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a ) Ta có : 4(x - 5) - 3(x + 7) = -19
<=> 4x - 20 - 3x - 21 = -19
=> x - 41 = -19
=> x = -19 + 41
=> x = 22
b) Ta có " 7(x - 3) - 5(3 - x) = 11x - 5
<=> 7x - 21 - 15 + 5x = 11x - 5
<=> 12x - 36 = 11x - 5
=> 12x - 11x = -5 + 36
=> x = 31
a) \(2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Leftrightarrow2x-10-3x-21=14\)
\(\Leftrightarrow-x-31=14\)
\(\Leftrightarrow-x=45\Leftrightarrow x=-45\)
b) \(5\left(x-6\right)-2\left(x+3\right)=12\)
\(\Leftrightarrow5x-30-2x-6=12\)
\(\Leftrightarrow3x-36=12\)
\(\Leftrightarrow3x=48\Leftrightarrow x=16\)
c) \(3\left(x-4\right)-\left(8-x\right)=12\)
\(\Leftrightarrow3x-12-8+x=12\)
\(\Leftrightarrow4x-20=12\)
\(\Leftrightarrow4x=32\Leftrightarrow x=8\)
d) \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)
\(\Leftrightarrow-21x+35+14x-28=28\)
\(\Leftrightarrow-7x+35=0\Leftrightarrow x=5\)
1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna
a) 8^2x+1 = 2^27 : 2^12 = 2^5
2^6x.3 = 2^5
Suy ra : 6x . 3 = 5
6x = 5:3=5/3
x = 5/3 : 6 = 5/18
(3x - 7)8 = (3x - 7)6
=> (3x - 7)8 - (3x - 7)6 = 0
=> (3x - 7)6.[(3x - 7)2 - 1) = 0
=> \(\orbr{\begin{cases}\left(3x-7\right)^6=0\\\left(3x-7\right)^2-1=0\end{cases}}\)=> \(\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)=> \(\orbr{\begin{cases}3x=7\\3x-7\in\left\{1;-1\right\}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{7}{3}\\3x\in\left\{8;6\right\}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x\in\left\{\frac{8}{3};2\right\}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{3};\frac{8}{3};2\right\}\)
\(\left(3x-7\right)^8=\left(3x-7\right)^6\)
TH1\(\left(3x-7\right)^2=\left(3x-7\right)^6:\left(3x-7\right)^6\)
\(\left(3x-7\right)^2=1\)
\(\Rightarrow3x-7=1\)
\(3x=8\)
\(x=\frac{8}{3}\)
TH2 \(3x-7=0\)
\(3x=7\)
\(x=\frac{7}{3}\)
Vậy \(x=\frac{8}{3};x=\frac{7}{3}\)
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