Đọc không ra. Học cách viết đề đi b

câu 2

<=> \(x^2+y^2-2xy=\left(x-y\right)^2\ge0\)

Mình ko rõ đề bài 

\(y=\frac{3x}{2}+\frac{1}{x}+1\)hay \(y=\frac{3x}{2}+\frac{1}{x+1}\)

ĐK: \(0\le x\le1\)

\(VT=\sqrt{x\left(x+1\right)}+\sqrt{x\left(1-x\right)}\le\frac{x+x+1+x+1-x}{2}=\frac{2x+2}{2}=x+1\)

Dấu "=" ko xảy ra 

tiếng anh mà như toán vậy

\(x^2+x\sqrt{2}+1>0\)

\(\Leftrightarrow\left(x+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}>0\)

\(\Leftrightarrow\left(x+\frac{1}{\sqrt{2}}\right)^2>-\frac{1}{2}\)

=> đpcm

\(x^2+x\sqrt{2}+1=x^2+2.x.\frac{\sqrt{2}}{2}+\left(\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}=x^2+2.x.\frac{\sqrt{2}}{2}+\frac{1}{2}+\frac{1}{2}\)

                                                              \(=\left(x+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}\)

Vì \(\left(x+\frac{\sqrt{2}}{2}\right)^2\ge0\left(\forall x\right)\)

Suy ra: \(\left(x+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}>0\)

Vậy \(x^2+x\sqrt{2}+1>0\)

\(y=\frac{3x}{2}+\frac{1}{x+1}=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\)

\(\Rightarrow y\ge2\sqrt{\frac{3\left(x+1\right)}{2}.\frac{1}{x+1}}-\frac{3}{2}=\sqrt{6}-\frac{3}{2}\)

Dấu "=" khi \(\left(x+1\right)^2=\frac{2}{3}\Rightarrow x=\frac{\sqrt{6}}{3}-1\)

\(y=\frac{x}{3}+\frac{5}{2x-1}=\frac{2x}{6}+\frac{5}{2x-1}=\frac{2x-1}{6}+\frac{5}{2x-1}+\frac{1}{6}\)

\(\Rightarrow y\ge2\sqrt{\frac{2x-1}{6}.\frac{5}{2x-1}}+\frac{1}{6}=\frac{\sqrt{30}}{3}+\frac{1}{6}\)

\(\Rightarrow P_{min}=\frac{\sqrt{30}}{3}+\frac{1}{6}\)

Dấu "=" xảy ra khi \(\left(2x-1\right)^2=30\Rightarrow x=\frac{\sqrt{30}+1}{2}\)