a/ \(-12\left(x-5\right)+7\left(3-x\right)=5\)
\(< =>-12x+60+21-7x=5\)
\(< =>-19x+81=5\)
\(< =>-19x=-76\)
\(< =>x=\frac{76}{19}\)
b/ 30(x+2)-6(x-5)-24x=100
<=>30x + 60 - 6x + 30 - 24x =100
<=> 90=100( vô lý)
c/ \(\left(x-1\right)\left(x^2+1\right)=0\)
\(< =>\hept{\begin{cases}x-1=0\\x^2+1=0\end{cases}}< =>\hept{\begin{cases}x=1\\x^2=-1\left(voly\right)\end{cases}}\)
d/ làm rồi mà
a. \(-12.\left(x-5\right)+7.\left(3-x\right)=5\)
\(-12x+60+21-7x=5\)
\(-19x+81=5\)
\(-19x=-76\)
\(x=4\)
b. \(30.\left(x+2\right)-6.\left(x-5\right)-24x=100\)
\(30x+60-6x+30-24x=100\)
\(\left(30x-6x-24x\right)+\left(60+30\right)=100\)
\(90=100\)(vô lí)
\(\Rightarrow x=\varnothing\)
c. \(\left(x-1\right)\left(x^2+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x^2=-1\left(loại\right)\end{cases}}}\)
\(\Rightarrow x=1\)
Câu d) chính là câu a) :D
a) - 12 . ( x - 5) + 7 . ( 3 - x ) =5
-12x-(-60)+21-7x=5
-12x+60+21-7x=5
-12x-7x=5-60-21
-19x=-76
19x=76
x=76:19
x=4
vậy x=4
b) 30 .( x+ 2 ) - 6 . ( x- 5) -24 .x = 100
30x-60-6x+30-24x=100
(30x-6x-24x)+(60+30)=100
90=100 (vô lí)
vậy \(x\in\varnothing\)
c) ( x - 1 ) . ( x 2+ 1) = 0
* x-1=0 * x2+1=0
x=0+1 x2=0-1
x=1 x2=-1 ( vô lí ,x2 \(\ge0\))
