hi

giúp mink vs help me

a) \(\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)

\(=\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}.\frac{x^3+y^3}{10x-10y}\)

\(=\frac{3\left(x^2-2xy+y^2\right)}{5\left(x^2-xy+y^2\right)}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{10\left(x-y\right)}\)

\(=\frac{3\left(x^2-2xy+y^2\right)}{5}.\frac{x+y}{10\left(x-y\right)}\)

\(=\frac{3\left(x-y\right)^2}{5}.\frac{x+y}{10\left(x-y\right)}\)

\(=\frac{3\left(x-y\right)}{5}.\frac{x+y}{10}\)

\(=\frac{3x^2-3y^2}{50}\)

c) \(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)-\frac{x^2-y^2}{\left(x-y\right)^2}\)

\(=\frac{2}{xy}:\frac{y-x}{xy}-\frac{\left(x+y\right)\left(x-y\right)}{\left(x-y\right)^2}\)

\(=\frac{2}{y-x}-\frac{x+y}{x-y}\)

\(=\frac{2}{y-x}+\frac{x+y}{y-x}\)

\(=\frac{x+y+2}{y-x}\)

Ta có

\(M=a+\frac{2a+b}{2-b}+\frac{2a-b}{2+b}+\frac{4a}{b^2-4}\)

\(=a-\frac{2a+b}{b-2}+\frac{2a-b}{2+b}+\frac{4a}{\left(b-2\right)\left(b+2\right)}\)

\(=\frac{a\left(b-2\right)\left(2+b\right)-\left(2a+b\right)\left(2+b\right)+\left(2a-b\right)\left(b-2\right)+4a}{\left(b-2\right)\left(2+b\right)}\)

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

\(=\frac{ab^2-8a-b^2}{\left(b-c\right)\left(b+2\right)}\)

Với \(b=\frac{a}{a+1}\)ta có

\(=\frac{a\cdot\frac{a^2}{a^2+2a+1}-8a-\frac{a^2}{a^2+2a+1}}{\left(\frac{a}{a+1}-2\right)\left(\frac{a}{a+1}+2\right)}\)

\(\frac{a\cdot\frac{a^2}{a^2+2a+1}-8a-\frac{a^2}{a^2+2a+1}}{\left(\frac{-a-1}{a+1}\right)\left(\frac{3a+1}{a+1}\right)}\)

\(=\frac{a\cdot\frac{a^2}{a^2+2a+1}-8a-\frac{a^2}{a^2+2a+1}}{\frac{1-3a}{a+1}}\)

\(=\frac{a\left(\frac{a^2}{a^2+2a+1}-8-\frac{a}{a^2+2a+1}\right)}{\frac{1-3a}{a+1}}\)

\(=\frac{a\left(\frac{-7a^2+15a+8}{a^2+2a+1}\right)}{\frac{1-3a}{a+1}}\)

tới đây tịt rồi ai làm tiếp đc k

Gọi O là giao điểm của AC và BD 

\(S_{\Delta ABC}=\frac{1}{2}AC.BO\)

\(S_{\Delta ADC=\frac{1}{2}AC.DO}\)

\(S_{\Delta ABC}+S_{\Delta ADC}=\frac{1}{2}AC.BO+\frac{1}{2}AC.BO\)

\(S_{\Delta BCD=\frac{1}{2}AC\left(BO+DO\right)}\)

\(=\frac{1}{2}AC.BD=\frac{1}{6}.6.3,6=10,8cm^2\)

Có: a³+b³=(a+b)(a²-ab+b²)

Mà a²-ab+b²=a²-ab+b²/4+3b²/4

              =(a-b/2)²+3b²/4 \(\ge0\forall x,y\)

và a+b>0

=> (a+b)(a²-ab+b²) \(\ge0\forall x,y\)

=> a³+b³ \(\ge0\forall x,y\)

thank you