\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)

=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

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

=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)

b/ Để R<-1   => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)

<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)

<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)

Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)   là sao vậy ạ?

Thì \(\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x}-3}=\frac{\sqrt{x}-3}{\sqrt{x}-3}=1\)

Điều kiện có 2 nghiệm phân biệt tự làm nha

Theo vi-et ta có:

\(\hept{\begin{cases}x_1+x_2=5\\x_1.x_2=m-2\end{cases}}\)

\(2\left(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\right)=3\)

\(\Leftrightarrow4\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{2}{\sqrt{x_1.x_2}}\right)=9\)

\(\Leftrightarrow4\left(\frac{5}{m-2}+\frac{2}{\sqrt{m-2}}\right)=9\)

Làm nốt nhé

Câu 1:

M=\(\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(4x^2-4x+1\right)+2014\)

=\(\left(\left(x+y\right)^2+2\left(x+y\right)+1\right)+\left(2x-1\right)^2+2014\)

=\(\left(x+y+1\right)^2+\left(2x-1\right)^2+2014\ge2014\)

\(\Rightarrow M\ge2014\Leftrightarrow minM=2014\)

\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\2x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,5\\y=1,5\end{cases}}\)

2/ \(S=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}=9\)

a,\(\frac{x}{\sqrt{x}+1}=\frac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\frac{1}{\sqrt{x}+1}\)

\(=\left(\sqrt{x}-1\right)+\frac{1}{\sqrt{x}-1}+2\ge2.\sqrt{\left(\sqrt{x}-1\right).\frac{1}{\sqrt{x}-1}+2}\ge4\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-1=\frac{1}{\sqrt{x}-1}\)

\(\Leftrightarrow\sqrt{x}-1=1\)

\(\Leftrightarrow\sqrt{x}=2\)

\(\Leftrightarrow x=4\left(t/m\right)\)

D_min = 4  <=> x=4

b,\(\frac{\sqrt{x-9}}{5x}\) 

\(\sqrt{x-9}=\sqrt{\frac{\left(x-9\right).9}{9}}=\frac{1}{3}.\sqrt{\left(x-9\right).9}\le\frac{1}{3}.\frac{x-9+9}{2}=\frac{x}{2}\)

\(\Rightarrow D\le\frac{x}{\frac{6}{5x}}=\frac{x}{30x}=\frac{1}{30}\)

Dấu "=" xảy ra \(\Leftrightarrow x-9=9\Leftrightarrow x=18\)

D_max=\(\frac{1}{30}\Leftrightarrow x=18\)

P/s : ko chắc lắm 

\(a)\)\(P=\frac{x}{\sqrt{x}+1}=\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}-\frac{2\sqrt{x}+2}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\)

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

\(P=\sqrt{x}+1+\frac{1}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\frac{1}{\sqrt{x}+1}}-2=2-2=0\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\sqrt{x}+1=\frac{1}{\sqrt{x}+1}\)\(\Leftrightarrow\)\(x=0\)

... 

ĐKXĐ : \(x\ne0\)

\(b)\)\(D=\frac{\sqrt{x-9}}{5x}\ge\frac{0}{5x}=0\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\sqrt{x-9}=0\)\(\Leftrightarrow\)\(x=9\)

...

Bạn xem lại ĐKĐB. Nếu $x\geq \frac{-1}{3}$ thì mình nghi ngờ $\sqrt{3x-1}$ của bạn viết là $\sqrt{3x+1}$Còn nếu đúng là $\sqrt{3x-1}$ thì ĐK cần là $x\geq \frac{1}{3}$.

ĐKXĐ: x \(\ge\)0; x \(\ne\)1 ; x \(\ne\)4

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

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

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

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

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

P = \(\frac{\sqrt{x}-1}{\sqrt{x}+2}\)

b) P < 0 <=> \(\frac{\sqrt{x}-1}{\sqrt{x}+2}< 0\)

Do \(\sqrt{x}+2>0\) => \(\sqrt{x}-1< 0\) => \(\sqrt{x}< 1\) => \(x< 1\)

kết hợp với đk => S = {x| \(0\le x< 1\)}

c) P = \(\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\ge-\frac{1}{2}\)

Do \(\sqrt{x}+2\ge2\) => \(-\frac{3}{\sqrt{x}+2}\ge-\frac{3}{2}\) => \(1-\frac{3}{\sqrt{x}+2}\ge-\frac{1}{2}\)

Dấu "=" xảy ra <=>  x = 0

Vậy MinP = -1/2 khi x = 0

Ta co:

\(M=\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{x+\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-2}=\frac{x}{\sqrt{x}-2}=8+\frac{\left(\sqrt{x}-4\right)^2}{\sqrt{x}-2}\ge8\)

Dau '=' xay ra khi \(x=16\)

Vay \(M_{min}=8\)khi \(x=16\)