Question

[linh katy]

b)-5 +/3x-1/+6=/-4/

c)(x-1)^2 = (x-1)^4

d) 5^-1 .25^x =125 (x thuộc Z)

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

Answer 1

[Nguyễn Lê Phước Thịnh]

b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)

\(\Leftrightarrow\left|3x+1\right|+1=4\)

\(\Leftrightarrow\left|3x+1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)

c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)

\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)

\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{0;1;2\right\}\)

d) Ta có: \(5^{-1}\cdot25^x=125\)

\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)

\(\Leftrightarrow5^{2x-1}=5^3\)

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=4\)

hay x=2

Vậy: x=2

Answer 2

[linh katy]

cảm ơn nhìu ak