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1.
$-7(5-x)-2(x-10)=15$
$-35+7x-2x+20=15$
$5x-15=15$
$5x=30$
$x=30:5=6$
2.
$3(x-4)-(8-x)=12$
$3x-12-8+x=12$
$4x-20=12$
$4x=12+20=32$
$x=32:4=8$
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a) -3.(x-5) + 6.(x+2) = 9
-3x + 15 + 6x + 12 = 9
3x + 27 = 9
3x = -18
x = -6
b) 7.(x-9) - 5.(6-x) = -6 + 11.x
7x - 63 - 30 + 5x = - 6 + 11.x
=> 12x - 11x = - 6 + 63 + 30
x = 87
c) 10.(x-7) - 8.(x+5) = 6.(-5) + 24
10x - 70 -8x - 40 = -6
2x - 110 = - 6
2x = 104
x = 52
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1) Ta có: \(\dfrac{3x+7}{11}=\dfrac{5x-7}{9}\)
\(\Leftrightarrow9\left(3x+7\right)=11\left(5x-7\right)\)
\(\Leftrightarrow27x+63=55x-77\)
\(\Leftrightarrow27x-55x=-77-63\)
\(\Leftrightarrow-28x=-140\)
hay x=5
Vậy: S={5}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3.\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=12\)
\(4x-20=12\)
\(4x=12+20\)
\(4x=32\)
\(x=32:4\)
\(x=8\)
\(b,4.\left(x-5\right)-\left(x+7\right)=-19\)
\(4x-20-x-7=-19\)
\(3x-27=-19\)
\(3x=-19+27\)
\(3x=8\)
\(x=\frac{8}{3}\)
\(c,7.\left(x-3\right)-5.\left(3-x\right)=11x-5\)
\(7x-21-15+5x=11x-5\)
\(12x-36=11x-5\)
\(12x-11x=-5+36\)
\(x=31\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)4x-5(-3+x)=7<=>4x+15-5x=7<=>4x-5x=7-15<=>-x=-8<=>x=8
b)-4(x+1)+8(x-3)=24
<=>-4x-4+8x-24=24
<=>-4x+8x=24+24+4
<=>4x=52<=>x=13
c)7(x-9)-5(6-x)=-6+11x
<=>7x-63-30+5x=-6+11x
<=>7x+5x-11x=-6+63+30
<=>x=87