Question

[Hải Yến]

Câu 1: Cho \(\frac{x}{x^2+x+1}\)=\(\frac{11}{133}\)

Tính A=\(\frac{x^2}{x^4+x^2+1}\)( 2 cách)

Câu 2: Cho x+y+z=4. Tính B=\(\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)

Câu 3: Cho G=\(\frac{a^2}{ab+b^2}+\frac{b^2}{ab-a^2}+\frac{-\left(a^2+b^2\right)}{ab}\)

a) Rút gọn G

b) Tính G khi \(\frac{a}{b}=\frac{a+1}{b+5}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

Câu 3: 

a: \(G=\dfrac{a^2}{b\left(a+b\right)}-\dfrac{b^2}{a\left(a-b\right)}+\dfrac{-\left(a^2+b^2\right)}{ab}\)

\(=\dfrac{a^3\left(a-b\right)-b^3\left(a+b\right)-\left(a^2+b^2\right)\left(a^2-b^2\right)}{ab\left(a-b\right)\left(a+b\right)}\)

\(=\dfrac{a^4-a^3b-ab^3-b^4-a^4+b^4}{ab\left(a-b\right)\left(a+b\right)}\)

\(=\dfrac{-ab\left(a^2+b^2\right)}{ab\left(a-b\right)\left(a+b\right)}=\dfrac{-a^2-b^2}{a^2-b^2}\)

b: \(\dfrac{a}{b}=\dfrac{a+1}{b+5}\)

nên ab+5a=ab+b

=>5a=b

\(G=\dfrac{-a^2-\left(5a\right)^2}{a^2-\left(5a\right)^2}=\dfrac{-a^2-25a^2}{a^2-25a^2}=\dfrac{-26}{-24}=\dfrac{13}{12}\)