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1) x - 2 = -6
x = -6 + 2
x = -4
2) -5 . x - ( -3 ) =13
-5 . x = 13 + ( -3 )
-5 . x = 10
x = 10 : ( -5 )
x = -2
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a) 3 . (x + 7) + 2 . (x - 5) = 26
3x + 21 + 2x - 10 = 26
5x = 26 - 21 + 10
5x = 15
x = 3
b) 15 . (x - 3) - 7 = 10 . (x -2) + 3
15x - 45 - 7 = 10x - 20 + 3
15x - 10x = -17 + 45 + 7
5x = 35
x = 7
c) 3 . (x + 5) - 2 . (x -7) = 18
3x + 15 - 2x + 14 = 18
x = 18 - 15 - 14
x = -11
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\(a,2.\left(x-5\right)-3.\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(x=-31-14\)
\(x=-45\)
\(b,5.\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36=12\)
\(3x=12+36\)
\(3x=48\)
\(x=16\)
\(c,-5.\left(2-x\right)+4.\left(x-3\right)=10.x-15\)
\(-10+5x+4x-12=10x-15\)
\(-6x-22=10x-15\)
\(-6x-10x=-15+22\)
\(-16x=7\)
\(x=-\frac{7}{16}\)
Câu d , e f tương tự nha
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3(x + 6) = 2(x - 5)
=> 3x + 18 = 2x - 10
=> 3x - 2x = -10 - 18
=> x = -28
vay_
c, |3x - 7| = 6
=> 3x - 7 = 6 hoac 3x - 7 = -6
=> 3x = 13 hoac 3x = 1
=> x = 13/3 hoac x = 1/3
vay_
3(x + 6) = 2(x - 5)
=> 3x + 18 = 2x - 10
=> 3x - 2x = -10 - 18
=> x = -28
Vậy x = -28
c, |3x - 7| = 6
=> 3x - 7 = 6 hoac 3x - 7 = - 6
=> 3x = 13 hoac 3x = 1
=> x = \(\frac{13}{3}\) hoac x = \(\frac{1}{3}\)
Vậy \(x\in\left\{\frac{3}{13};\frac{1}{3}\right\}\)
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a)2(x - 2) + 7.(3 - x) = -2.(3x - 18)
=> 2x - 4 + 21 - 7x = -6x - 36
=> 2x - 7x + 6x = 36 - 21 + 4
=> x = 19
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\(-2.\left(x+\frac{1}{3}\right)-5.\left(x+\frac{1}{3}\right)=\frac{1}{2}.x\) \(x\)
<=> \(\left(x+\frac{1}{3}\right).\left(-2-5\right)=\frac{1}{2}.x\)
<=> \(\left(x+\frac{1}{3}\right).\left(-7\right)=\frac{1}{2}.x\)
<=> \(-7x-\frac{7}{3}=\frac{1}{2}.x\)
<=> \(-7x-\frac{1}{2}.x=\frac{7}{3}\)
<=> \(\left(-7-\frac{1}{2}\right).x=\frac{7}{3}\)
<=> \(\frac{-15}{2}.x=\frac{7}{3}\)
<=> \(x=\frac{7}{3}:\frac{-15}{2}=\frac{-14}{45}\)
\(-7x-\frac{7}{3}=\frac{1}{2}.x\)
<=> \(-7x-\frac{1}{2}x=\frac{7}{3}\)
<=> \(\left(-7-\frac{1}{2}\right).x=\frac{7}{3}\)
<=> \(\frac{-15}{2}.x=\frac{7}{3}\)
<=> \(x=\frac{7}{3}:\frac{-15}{2}=\frac{-14}{45}\)
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a) A = |x - 3| + 10
Vì |x - 3| >= 0
=> A = |x - 3| + 10 >= 10
A = 10 <=> |x - 3| = 0=> x - 3 = 0 => x = 3
Vậy: Amin = 10 <=> x = 3
b) B = -7 + (x - 1)2
Vì (x - 1)2 >= 0
=> B = -7 + (x - 1)2 >= -7
B = -7 <=> (x - 1)2 = 0 => x - 1 = 0 => x = 1
Vậy: Bmin = -7 <=> x = 1