`5)`
`@`\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{2}{ab}-\dfrac{2}{ab}+\dfrac{1}{\left(a+b\right)^2}}\)
\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2-\dfrac{2}{ab}+\dfrac{1}{\left(a+b\right)^2}}\)
\(=\sqrt{\left(\dfrac{a+b}{ab}\right)^2-2.\dfrac{a+b}{ab}.\dfrac{1}{a+b}+\dfrac{1}{\left(a+b\right)^2}}\)
\(=\sqrt{\left(\dfrac{a+b}{ab}-\dfrac{1}{a+b}\right)^2}\)
\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\right)^2}\)
\(=\left|\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\right|\)
`@`\(\sqrt{a^2+b^2+\dfrac{\left(ab\right)^2}{\left(a+b\right)^2}}=\sqrt{a^2+b^2+2ab-2ab+\dfrac{\left(ab\right)^2}{\left(a+b\right)^2}}\)
\(=\sqrt{\left(a+b\right)^2-2ab+\dfrac{\left(ab\right)^2}{\left(a+b\right)^2}}\)
\(=\sqrt{\left(a+b\right)^2-2.\dfrac{ab}{a+b}.\left(a+b\right)+\dfrac{\left(ab\right)^2}{\left(a+b\right)^2}}\)
\(=\sqrt{\left(a+b-\dfrac{ab}{a+b}\right)^2}\)
\(=\left|a+b-\dfrac{ab}{a+b}\right|\)
`6)`\(\left(\sqrt{x^2+2022}+x\right)\left(\sqrt{y^2+2022}+y\right)=2022\)
\(\Leftrightarrow\left(\sqrt{x^2+2022}+x\right)\left(\sqrt{y^2+2022}+y\right).\left(\sqrt{y^2+2022}-y\right)=2022\left(\sqrt{y^2+2022}-y\right)\)
\(\Leftrightarrow\left(\sqrt{x^2+2022}+x\right)\left(y^2+2022-y^2\right)=2022\left(\sqrt{y^2+2022}-y\right)\)
\(\Leftrightarrow\left(\sqrt{x^2+2022}+x\right)=\left(\sqrt{y^2+2022}-y\right)\)
\(\Leftrightarrow\left(x+y\right)+\sqrt{x^2+2022}-\sqrt{y^2+2022}=0\)
\(\Leftrightarrow\left(x+y\right)+\dfrac{\left[\left(x^2-2022^2\right)-\left(y^2-2022^2\right)\right]}{\sqrt{x^2+2022}+\sqrt{y^2+2022}}=0\)
\(\Leftrightarrow\left(x+y\right)+\dfrac{x^2-y^2}{\sqrt{x^2+2022}+\sqrt{y^2+2022}}=0\)
\(\Leftrightarrow\left(x+y\right)+\dfrac{\left(x-y\right)\left(x+y\right)}{\sqrt{x^2+2022}+\sqrt{y^2+2022}}=0\)
\(\Leftrightarrow\left(x+y\right)\left(1+\dfrac{x-y}{\sqrt{x^2+2022}+\sqrt{y^2+2022}}\right)=0\) (1)
Ta có: \(\left|x\right|+\left|y\right|>\left|x-y\right|\)
\(\sqrt{x^2+2022}+\sqrt{y^2+2022}>\left|x\right|+\left|y\right|\)
\(\Rightarrow\dfrac{x-y}{\sqrt{x^2+2022}+\sqrt{y^2+2022}}< 1\)
\(\left(1\right)\Rightarrow x+y=0\)
\(P=x^3+y^3\)
\(P=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(P=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)
\(P=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(P=0^2-3xy.0\)
\(P=0\)
