-\frac{3}{2}\le x\le\frac{3}{2}\)

<=> \(-1\le x\le\frac{3}{2}\)

b) Ta có: VT = |x + 1| + |x - 2| + |x - 3| + |x - 5| = (|x + 1| + |5 - x|) + (|x - 2| + |3 - x|) \(\ge\)|x + 1 + 5 - x| + |x - 2 + 3 - x| = |6| + |1| = 7

VP = 7

Dấu "=" xảy ra<=> \(\hept{\begin{cases}\left(x+1\right)\left(5-x\right)\ge0\\\left(x-2\right)\left(3-x\right)\ge0\end{cases}}\) <=> \(\hept{\begin{cases}-1\le x\le5\\2\le x\le3\end{cases}}\) <=> \(2\le x\le3\)