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a) (3x-1)3 = 27 = 33
=> 3x - 1 = 3
3x = 4
x = 4/3
b) 32x-1 = 27 = 33
=> 2x - 1 = 3
2x = 4
x = 2
c) 68-2x = 36 = 62
=> 8 -2x = 2
=> 2x = 6
x = 3
d) (3x-2)10 = (3x-2)7
=> (3x-2)10 - (3x-2)7 = 0
(3x-2)7. [(3x-2)3 - 1] = 0
=> (3x-2)7 = 0 => 3x-2 = 0 => 3x = 2 => x = 2/3
(3x-2)3 - 1 = 0 => (3x-2)3 = 1 => 3x - 2 = 1 => 3x = 3 => x = 1
KL:...
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\(d,\left(6x-2\right)^3=27\)
\(\Leftrightarrow\left(6x-2\right)^3=3^3\)
\(\Leftrightarrow6x-2=3\)
\(\Leftrightarrow6x=5\)
\(\Leftrightarrow x=\frac{5}{6}\)
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a) \(175-\left(-25\right)+62-\left(1200+62\right)\)
\(=\left(175+25\right)+\left(62-62\right)-1200\)
\(=200-1200\)
\(=-1000\)
b) \(\left(-25\right)\cdot8\cdot\left(-4\right)\cdot\left(-125\right)\cdot3\)
\(=3\left[\left(-25\right)\cdot\left(-4\right)\right]\left[\left(-125\right)\cdot8\right]\)
\(=3\cdot100\cdot\left(-1000\right)\)
\(=-300000\)
c) \(27\cdot\left(-17\right)+17\cdot\left(-73\right)\)
\(=27\cdot\left(-17\right)+73\cdot\left(-17\right)\)
\(=\left(-17\right)\cdot\left(27+73\right)\)
\(=\left(-17\right)\cdot100\)
\(=-1700\)
d) \(512\cdot\left(2-128\right)-128\cdot\left(-512\right)\)
\(=512\cdot\left(2-128\right)+128\cdot512\)
\(=512\left(2-128+128\right)\)
\(=512\cdot2\)
\(=1024\)
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a) \(\left(2x+3\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(2x+3\right)^2=\left(\frac{3}{5}\right)^2\) hoặc \(\left(2x+3\right)^2=\left(-\frac{3}{5}\right)^2\)
\(\Rightarrow2x+3=\frac{3}{5}\) hoặc \(2x+3=-\frac{3}{5}\)
\(\Rightarrow2x=-\frac{12}{5}\) hoặc \(2x=-\frac{18}{5}\)
\(\Rightarrow x=-\frac{6}{5}\) hoặc \(x=-\frac{9}{5}\)
Vậy.......
b) \(\left(3x-1\right)^3=\frac{1}{27}\)
\(\left(3x-1\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{1}{3}\)
\(\Rightarrow3x=\frac{4}{3}\)
\(\Rightarrow x=\frac{4}{9}\)
Vậy.......
a) (2x+3)2=\(\frac{9}{25}\)
2x+3=\(\sqrt{\frac{9}{25}}\)=\(\frac{3}{5}\)
2x=\(\frac{-12}{5}\)
x=\(\frac{-6}{5}\)
b) (3x-1)3=\(\frac{-1}{27}\)
3x-1=\(\frac{-1}{3}\)
3x=\(\frac{2}{3}\)
x=\(\frac{2}{9}\)
3x-1+|-27|=62.|3|
3x-1+27=62.3
3x+26=186
3x=186-26
3x=160
x=160:3
x=160/3
HT