Với a + b + c = 0 , ta có :

\(A=\frac{ab}{a^2+b^2-c^2}\)\(+\frac{bc}{b^2+c^2-a^2}\)\(+\frac{ca}{c^2+a^2-b^2}\)

\(\Leftrightarrow\frac{ab}{\left(a+b\right)^2-2ab-c^2}\)\(+\frac{bc}{\left(b+c\right)^2-2ab-a^2}\)\(+\frac{ca}{\left(c+a\right)^2-2ca-b^2}\)

\(\Leftrightarrow A=\frac{ab}{\left(a+b+c\right)\left(a+b-c\right)-2ab}\)\(+\frac{bc}{\left(b+c-a\right)\left(b+c+a\right)-2ab}\)\(+\frac{ac}{\left(a+c+b\right)\left(c+a-b\right)-2ca}\)

\(\Leftrightarrow A=\frac{ab}{-2ab}\)\(+\frac{bc}{-2bc}\)\(+\frac{ac}{-2ac}\)

\(\Leftrightarrow A=\frac{-1}{2}\)\(+\frac{-1}{2}\)\(+\frac{-1}{2}\)

\(\Leftrightarrow A=\frac{-3}{2}\)

Trả lời :

*Tự vẽ hình.

a, +) Do ABDE là hình vuông (gt) => AE = AB                

+) Do ACFH là hình vuông (gt) => AC = AH (tính chất)

+) \(\widehat{HAB}=\widehat{BAC}=90^o\)mà \(\widehat{HAB}+\widehat{BAC}=\widehat{BAH}\);\(\widehat{EAB}+\widehat{BAC}=\widehat{EAC}\)

=> \(\widehat{BAH}=\widehat{EAC}\)

Xét \(\Delta EAC\)và\(\Delta BAH\)có : AE = AB (cmt) ; AC = AH (cmt) ; \(\widehat{BAH}=\widehat{EAC}\)(cmt)

=> \(\Delta EAC\)=\(\Delta BAH\)

\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+2x\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow-2x+8=15\)

\(\Leftrightarrow-2x=7\)

\(\Leftrightarrow x=-\frac{7}{2}\)

x2 là gì vậy ạ??