Bài 5:
\(\dfrac{1}{b^2+\left(c-a\right)\left(c+a\right)}=\dfrac{1}{b^2+b\left(a-c\right)}=\dfrac{1}{b\left(a+b-c\right)}=\dfrac{1}{b\left(-c-c\right)}=\dfrac{1}{-2bc}\\ \Leftrightarrow A=\sum\dfrac{1}{b^2+c^2-a^2}=\sum\dfrac{1}{-2bc}=\dfrac{a+b+c}{-2abc}=0\)
Bài 6:
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\\ \Leftrightarrow\dfrac{a+b}{ab}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\\ \Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\\ \Leftrightarrow\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{ac+bc+ca}\right)=0\\ \Leftrightarrow\left(a+b\right)\cdot\dfrac{ab+bc+ca+ab}{ab\left(ac+bc+ca\right)}=0\\ \Leftrightarrow\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{ab\left(ab+bc+ca\right)}=0\\ \Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\\ \Leftrightarrow\left(2021-c\right)\left(2021-a\right)\left(2021-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2021\\b=2021\\c=2021\end{matrix}\right.\)
Vì vai trò của a,b,c trong P là như nhau nên \(P=\left(2021-2021\right)^{2021}\left(b-2021\right)^{2021}\left(c-2021\right)^{2021}=0\)
Bài 7:
\(x=\sqrt{2}+1-\sqrt{2}+1=2\\ \Leftrightarrow P=\left(5\cdot8-30\cdot2+21\right)^{2021}=1^{2021}=1\)
Bài 8:
\(xy+yz+zx=1\Leftrightarrow\left\{{}\begin{matrix}1+x^2=\left(x+y\right)\left(x+z\right)\\1+y^2=\left(y+x\right)\left(y+z\right)\\1+z^2=\left(z+x\right)\left(z+y\right)\end{matrix}\right.\\ \Leftrightarrow P=\sum x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=\sum x\sqrt{\dfrac{\left(y+z\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}\\ \Leftrightarrow P=\sum x\sqrt{\left(y+z\right)^2}=\sum x\left(y+z\right)=2\left(xy+yz+zx\right)=2\)
