a) Ta có:\(VT=a^2-2ab+b^2\)
\(=a^2+2ab+b^2-4ab=\left(a+b\right)^2-4ab=VP\)
Vậy ...
b) \(a+b=7\Rightarrow\left(a+b\right)^2=7^2=49\)
\(\Leftrightarrow a^2+2ab+b^2=49\)
\(\Leftrightarrow a^2+24+b^2=49\)
\(\Leftrightarrow a^2+b^2=25\)
\(\Leftrightarrow\left(a-b\right)^2+2ab=25\)
\(\Leftrightarrow\left(a-b\right)^2=25-24=1\)
\(\Leftrightarrow\left[{}\begin{matrix}a-b=1\\a-b=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(a-b\right)^{2017}=1\\\left(a-b\right)^{2017}=-1\end{matrix}\right.\)
Vậy (a - b)2017 = 1 hoặc (a - b)2017 = -1.
Câu 1:
Ta có:
\(VP=\left(a+b\right)^2-4ab=a^2+2ab+2b^2-4ab\)
\(=a^2-2ab+b^2=\left(a-b\right)^2=VT\)
Vậy \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)(đpcm)
Chúc bạn học tốt!!!
Sửa đề : Tính \(\left(a-b\right)^2\) biết \(a+b=7\) và \(ab=12\)
Ta có :\(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
\(\Rightarrow\left(a-b\right)^2=7^2-4.12\)
\(=1\)