a) ĐKXĐ: \(\hept{\begin{cases}x-9\ne0\\\sqrt{x}\ge0\\\sqrt{x}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne9\\x\ge0\\x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne9\\x>0\end{cases}}}\)

\(A=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(\Leftrightarrow A=\left(\frac{x+3}{x-9}+\frac{\sqrt{x}-3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(\Leftrightarrow A=\frac{x+\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)

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

b) \(x=\sqrt{6+4\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)

\(\Leftrightarrow x=\sqrt{4+4\sqrt{2}+2}-\sqrt{2+2\sqrt{2}+1}\)

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

\(\Leftrightarrow x=\left|2+\sqrt{2}\right|-\left|\sqrt{2}+1\right|\)

\(\Leftrightarrow x=2+\sqrt{2}-\sqrt{2}-1=1\left(TM\right)\)

Vậy với x= 1 thì giá trị của biểu thức \(A=\frac{\left(1+1\right)\left(1-3\right)}{1-9}=\frac{2.\left(-2\right)}{-8}=\frac{-4}{-8}=\frac{1}{2}\)

c)

Ta có :

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

+)  \(\frac{1}{A}\)nguyên 

\(\Leftrightarrow1+\frac{2}{\sqrt{x}+1}\)nguyên

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)\)

\(\Leftrightarrow x=1\)

Vậy ..............

ĐKXĐ: x \(\ge\)0; x \(\ne\)1

a) P = \(\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

P = \(\left(\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{5}{x+2\sqrt{x}-\sqrt{x}-2}\right):\frac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

P = \(\frac{2\sqrt{x}+4-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\)

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

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

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

<=> \(2x+\sqrt{x}-\sqrt{x}-1=0\)

<=> \(x=\frac{1}{2}\)(tm)

c)Với đk: x \(\ge\)0 và x \(\ne\)1

 \(x-2\sqrt{x-1}=0\) (đk: \(x\ge1\))

<=> \(x-1-2\sqrt{x-1}+1=0\)

<=> \(\left(\sqrt{x-1}-1\right)^2=0\)

<=> \(\sqrt{x-1}-1=0\)

<=> \(\sqrt{x-1}=1\)

<=> \(\left(\sqrt{x-1}\right)^2=1\)

<=> \(\left|x-1\right|=1\)

<=> \(\orbr{\begin{cases}x=0\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)

Với x = 2 => P = \(\frac{2\sqrt{2}+1}{\sqrt{2}+1}=\frac{\left(2\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{4-2\sqrt{2}+\sqrt{2}-1}{2-1}=3-\sqrt{2}\)

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

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

<=> \(2x-2\sqrt{x}-1=0\)<=> \(2\left(x-\sqrt{x}+\frac{1}{4}\right)-\frac{3}{4}=0\)

<=>  \(2\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{4}\) <=> \(\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{8}\)....(tiếp tự lm)

a) \(ĐKXĐ:\hept{\begin{cases}a>0\\b>0\\a\ne b\end{cases}}\)

\(A=\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)\)

\(\Leftrightarrow A=\frac{a+\sqrt{ab}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\frac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}-\frac{b}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{a+b}{\sqrt{ab}}\right)\)

\(\Leftrightarrow A=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\frac{a\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-b\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right)\left(a-b\right)}{\sqrt{ab}\left(a-b\right)}\)

\(\Leftrightarrow A=\left(\sqrt{a}-\sqrt{b}\right)\cdot\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{a^2-a\sqrt{ab}-b\sqrt{ab}-b^2-a^2+b^2}\)

\(\Leftrightarrow A=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{-a\sqrt{ab}-b\sqrt{ab}}\)

\(\Leftrightarrow A=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{-\sqrt{ab}\left(a+b\right)}\)

\(\Leftrightarrow A=\frac{-\sqrt{a}-\sqrt{b}}{a+b}\)

b) Thay \(a=6-2\sqrt{5}\)và \(b=5\)vào A ta được :

\(A=\frac{-\sqrt{6-2\sqrt{5}}-\sqrt{5}}{6-2\sqrt{5}+5}=\frac{-\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}}{1-2\sqrt{5}}=\frac{1-2\sqrt{5}}{1-2\sqrt{5}}=1\)

Vậy ...

Bài 2 :

b) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\) (1)

ĐKXĐ : \(x\ge1\)

Pt(1) tương đương :

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

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

\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\) (*)

Xét \(x\ge2\Rightarrow\sqrt{x-1}-1\ge0\)

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

Khi đó pt (*) trở thành :

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

\(\Leftrightarrow2\sqrt{x-1}=2\)

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\) ( Thỏa mãn )

Xét \(1\le x< 2\) thì \(x\ge2\Rightarrow\sqrt{x-1}-1< 0\)

Nên : \(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\). Khi đó pt (*) trở thành :

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

\(\Leftrightarrow2=2\) ( Luôn đúng )

Vậy tập nghiệm của phương trình đã cho là \(S=\left\{x|1\le x\le2\right\}\)

Bài 1 : 

a) ĐKXĐ : \(-1\le a\le1\)

Ta có : \(Q=\left(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}\right)\)

\(=\left(\frac{3+\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right)\cdot\frac{\sqrt{1-a^2}}{3}\)

\(=\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{3}\)

\(=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\)

Vậy \(Q=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\) với \(-1\le a\le1\)

b) Với \(a=\frac{\sqrt{3}}{2}\) thỏa mãn ĐKXĐ \(-1\le a\le1\)nên ta có :

\(\hept{\begin{cases}1-a=1-\frac{\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{\left(\sqrt{3}-1\right)^2}{2^2}\\1-a^2=1-\frac{3}{4}=\frac{1}{4}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\sqrt{1-a}=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2^2}}=\left|\frac{\sqrt{3}-1}{2}\right|=\frac{\sqrt{3}-1}{2}\\\sqrt{1-a^2}=\frac{1}{2}\end{cases}}\)

Do đó : \(Q=\frac{\left(3+\frac{1}{2}\right)\cdot\frac{\sqrt{3}-1}{2}}{3}=\frac{5\sqrt{3}-5}{12}\)

a)

Ta có:\(\left|P\right|>P\Leftrightarrow P< 0\) tức là \(\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\) 

b)

\(\left|P\right|=P\Leftrightarrow P\ge0\Leftrightarrow\sqrt{x}-2>0\Leftrightarrow x>4\)

Ta luôn có 1 tính chất rất quan trọng sau:     \(|P|\ge P\forall P\inℝ\)        (*)

DẤU "=" XẢY RA <=>    \(P\ge0\)

** Mà để    \(|P|>P\Rightarrow P< 0\)      (KO CÓ DẤU "=" XẢY RA)

=>    \(\frac{3\sqrt{x}}{\sqrt{x}-2}< 0\)

Mà do     \(x>0\Rightarrow3\sqrt{x}>0\)

=>   \(\sqrt{x}-2< 0\)

=>     \(0< x< 4\)     VÀ        \(x\ne1\)   (ĐKXĐ)

** ĐỂ      \(|P|=P\)

=> DẤU "=" CỦA BĐT (*) PHẢI XẢY RA

<=>    \(P\ge0\)      DO:    \(x>0\Rightarrow P>0\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-2}>0\)

MÀ:     \(\sqrt{x}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)

Vậy x > 4 thì     \(|P|=P\)

\(A=\left(x^3+y^3+xy\left(x+y\right)\right)-xy\left(x+y\right)+xy\)

=>    \(A=\left(x+y\right)\left(x^2+y^2\right)-xy.1+xy\)

=>   \(A=x^2+y^2-xy+xy\)

=>    \(A=x^2+y^2\ge\frac{\left(x+y\right)^2}{2}=\frac{1^2}{2}=\frac{1}{2}\)

DẤU "=" XẢY RA <=>    \(x=y\). MÀ    \(x+y=1\)

=> A min \(=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\).

\(B=x^2-2x+1+x^2-6x+9\)

=>   \(B=2x^2-8x+10\)

=>   \(B=2\left(x^2-4x+4\right)+2\)

=>   \(B=2\left(x-2\right)^2+2\)

CÓ:    \(2\left(x-2\right)^2\ge0\forall x\Rightarrow2\left(x-2\right)^2+2\ge2\)

=>   \(B\ge2\)

DẤU "=" XẢY RA <=>    \(2\left(x-2\right)^2=0\Leftrightarrow x=2\)

VẬY B MIN = 2 <=>    \(x=2\)

Mình nghĩ đề là tìm min chứ?

Ta có: \(\Delta=m^2+2m+49=\left(m+1\right)^2+48>0\left(\forall m\right)\) (*)

Từ (*) ta thấy phương trình trên có hai nghiệm phân biệt nên ta có thể giả sử:

\(\hept{\begin{cases}x_1=\frac{-\left(m-7\right)+\sqrt{m^2+2m+49}}{8}\\x_2=\frac{-\left(m-7\right)-\sqrt{m^2+2m+49}}{8}\end{cases}}\)

\(\Rightarrow F=\left|x_1-x_2\right|=\left|\frac{1}{4}\sqrt{m^2+2m+49}\right|=\frac{1}{4}\sqrt{\left(m+1\right)^2+48}\ge\frac{1}{4}\cdot\sqrt{48}=\sqrt{3}\)

Dấu '=' xảy ra khi m=-1

Vậy \(m=-1\)  thì F đạt giá trị nhỏ nhất tại \(\sqrt{3}\)

Mình nghĩ nên bổ sung x nguyên

\(A=\frac{3\sqrt{x}-2}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)-5}{\sqrt{x}+1}=3-\frac{5}{\sqrt{x}+1}\)

Để A nguyên thì \(\frac{5}{\sqrt{x}+1}\in Z\Leftrightarrow\sqrt{x}+1\in\left\{1;5\right\}\Leftrightarrow x=0\)

Thay x = 0 vào thì ta có A âm vậy ............