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\(8^4.2^3.16^2=\left(2^3\right)^4.2^3.\left(2^4\right)^2\)
\(=2^{12}.2^3.2^8\)
\(=2^{12+3+8}\)
\(=2^{23}\)
Chúc bạn học tốt !
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\(5x+x=39-3^{11}:3^9\)
\(x\left(5+1\right)=39-3^2\)
\(6x=39-9\)
\(6x=30\)
\(x=30:6\)
\(x=5\)
Vậy x = 5
Nhớ k cho mình nhé! Thank you!!!
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\(4^{2x-3+7}.4^3=4^{3.23}\)
\(4^{2x-10}.4^3=4^{69}\)
\(4^{2x-10}=4^{69}:4^3\)
\(4^{2x-10}=4^{66}\)
\(\Rightarrow\)\(2x-10=66\)
\(2x=66+10\)
\(2x=76\)
\(\Rightarrow x=76:2\Rightarrow x=38.\)
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\(8^4.16^5\)
\(=\left(2^3\right)^4.\left(2^4\right)^5\)
\(=2^{12}.2^{20}=2^{32}\)
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\(\left(3x-2\right)^3=2.32\)
\(\Leftrightarrow\left(3x-2\right)^3=64\)
\(\Leftrightarrow\left(3x-2\right)^3=4^{^3}\)
\(\Rightarrow3x-2=4\)
\(\Leftrightarrow3x=6\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
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à mik nhầm toán nhé
Ta có: \(\left(x+1\right)^2=\left(x+1\right)^5\)
\(\Leftrightarrow\left(x+1\right)^5-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left[\left(x+1\right)^3-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^2=0\\\left(x+1\right)^3-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^3=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x+1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)