2,5 + 7,5. x = – 5
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\(\dfrac{1}{6}x-\dfrac{3}{8}=\dfrac{1}{4}\)
=>\(\dfrac{1}{6}x=\dfrac{1}{4}+\dfrac{3}{8}=\dfrac{5}{8}\)
=>\(x=\dfrac{5}{8}:\dfrac{1}{6}=\dfrac{5}{8}\cdot6=\dfrac{30}{8}=\dfrac{15}{4}\)
\(121212:\left(x-2020\right)=12\)
=>x-2020=121212:12=10101
=>x=10101+2020=12121
\(125\%\cdot\left(-\dfrac{1}{2}\right)^2\cdot\left(\dfrac{15}{16}+1.5\right)+2016^0\)
\(=\dfrac{5}{4}\cdot\dfrac{1}{4}\cdot\left(\dfrac{15}{16}+\dfrac{24}{16}\right)+1\)
\(=\dfrac{5}{16}\cdot\dfrac{39}{16}+1=\dfrac{195}{256}+1=\dfrac{451}{256}\)
ĐKXĐ: x<>-1
\(\dfrac{x+1}{2}=\dfrac{8}{x+1}\)
=>\(\left(x+1\right)^2=2\cdot8=16\)
=>\(\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)
a; \(\dfrac{x+1}{2}\) = \(\dfrac{8}{x+1}\) (\(x\ne\) -1)
(\(x\) + 1)2 = 8.2
(\(x\) + 1)2 = 16
\(\left[{}\begin{matrix}x+1=-4\\x+1=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
Vậy \(x\in\) {-5; 3}
\(M=1+2+2^2+2^3+...+2^{2022}+2^{2023}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{2022}+2^{2023}\right)\)
\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{2022}\left(1+2\right)\)
\(=3\left(1+2^2+...+2^{2022}\right)⋮3\)
\(\dfrac{2x+1}{2}=\dfrac{8+x}{x}\)
x (2x +1) = 2 (8 +x)
2x2 + x = 16 + 2x
2x2 -x - 16 = 0
=> \(x=\dfrac{\sqrt{129}+1}{4}\)
\(x=\dfrac{-\sqrt{129}+1}{4}\)
2,5+7,5x=-5
=>7,5x=-5-2,5=-7,5
=>x=-1
2,5 + 7,5\(x\) = -5
7,5\(x\) = -5 - 2,5
7,5\(x\) = -7,5
\(x\) = -7,5 : 7,5
\(x\) = -1
Vậy \(x=-1\)