1)
a) \(2xy^2\left(x^2-2y\right)=2xy^2x^2-2xy^2\cdot2y=2x^3y^2-4xy^3\)
b) \(\left(x-3\right)\left(x+3\right)=x^2-3^2=x^2-9\)
c) \(\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
2)
a) \(2\left(x-4\right)-3\left(2x+7\right)=5\left(x-3\right)+12\) (1)
\(\Leftrightarrow2x-8-6x-21=5x-15+12\)
\(\Leftrightarrow2x-6x-5x=-15+12+8+21\)
\(\Leftrightarrow-9x=26\)
\(\Leftrightarrow x=-\dfrac{26}{9}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{-\dfrac{26}{9}\right\}\)
b) \(x\left(x+2\right)-x=2\) (2)
\(\Leftrightarrow x^2+2x-x=2\)
\(\Leftrightarrow x^2+x=2\)
\(\Leftrightarrow x^2+x-2=2-2\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{-2;1\right\}\)
3)
\(2^m+2^n=2048\)
\(\Leftrightarrow2^m+2^n=2^8\)
\(\Leftrightarrow2^n\left(2^{m-n}-1\right)=2^8\)
Nếu:
♦ m - n = 0 (vô lý)
♦ m - n > 0:
\(\Rightarrow2^{m-n}-1\) lẻ mà \(2^8\) chẵn suy ra \(2^{m-n}-1=1\Rightarrow m=n+1\)
\(\Rightarrow2^n=2^8\Rightarrow n=8;m=9\)
Vậy \(n=8;m=9\)
