\(C=x^3+y^3+3xy\) biet x+y \(=\)2

B\(=x^2+y^2-2xy\)

ĐKXĐ: x > 0

Ta có: \(M=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\Leftrightarrow\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\frac{9}{2}\Rightarrow\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{9}{2}\)

\(\Rightarrow2\left(x+2\sqrt{x}+1\right)=9\sqrt{x}\)

\(\Rightarrow2x-5\sqrt{x}+2=0\)

\(\Rightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=4\left(n\right)\\x=\frac{1}{4}\left(n\right)\end{cases}}}\)

                                                                Vậy x = {1/4 ; 4}

900 cm2

ai tích mình mình tích lại

3x\(\left(x-a\right)+5a\left(a-x\right)\)

2x \((x+1)+(x^2+2x+1)\)

Ta phải có : \(\hept{\begin{cases}\sqrt{x-2}>0\\\sqrt{6-x}>0\end{cases}\Leftrightarrow2< x< 6}\)