a)VT=\(\left(a+b\right)^2=a^2+2ab+b^2\)(1)VP=\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab\)(2)
từ (1) và (2)\(\Rightarrow\)VT=VP.Vậy \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\left(đpcm\right)\)
a) Ta có \(VP=\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2=VT\)
\(\Rightarrow\)đpcm
b) Ta có \(VP=\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab\)
\(=a^2-2ab+b^2=\left(a-b\right)^2=VT\)
\(\Rightarrow\)đpcm
a, Ta có:
\(\left(a-b\right)^2+4ab\)
\(=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2=VT\)
=>đpcm
b, ta có:
\(Vp=\left(a+b\right)^2-4ab\)
\(=a^2+2ab+b^2-4ab\)
\(=a^2-2ab+b^2=\left(a-b\right)^2=VT\)
=>đpcm
a) \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
VP: \(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2\)
b) \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
VP: \(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2\)
a) Ta có VP=(a−b)2+4ab=a2−2ab+b2+4ab
=a2+2ab+b2=(a+b)2=VT
b) Ta có VP=(a+b)2−4ab=a2+2ab+b2−4ab
=a2−2ab+b2=(a−b)2=VT
