Xét x\(\ge\) 2 ta có:

x-1+x-2=1

=>2x-3=1

=>x=3/2(loại)

Xét 1\(\le x<2\)

x-1-x+2=1

1=1(thỏa mãn với mọi 1\(\le x<2\)

Xét x<1

-x+1-x+2=1

-2x+3=1

x=-1(TMĐK)

Vậy với x=-1 và1\(\le x<2\)

thì |x-1|+|x-2|=1

tìm x hả?

kết bạn đi mình giải cho

\(\left(x+1\right)^3+\left(x-2\right)^3-2x^2\left(x-1,5\right)=3\)

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8-2x^3+3x^2=3\)

\(\Leftrightarrow15x=10\Leftrightarrow x=\dfrac{2}{3}\)

Mình làm hơi tắt phần sau chút bạn nhé !

\(6,\Rightarrow6x^2+12x+6-2x^3-6x^2-6x-2+2x^3-2=1\\ \Rightarrow6x=-1\Rightarrow x=-\dfrac{1}{6}\\ 7,\Rightarrow\left(x-2\right)\left(x-2-x-2\right)=0\\ \Rightarrow-4\left(x-2\right)=0\\ \Rightarrow x-2=0\Rightarrow x=2\\ 8,\Rightarrow\left(x-2\right)^2-25=0\\ \Rightarrow\left(x-2-5\right)\left(x-2+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

x^2+x(6-2x) = 3x(x+1)-4(x^2-1)

x^2+6x-2x^2=3x^2-4x^2+4

6x-x^2=4-x^2

6x=4

x=3/2

ta có  x^2 +x(6-2x) = 3x(x+1)-4(x^2-1)

hay: x^2+6x-2x^2=3x^2+3x-4x^2+4

=> x^2 + 6x -2x^2 - 3x^2 - 3x +4x -4 =0

=>3x - 4 = 0

=>3x=4

=>x=4/3

c)
\(x^3-3.x^2.6+3.x.6^2-6^3=0\)
\(\left(x-6\right)^3=0\)
x-6=0
x=6
d)
\(x^3-3.x^2.1+3.x.1^2-1-x^3-3x-2=0\)
\(x^3-3x^2+3x-1-x^3-3x^2-2=0\)
\(-6x^2-3=0\)
\(-3\left(2x^2+1\right)=0\)
\(2x^2+1=0\)
2x²=-1
x²=1/2
x=\(\dfrac{\sqrt{2}}{2}\)

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4

c: Ta có: \(x^3-12x^2+48x-64=0\)

\(\Leftrightarrow x-4=0\)

hay x=4