\(2a^2+2b^2=5ab\)

<=>   \(2a^2+2b^2-5ab=0\)

<=>  \(2a^2-4ab-ab+2b^2=0\)

<=>   \(2a\left(a-2b\right)-b\left(a-2b\right)=0\)

<=>  \(\left(2a-b\right)\left(a-2b\right)=0\)

<=>  \(\orbr{\begin{cases}2a-b=0\left(L\right)\\a-2b=0\end{cases}}\)

=>  \(a=2b\)

=>  \(A=\frac{a+2b}{2a-b}=\frac{2b+2b}{2.2b-b}=\frac{4b}{3b}=\frac{4}{3}\)

\(S=\left(\frac{c}{a+b}+1\right)+\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)-3\)

\(=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}-3\)

\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)

\(=2001\cdot\frac{1}{10}-3=\frac{1971}{10}\)

ta có\(17x^2+10y^2-14xy+8x-24y+20=\)\(16+\left(8x-24y\right)+\left(x^2-6xy+9y^2\right)+\left(16x^2-8xy+y^2\right)+20-16=\)

\(4^2+8\left(x-3y\right)+\left(x-3y\right)^2+\left(4x-y\right)^2+4\)\(=\left(4+x-3y\right)^2+\left(4x-y\right)+4>0\)(luôn đúng)

bài này ở sbt vật lí 8

a)\(\frac{3xy+6}{6xy+12}=\frac{1}{2}\Leftrightarrow\left(3xy+6\right)\cdot2=\left(6xy+12\right)\cdot1\)

                                    \(\Leftrightarrow6xy+12=6xy+12\)

Vậy.......

b)\(\frac{x^2-xy}{5y^2-5xy}=\frac{x}{5y}\Leftrightarrow\left(x^2-xy\right)\cdot5y=\left(5y^2-5xy\right)\cdot x\)

                                          \(\Leftrightarrow5x^2y-5xy^2=5xy^2-5x^2y\)

Vậy.....