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)

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

Để \(A\in Z\)thì \(\frac{1}{\sqrt{x}-3}\in Z\)

=> \(\sqrt{x}-3\inƯ_{\left(1\right)}\)

=>\(\sqrt{x}-3\in\left(1;-1\right)\)

=>\(\sqrt{x}\in\left(4;2\right)\)

=>\(x\in\left(-2;2\right)\)

Vậy...

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

Vì \(1\inℤ\)nên \(A\inℤ\)thì \(\frac{5}{\sqrt{x}-3}\inℤ\)

\(\Rightarrow\sqrt{x}-3\inƯ_{\left(5\right)}=\left(\pm1;\pm5\right)\)

Bảng 

VẬy với x=2;x=4;x=8 thì \(A\inℤ\)

với x\(\inℤ\)

\(A=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{45\sqrt{x}-11}{\left(\sqrt{x}+3\right)(\sqrt{x}-1)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{45\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{37\sqrt{x}-5x-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

A.2

......

Chúc học tốt

sao tôi lại thấy tên tôi nhỉ ?

Machi!Rồi bạn trong đội tuyển văn không?