ta thấy 1+x>= 2 căn x

=> 2 căn x/1+x bé hơn hoặc = 1

hok tốt

dấu = xảy ra khi x=-1

ĐKXĐ: x > 0

Áp dụng bđt Cô-si có \(x+1\ge2\sqrt{x}\)

                              \(\Rightarrow\frac{2\sqrt{x}}{1+x}\le1\)

Dấu "=" tại x = 1 (T/m ĐKXĐ)

a) Với x = 25 thì \(N=\frac{\sqrt{25}+1}{\sqrt{25}}=\frac{6}{5}\)

b) Ta có   \(M=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\)

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

Suy ra \(S=M.N=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

Do \(x=0\) không phải nghiệm

\(x^2+3x+1=0\Leftrightarrow x+3+\frac{1}{x}=0\Leftrightarrow x+\frac{1}{x}=-3\)

\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2=9\Rightarrow x^2+\frac{1}{x^2}=7\)

Đặt \(x_n=x^n+\frac{1}{x^n}\Rightarrow x_1=-3;x_2=7\)

\(x_1x_n=\left(x+\frac{1}{x}\right)\left(x^n+\frac{1}{x^n}\right)=x^{n+1}+\frac{1}{x^{n+1}}+x^{n-1}+\frac{1}{x^{n-1}}=x_{n+1}+x_{n-1}\)

\(\Rightarrow x_{n+1}=x_1x_n-x_{n-1}=-3x_n-x_{n-1}\)

Cho \(n=2\Rightarrow x_3=x^3+\frac{1}{x^3}=-3.x_2-x_1=-18\)

\(n=3\Rightarrow x_4=x^4+\frac{1}{x^4}=-3x_3-x_2=47\)

\(n=4\Rightarrow x_5=x^5+\frac{1}{x^5}=-3x_4-x_3=-123\)

\(n=5\Rightarrow x_6=x^6+\frac{1}{x^6}=-3x_5-x_4=322\)

Thay vào và tính, kết quả rất to

Ta có: 

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

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

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

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

\(P=\left(-\sqrt{x}\right)\left(\sqrt{x}-1\right)\)

\(P=\sqrt{x}-x\)

b) Để \(P>0\) thì \(\sqrt{x}-x>0\)

• \(\sqrt{x}-x>0\)

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

Suy ra: TH₁: \(\sqrt{x}< 0\) và \(1-\sqrt{x}< 0\) (Loại) vì \(\sqrt{x}\ge0\)

            TH₂:\(\sqrt{x}>0\)  và \(1-\sqrt{x}>0\) (Nhận)

Ta có \(\sqrt{x}>0\) và \(1-\sqrt{x}>0\) để \(P>0\)

• \(\sqrt{x}>0\) \(\Rightarrow x>0\)
• \(1-\sqrt{x}>0\) \(\Rightarrow\sqrt{x}< 1\) \(\Rightarrow x< 1\)

Vậy để \(P>0\) thì \(0< x< 1\)

c)\(P=\sqrt{x}-x\)

\(P=-\left(x-\sqrt{x}\right)\)

\(P=-\left(\left(\sqrt{x}\right)^2-2.\frac{1}{2}.\sqrt{x}+\frac{1}{4}-\frac{1}{4}\right)\)

\(P=-\left(\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\right)\)

\(P=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\)

Nên \(-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{x}-\frac{1}{2}=0\) \(\Rightarrow x=\frac{1}{4}\)

Vậy GTLN của \(P\) là \(\frac{1}{4}\) khi \(x=\frac{1}{4}\)

a. Ta có:\(\frac{x}{y}\sqrt{\frac{y^2}{x^4}=}\) \(\frac{x}{y}.\frac{\left|y\right|}{x^2}=\frac{x.y}{x^2y}\)\(=\frac{1}{x}\)(Vì \(x\ne0;y>0\))

b \(3x^2\sqrt{\frac{8}{x^2}}=3x^2\frac{2\sqrt{2}}{\left|x\right|}=\frac{6x^2\sqrt{2}}{-x}=-6x\sqrt{2}\)( Vì \(x< 0\))