Tìm x, biết:a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)b) \(\left[\left(-0,5\right)^3\right]^x=\dfrac{1}{64}\)c) \(2020^{\left(x-2\right).\left(2x+3\right)}=1\)d) \(\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}\) với \(x\in...
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Tìm x, biết:
a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
b) \(\left[\left(-0,5\right)^3\right]^x=\dfrac{1}{64}\)
c) \(2020^{\left(x-2\right).\left(2x+3\right)}=1\)
d) \(\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}\) với \(x\in Z\)
e) \(\dfrac{3}{4}\sqrt{x}-\dfrac{1}{2}=\dfrac{1}{3}\)
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)