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x2 - 3x - 2x2 + 2x = -x2 + 5x
x2 - 3x - 2x2 + 2x + x2 - 5x = 0
(x2 - 2x2 + x2) + (-3x + 2x - 5x) = 0
-6x = 0
x = 0
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Ta có:
\(5x+x=30\)
\(\Leftrightarrow x\left(5+1\right)=30\)
\(\Leftrightarrow6x=30\)
\(\Rightarrow x=5\)
P/s tham khảo nha
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a) x2 + 5x = 0
=> x.(x + 5) = 0
=> x = 0 hoặc x + 5 = 0
=> x = 0 hoặc x = -5
b) 1/21 + 1/28 + 1/36 + ... + 2/x(x + 1) = 2/9
=> 2/42 + 2/56 + 2/72 + ... + 2/x(x + 1) = 2/9
=> 2 × [1/6×7 + 1/7×8 + 1/8×9 + ... + 1/x(x + 1)] = 2/9
=> 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + ... + 1/x - 1/x + 1 = 2/9 : 2
=> 1/6 - 1/x + 1 = 2/9 × 1/2 = 1/9
=> 1/x + 1 = 1/6 - 1/9
=> 1/x + 1 = 3/18 - 2/18
=> 1/x + 1 = 1/18
=> x + 1 = 18
=> x = 17
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\(4x=29-9\Leftrightarrow4x=20\Leftrightarrow x=20:5\Leftrightarrow x=4\)
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a)\(x-15\%x=\frac{1}{3}\)
\(x.\left(1-15\%\right)=\frac{1}{3}\)
\(x.\frac{-280}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{-280}{3}\)
\(x=\frac{-1}{280}\)
Vậy \(x=\frac{-1}{280}\)
b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)
\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)
\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)
\(-\frac{17}{10}x=\frac{-61}{30}\)
\(x=\frac{-61}{30}:\frac{-17}{10}\)
\(x=\frac{61}{51}\)
Vậy \(x=\frac{61}{51}\)
\(5^{x+1}=125\)
\(\Rightarrow5^x.5=125\)
\(\Rightarrow5^x=125:5=25\)
\(\Rightarrow5^x=5^2\Leftrightarrow x=2\)
5^x+1=5^3
x+1=3=> x=2