Bài 2: 

\(\Leftrightarrow3\sqrt{x+5}-2\sqrt{x+5}=7\)

\(\Leftrightarrow\sqrt{x+5}=7\)

=>x+5=25

hay x=18

a) ĐKXĐ: a\(\ge\)0, a\(\ne\)1

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

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

A=\(\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-1\right)}.\dfrac{\sqrt{a}+1}{\sqrt{a}}\)=\(\dfrac{2}{a-1}\)

b) Để A\(\in\)Z\(\Rightarrow\)x-1\(\in\) Ư(2)=\(\left\{-1,1,-2,2\right\}\)

vì x\(\ge\)0,x\(\ne\)1 nên x\(\in\)\(\left\{-1,0,2,3\right\}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b: Ta có: \(A=\dfrac{3x+2\sqrt{x}-5}{x+\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{1-\sqrt{x}}\)

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

\(=\dfrac{3x+4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}\)

Sửa đề: \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{3\sqrt{x}+1}{x-1}\)

a: ĐKXĐ: x>=0; x<>1

b: \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{3\sqrt{x}+1}{x-1}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2+\left(\sqrt{x}+1\right)^2-3\sqrt{x}-1}{x-1}\)

\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}\)

\(=\dfrac{2x-3\sqrt{x}+1}{x-1}=\dfrac{\left(\sqrt{x}-1\right)\cdot\left(2\sqrt{x}-1\right)}{x-1}\)

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

a) ĐKXĐ: \(x\ge0,x\ne1\)

b) \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{3\sqrt{x}+1}{\sqrt{x}-1}\)

\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1-3\sqrt{x}-1}{\sqrt{x}-1}\)

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

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

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

\(A=\dfrac{-x-4\sqrt{x}+1}{x-1}\)

Bạn xem lại đề giúp mình !

giải giúp mình với ạ

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$
\(A=\frac{5\sqrt{x}+3x}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{(3\sqrt{x}-1)(\sqrt{x}+3)}{(\sqrt{x}-1)(\sqrt{x}+3)}+\frac{7(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-1)}\)

\(=\frac{5\sqrt{x}+3x-(3x+8\sqrt{x}-3)+(7\sqrt{x}-7)}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{4(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+3)}=\frac{4}{\sqrt{x}+3}\)

Dễ thấy $A>0$

$\sqrt{x}+3\geq 3\Rightarrow A\leq \frac{4}{3}$

Vậy $0< A\leq \frac{4}{3}$. 

$A$ nguyên $\Leftrightarrow A=1\Leftrightarrow \frac{4}{\sqrt{x}+3}=1$

$\Leftrightarrow \sqrt{x}=1\Leftrightarrow x=1$ (trái đkxđ)

Vậy không tồn tại $x$ để $A$ nguyên.

a) ĐKXĐ: \(a>0;a\ne1\)

b) ta có:

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

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

\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)

\(=\sqrt{a}\left(\sqrt{a}-1\right)\)

c) ta có:

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

\(=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu "=" xảy ra khi : \(a=\frac{1}{4}\)

Vậy min P =-1/4 khi a=1/4

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a) ĐKXĐ: \(a\ne1;a\ne0\))

\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a+1}}{a}\)

    \(=\left(\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}.\left(\sqrt{a}+1\right)}\right):\frac{\sqrt{a+1}}{a}\)

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

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