1/ \(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x^2-y^2\right)-4y^2+10\)

\(=x^2-2xy+y^2+x^2+2xy+y^2-2x^2+2y^2-4y^2+10\)

\(=10\)

2/ \(5a^2+b^2=6ab\Leftrightarrow\left(5a^2-5ab\right)+\left(b^2-ab\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(5a-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\5a=b\end{cases}}\)

Với a = b thì

\(M=\frac{a-b}{a+b}=\frac{a-a}{a+a}=0\)

Với 5a = b thì

\(M=\frac{a-b}{a+b}=\frac{a-5a}{a+5a}=\frac{-4}{6}=\frac{-2}{3}\)

1.(x-y)²+(x+y)²-2(x²-y²)-4y²+10

=x²-2xy+y²+x²+2xy+y²-2x²+2y²-4y²+10

=x²+x-2x²-2xy+2xy+y²+y²+2y²-4y²+10

=10

=>dpcm

2.Ta co : 5a²+b²=6ab

5a²+b²-6ab=0

5a²+b²-5ab-ab=0

5a²-5ab+b²-ab=0

5a(a-b)+b(b-a)=0

5a(a-b)-b(a-b)=0

(a-b)(5a-b)=0

Ta lai co : a-b=0 \(\Rightarrow\)a=b

Va : 5a-b=0 \(\Rightarrow\)5a=b

Thay : a=b vao M

\(\Rightarrow M=\frac{a-b}{a+b}=\frac{b-b}{b+b}=\frac{0}{2b}=0\)

Thay : 5a=b vao M

\(\Rightarrow M=\frac{a-b}{a+b}=\frac{a-5a}{a+5a}=-\frac{4a}{6a}=-\frac{4}{6}=-\frac{2}{3}\)

435 đường thẳng

435 đường thẳng nha

\(\frac{a^2-2ab}{a^2b}.P=\frac{a^2b-4b^3}{3ab^2}\)

\(P=\frac{a^2b-4b^3}{3ab^2}:\frac{a^2-2ab}{a^2b}\)

\(P=\frac{a^2b-4b^3}{3ab^2}.\frac{a^2b}{a^2-2ab}\)

\(P=\frac{b\left(a^2-4b^2\right)}{3ab^2}.\frac{a^2b}{a\left(a-2b\right)}\)

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

\(P=\frac{b\left(a+2b\right)}{3b}.\frac{a}{a}\)

\(P=\frac{a+2b}{3}\)

P=\(\frac{a^2b.b\left(a^2-4b^2\right)}{3ab^2.a\left(a-2b\right)}=\frac{a^2b^2\left(a-2b\right)\left(a+2b\right)}{3a^2b^2\left(a-2b\right)}\)

=> P=\(\frac{a+2b}{3}\)

Xem lại cái đề nhé

Đáp số bằng 1. Chắc chắn đấy. Hong Pham thi vòng 9 gặp bài này phải ko ?