Question

a) Chứng minh \(\left(a+b\right)^2=\left(a+b\right)^2+4ab\)

b) Cho \(a+b=9\), \(ab=20\)

Tính\(\left(a-b\right)^{2011}\)

Answer 1

a) mk chỉnh đề:

Chứng minh:  \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)   (1)

                hoặc   \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\) (2)

            BÀI LÀM

TH1:

\(VP=\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2=VP\)  (đpcm)

TH2:

\(VP=\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2=VT\)  (đpcm)

b)  \(a+b=9\)\(\Rightarrow\)\(a=9-b\)

Ta có:  \(ab=20\)\(\Rightarrow\)\(\left(9-b\right).b=20\)

\(\Leftrightarrow\)\(b^2-9b+20=0\)

\(\Leftrightarrow\)\(\left(b-4\right)\left(b-5\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}b=4\\b=5\end{cases}}\)

Nếu  \(b=4\)thì:  \(a=5\)\(\Rightarrow\)\(\left(a-b\right)^{2011}=\left(5-4\right)^{2011}=1\)

Nếu  \(b=5\)thì  \(a=4\)\(\Rightarrow\)\(\left(a-b\right)^{2011}=\left(4-5\right)^{2011}=-1\)

Answer 2

a, sửa đề CM: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

\(VP=\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2=VT\left(đpcm\right)\)

b, \(a+b=9\Leftrightarrow\left(a+b\right)^2=81\Leftrightarrow\left(a-b\right)^2+4ab=81\Leftrightarrow\left(a-b\right)^2=81-4.20=1\Leftrightarrow a-b=\pm1\)

Với \(a-b=1\Rightarrow\left(a-b\right)^{2011}=1\)

Với \(a-b=-1\Rightarrow\left(a-b\right)^{2011}=-1\)