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ĐKXĐ : \(\hept{\begin{cases}ab-2\ne0\\ab+2\ne0\\a^4b^4\ne0\end{cases}}\Rightarrow ab\ne\pm2;a\ne0;b\ne0\)
\(P=\left(\frac{1}{ab-2}+\frac{1}{ab+2}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\left(\frac{2ab}{a^2b^2-4}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\left(\frac{4a^3b^3}{a^4b^4-16}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\frac{8a^5b^5}{a^8b^8-16^2}.\frac{a^4b^4+16}{a^4b^4}=\frac{8a^5b^5\left(a^4b^4+16\right)}{\left(a^4b^4-16\right)\left(a^4b^4+16\right).a^4b^4}\)
\(=\frac{8ab}{a^4b^4-16}\)
b) Khi \(\frac{a^2+4}{b^2+9}=\frac{a^2}{9}\)
=> (a2 + 4).9 = a2(b2 + 9)
=> 9a2 + 36 = a2b2 + 9a2
=> a2b2 = 36
=> (ab)2 = 36
=> \(\orbr{\begin{cases}ab=6\left(tm\right)\\ab=-6\left(tm\right)\end{cases}}\)
Khi ab = 6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.6}{6^4-16}=\frac{48}{1280}=\frac{3}{80}\)
Khi ab = -6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.\left(-6\right)}{\left(-6\right)^4-16}=-\frac{3}{80}\)
a(a + 2) + b(b - 2) - 2ab
= a2 + 2a + b2 - 2b - 2ab
= (a2 - 2ab + b2) +(2a - 2b)
= (a - b)2 + 2(a - b)
= 72 + 2.7
= 49 + 14 =63
\(a\left(a+2\right)+b\left(b-2\right)-2ab=a^2+2a+b^2-2b-2ab\)
\(=\left(a^2-2ab+b^2\right)+\left(2a-2b\right)=\left(a-b\right)^2+2\left(a-b\right)\)
Với \(a-b=7\)thì biểu thức có giá trị là: \(7^2-7=49-7=42\)
\(a^2+2ab+b^2+4a+4b+2015\\ =\left(a+b\right)^2+4\left(a+b\right)+2015\\ =\left(a+b\right)\left(a+b+4\right)+2015\\ =1.\left(1+4\right)+2015\\ =5+2015\\ =2020\)
\(A=\left(a+b\right)^2+4\left(a+b\right)+2015=2020\)