a/ \(13x-1=10x+5\)
\(\Rightarrow13x-10x=5+1\)
\(\Rightarrow3x=6\Rightarrow x=\dfrac{6}{3}=2\)
Vậy x = 2
b/ \(3\cdot\left|x-1\right|-1=5\cdot\left|x-1\right|-7\)
\(\Rightarrow3\cdot\left|x-1\right|-5\cdot\left|x-1\right|=-7+1\)
\(\Rightarrow\left(3-5\right)\cdot\left|x-1\right|=-6\)
\(\Rightarrow\left|x-1\right|=-6:\left(-2\right)=3\)
\(\Rightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Vậy...................
c/ \(x^2+8x+15=0\)
\(\Leftrightarrow x^2+3x+5x+15=0\)
\(\Leftrightarrow x\left(x+3\right)+5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=-3\end{matrix}\right.\)
Vậy............
a) 13x - 1 = 10x + 5
13x - 10x = 5 + 1
3x = 6
x = 2
Vậy x = 2 là giá trị cần tìm
b) 3|x - 1| - 1 = 5|x - 1| - 7
-1 + 7 = 5|x - 1| - 3|x - 1|
6 = 2|x - 1|
3 = |x - 1|
=> \(\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\)
TH1 : x - 1 = 3
x = 3 + 1
x = 4
TH1 : x - 1 = -3
x = -3 + 1
x = -2
Vậy \(\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\) là giá trị cần tìm
c)