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126-2.(x-1)=20 120+3.(x-3)=180
2.(x-1)=126-20 3.(x-3)=180-120
2.(x-1)=106 3.(x-3)=60
x-1=106:2 x-3=60:3
x-1=53 x-3=20
x=53+1 x=20+3
x=54 x=23
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a, 7\(x\).(\(x\) - 10) = 0
\(\left[{}\begin{matrix}7x=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
Vậy \(x\in\) {0; 10}
b, 17.(3\(x\) - 6).(2\(x\) - 18) = 0
\(\left[{}\begin{matrix}3x-6=0\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=6\\2x-18=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=6:3\\x=18:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=9\end{matrix}\right.\)
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a) (x-1).(x+2)=0
=> +)x-1=0=>x=1
+)x+2=0=>x=-2
vậy x thuộc {1;-2)
b) (x+4).(4-x)=0
suy ra: +) x+4=0=>x=-4
+)4-x=0=>x=4
vậy x thuộc {-4;4}
c) (x+4)(-3x+9)=0
suy ra : +) x+4= 0=>x=-4
+)-3x+9=0=>x=3
vậy x thuộc {-4;3)
d) (2x-4)(x+3)=0
suy ra : +) 2x-4=0=>x=2
+)x+3=0=>x=-3
vậy x thuộc {2;-3}
e) (x2-9).(2x+10)=0
suy ra : +) x2-9=0=>x=9/2
+) 2x+10=0=>x=-5
Vậy x thuộc {9/2;-5}
g) (4-x).x2=0
suy ra : +)4-x=0 => x=4
+) x.2=0=> x=0
Vậy x thuộc {4;0}
HT
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a) -45:5(-3-2x)=3
5(-3-2x)=-45:3
5(-3-2x)=-15
-3-2x=-15:5
-3-2x=-3
2x=(-3)-(-3)
2x=-6
x=-6:2
x=-3
𝑥=5
𝑥=−3
ht
( 2x - 10 ) ( x + 3 ) = 0
<=>\(\hept{\begin{cases}2x-10=0\\x+3=0\end{cases}}\)
<=>\(\hept{\begin{cases}2x=10\\x=-3\end{cases}}\)
<=>\(\hept{\begin{cases}x=5\\x=-3\end{cases}}\)
Vậy x \(\in\){5;-3}