a) Ta có: (4x-2)(x+5)=0
⇔2(2x-1)(x+5)=0
mà 2≠0
nên \(\left[{}\begin{matrix}2x-1=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(ktm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)
Vậy: x=-5
b) Ta có: 2x-9=-8-9
⇔2x-9=-17
hay 2x=-8
⇔x=-4(tm)
Vậy: x=-4
c) Ta có: 3|x-1|-27=0
⇔3|x-1|=27
⇔|x-1|=9
⇔\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=-8\left(tm\right)\end{matrix}\right.\)
Vậy: x∈{-8;10}
d) Ta có: 5(3x+8)-7(2x+3)=16
⇔15x+40-14x-21-16=0
⇔x+3=0
hay x=-3(tm)
Vậy: x=-3
a)\(\left(4x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x-2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)
Vậy...
b)\(2x-9=-8-9\\ \Leftrightarrow2x-9=-17\\ \Leftrightarrow2x=-8\\ \Leftrightarrow x=-4\)
Vậy...
c)\(2\left|x-1\right|-27=0\\ \Leftrightarrow\left|x-1\right|=\frac{27}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{27}{2}\\x-1=-\frac{27}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{2}\\x=-\frac{25}{2}\end{matrix}\right.\)
Vậy...
a) (4x – 2)(x + 5) = 0
⇔ \(\left[{}\begin{matrix}4x-2=0\\x+5=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}4x=2\\x=-5\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=-5\end{matrix}\right.\)
Vậy ...
b) 2x – 9 = -8 – 9
⇔ 2x = - 8
⇔ x = - 4
Vậy x = - 4
c) 3.| x -1 | - 27 = 0
⇔ 3 . | x - 1| = 27
⇔ | x - 1| = 9
⇔ \(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)
Vậy x ∈ { 10 ; - 8}
d) 5.(3x + 8) –7.(2x + 3) = 16
⇔ 15x + 40 - 14x - 21 = 16
⇔ x + 19 = 16
⇔ x = - 3
Vậy ..
