|x+1|-3|5-x|=2(x-1)-3x
=>|x+1|-3|x-5|=2(x-1)-3x
Với x\(\ge\)5 =>x+1-3.(x-5)=2.(x-1)-3x
=>x+1-3x+15=2x-2-3x
=>-2x+16=-x-2
=>-2x+x=-2-16
=>-x=-18
=>x=18(nhận)
Với \(-1\le x\le5\) =>(x+1)-3.[-(x-5)]=2(x-1)-3x
=>x+1+3x+15=2x-2-3x
=>4x+16=-x-2
=>4x+x=-2-16
=>5x=-18
=>x=-18/5(loại)
Với x<-1 =>-(x+1)-3.[-(x-5)]=2(x-1)-3x
=>-x-1+3x-15=2x-2-3x
=>2x-16=-x-2
=>3x=-2+16
=>3x+14
=>x=14/3 (loại)
Vậy x=18
Tìm x: \(\left(x-2\right)^3+3\left(2x-3\right)^2=\left(3+x\right)^3-5\left(1+3x\right)^2-\left(x-1\right)\left(3-x\right)\)
Tìm x: \(\frac{\left(x-2\right)^2}{2}-4\frac{1}{3}\left(x+3\right)^2=\frac{1}{4}\left(x-1\right)\left(x-2\right)-2\left(3x-1\right)\left(2+3x\right)\)
Tìm x: \(\frac{\left(x-3\right)^2}{3}-\frac{\left(2x+1\right)\left(1-3x\right)}{2}=\frac{x\left(x+2\right)}{3}-\left(1+x\right)^2\)
Tìm x: \(\left(x-3\right)^3+2\left(2x-1\right)^2-2\left(x-1\right)=\left(2+x\right)^3-3\left(1+3x\right)^2\)
Tìm x: \(3\left(x-2\right)^2-1\frac{1}{4}\left(1+3x\right)^2=1\frac{1}{3}\left(x-1\right)\left(3-x\right)-1\frac{1}{6}x\left(x-3\right)\)
Tìm x: \(\left(2x-1\right)^3-3\left(x+2\right)\left(x-3\right)=\left(3+2x\right)^3-3x\left(x+1\right)\)
Tìm x: \(\frac{\left(2-3x\right)^2}{3}-\frac{\left(1+2x\right)^2}{2}=\frac{3}{4}-2\left(x-1\right)\left(x+2\right)+x\left(1+x\right)\)
Tìm x: \(3\left(x-2\right)^2-1\frac{1}{4}\left(1+3x\right)^2=1\frac{1}{3}\left(x-1\right)\left(3+x\right)-1\frac{1}{6}x\left(x-3\right)\)
Tìm x: \(\frac{\left(2x-1\right)^2}{3}-\frac{1}{2}\left(1+3x\right)^2=\frac{1}{3}\left(x-1\right)\left(2-3x\right)\)
