a: \(=4a-4\sqrt{10a}-9\sqrt{10a}=4a-13\sqrt{10a}\)

b: \(=\sqrt{x}\left(4-\sqrt{2}\right)\cdot\sqrt{x}\left(1-\sqrt{2}\right)\)

\(=x\cdot\left(4-4\sqrt{2}-\sqrt{2}+2\right)\)

\(=\left(6-5\sqrt{2}\right)x\)

c: \(=\dfrac{2}{2x-1}\cdot x\sqrt{5}\cdot\left(2x-1\right)=2x\sqrt{5}\)

\(M=\left[\frac{\sqrt{x}\left(2\sqrt{x}+3\right)}{2x+2\sqrt{x}+3\sqrt{x}+3}+\frac{2}{\sqrt{x}+1}\right].\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\left[\frac{\sqrt{x}\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+3\right)}+\frac{2}{\sqrt{x}+1}\right].\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}+1}.\frac{\sqrt{x}+2018}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}+2018}{\sqrt{x}+1}\)

\(\frac{\sqrt{x}+2018}{\sqrt{x}+1}=1+\frac{2017}{\sqrt{x}+1}\le2018\)

Dấu "=" xảy ra \(\Leftrightarrow\)

... 

Giúp mình nhanh nhé, mai cô kt r

Ai bik ko trả lời với ạ

Bài 1 :

\(P=2x+y+\frac{30}{x}+\frac{5}{y}\)

\(=\frac{10x}{5}+\frac{5y}{5}+\frac{30}{x}+\frac{5}{y}\)

\(=\frac{6x}{5}+\frac{4x}{5}+\frac{y}{5}+\frac{4y}{5}+\frac{30}{x}+\frac{5}{y}\)

\(=\left(\frac{6x}{5}+\frac{30}{x}\right)+\left(\frac{4x}{5}+\frac{4y}{5}\right)+\left(\frac{y}{5}+\frac{5}{y}\right)\)

Áp dụng bất đẳng thức Cô - si cho 2 số không âm

\(\frac{6x}{5}+\frac{30}{x}\ge2\sqrt{\frac{6x}{5}.\frac{30}{x}}=2\sqrt{36}=2.6=12\left(1\right)\)

\(\frac{y}{5}+\frac{5}{y}\ge2\sqrt{\frac{y}{5}.\frac{5}{y}}=2\left(2\right)\)

Theo đề bài ta có : \(x+y\ge10\) suy ra

\(\frac{4x}{5}+\frac{4y}{5}=\frac{4\left(x+y\right)}{5}\ge\frac{4.10}{5}=8\left(3\right)\)

Cộng (1) ; (2) và (3) vế với vế ta được :
\(\frac{6x}{5}+\frac{30}{x}+\frac{y}{5}+\frac{5}{y}+\frac{4x}{5}+\frac{4y}{5}\ge12+2+8=22\)

Dấu " = " xay ra \(\Leftrightarrow\left\{{}\begin{matrix}\frac{6x}{5}=\frac{30}{x}\\\frac{y}{5}=\frac{5}{y}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=25\\y^2=25\end{matrix}\right.\)

Vì x ; y dương nên \(\left(x;y\right)=\left(5;5\right)\)

Bài 2 :

Đặt \(x=a+b=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)

\(\Leftrightarrow x^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

\(\Leftrightarrow x^3=2+\sqrt{5}+2-\sqrt{5}+\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}.x\)

\(\Leftrightarrow x^3=4+\sqrt[3]{4-5}.x\)

\(\Leftrightarrow x^3=4-3x\)

\(\Leftrightarrow x^3+3x-4=0\)

\(\Leftrightarrow x^3-x^2+x^2-x+4x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)+x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+4\right)=0\)

Vì \(x^2+x+4=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{15}{4}=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\left(\forall x\right)\)

Nên \(x-1=0\Leftrightarrow x=1\)

Vậy \(x=a+b=1\)

\(\Rightarrow\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1\left(đpcm\right)\)

Chúc bạn học tốt !!