Question

[Vũ Văn Thắng]

Giải hộ mình 😘

Answer 1

[Phạm Thành Đông]

Câu 5:

Ta có:

\(\frac{1}{a^2+1}+\frac{1}{b^2+1}=\frac{2}{1+ab}\left(ĐKXĐ:a,b\in R\right)\)

\(\Leftrightarrow\frac{1}{a^2+1}+\frac{1}{b^2+1}-\frac{2}{1+ab}=0\)

\(\Leftrightarrow\left(\frac{1}{a^2+1}-\frac{1}{1+ab}\right)+\left(\frac{1}{b^2+1}-\frac{1}{1+ab}\right)=0\)

\(\Leftrightarrow\left[\frac{1+ab}{\left(a^2+1\right)\left(1+ab\right)}-\frac{a^2+1}{\left(a^2+1\right)\left(1+ab\right)}\right]\)\(+\left[\frac{1+ab}{\left(b^2+1\right)\left(1+ab\right)}-\frac{b^2+1}{\left(1+ab\right)\left(b^2+1\right)}\right]\)\(=0\)

\(\Leftrightarrow\frac{1+ab-a^2-1}{\left(a^2+1\right)\left(1+ab\right)}+\frac{1+ab-b^2-1}{\left(b^2+1\right)\left(1+ab\right)}=0\)

\(\Leftrightarrow\frac{ab-a^2}{\left(a^2+1\right)\left(1+ab\right)}+\frac{ab-b^2}{\left(b^2+1\right)\left(1+ab\right)}=0\)

\(\Leftrightarrow\frac{\left(ab-a^2\right)\left(b^2+1\right)}{\left(a^2+1\right)\left(1+ab\right)\left(b^2+1\right)}+\frac{\left(ab-b^2\right)\left(a^2+1\right)}{\left(b^2+1\right)\left(1+ab\right)\left(a^2+1\right)}=0\)

\(\Leftrightarrow\frac{\left(ab-a^2\right)\left(b^2+1\right)+\left(ab-b^2\right)\left(a^2+1\right)}{\left(a^2+1\right)\left(1+ab\right)\left(b^2+1\right)}=\frac{0}{\left(a^2+1\right)\left(1+ab\right)\left(b^2+1\right)}\)

\(\Rightarrow\left(ab-a^2\right)\left(b^2+1\right)+\left(ab-b^2\right)\left(a^2+1\right)=0\)

\(\Leftrightarrow ab^3+ab-a^2b^2-a^2+a^3b+ab-a^2b^2-b^2=0\)

\(\Leftrightarrow ab^3+a^3b-2a^2b^2+2ab-a^2-b^2=0\)

\(\Leftrightarrow\left(ab^3+a^3b-2a^2b^2\right)-\left(a^2-2ab+b^2\right)=0\)

\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a-b\right)^2=0\)

\(\Leftrightarrow ab\left(a-b\right)^2-\left(a-b\right)^2=0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}a-b=0\\ab-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=b\\ab=1\end{cases}}\Leftrightarrow a=b=\pm1\)

Lại có:

\(M=\frac{1}{a^{2021}+1}+\frac{1}{b^{2021}+1}\left(ĐK:a\ne-1;b\ne-1\right)\)

Mà ta có được \(a=b=\pm1\)nên thay \(a=b=1\)vào biểu thức M, ta được:

\(M=\frac{1}{1^{2021}+1}+\frac{1}{1^{2021}+1}=\frac{1}{1+1}+\frac{1}{1+1}=\frac{1}{2}+\frac{1}{2}=1\)

Vậy \(M=1\).