a)     \(ĐKXĐ:x\ne-3;x\ne2\)

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

\(P=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

\(P=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

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

vậy \(P=\frac{x-4}{x-2}\)

\(P=\frac{-3}{4}\) \(\Leftrightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)

\(\Leftrightarrow4\left(x-4\right)=-3.\left(x-2\right)\)

\(\Leftrightarrow4x-16=-3x+6\)

\(\Leftrightarrow7x=22\)

\(\Leftrightarrow x=\frac{22}{7}\)

c) \(P\in Z\Leftrightarrow\frac{x-4}{x-2}\in Z\)

\(\frac{x-2-6}{x-2}=1-\frac{6}{x-2}\in Z\)

mà \(1\in Z\Rightarrow\left(x-2\right)\inƯ\left(6\right)\in\left(\pm1;\pm2;\pm3;\pm6\right)\)

mà theo ĐKXĐ:  \(\Rightarrow\in\left(\pm1;-2;3;\pm6\right)\)

thay mấy cái kia vào rồi tìm \(x\)

d) \(x^2-9=0\Rightarrow x^2=9\Rightarrow x=\pm3\)

khi \(x=3\Rightarrow P=\frac{3-4}{3-2}=-1\)

khi \(x=-3\Rightarrow P=\frac{-3-4}{-3-2}=\frac{-7}{-5}=\frac{7}{5}\)

AI GIẢI HỘ MÌNH K CHO Ạ!!!

1)  a) Căn thức có nghĩa \(\Leftrightarrow4-2x\ge0\Leftrightarrow2x\le4\Leftrightarrow x\le2\)

b) Thay x = 2 vào biểu thức A, ta được: \(A=\sqrt{4-2.2}=\sqrt{0}=0\)

Thay x = 0 vào biểu thức A, ta được: \(A=\sqrt{4-2.0}=\sqrt{4}=2\)

Thay x = 1 vào biểu thức A, ta được: \(A=\sqrt{4-2.1}=\sqrt{2}\)

Thay x = -6 vào biểu thức A, ta được: \(A=\sqrt{4-2.\left(-6\right)}=\sqrt{16}=4\)

Thay x = -10 vào biểu thức A, ta được: \(A=\sqrt{4-2.\left(-10\right)}=\sqrt{24}=2\sqrt{6}\)

c) \(A=0\Leftrightarrow\sqrt{4-2x}=0\Leftrightarrow4-2x=0\Leftrightarrow x=2\)

\(A=5\Leftrightarrow\sqrt{4-2x}=5\Leftrightarrow4-2x=25\Leftrightarrow x=\frac{-21}{2}\)

\(A=10\Leftrightarrow\sqrt{4-2x}=10\Leftrightarrow4-2x=100\Leftrightarrow x=-48\)

2) a) P xác định \(\Leftrightarrow x\ge0\)và \(2\sqrt{x}-3\ne0\Leftrightarrow\sqrt{x}\ne\frac{3}{2}\Leftrightarrow x\ne\frac{9}{4}\)

b) Thay x = 4 vào P, ta được: \(P=\frac{9}{2\sqrt{4}-3}=\frac{9}{1}=9\)

Thay x = 100 vào P, ta được: \(P=\frac{9}{2\sqrt{100}-3}=\frac{9}{17}\)

c) P = 1 \(\Leftrightarrow\frac{9}{2\sqrt{x}-3}=1\Leftrightarrow2\sqrt{x}-3=9\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)

P = 7 \(\Leftrightarrow\frac{9}{2\sqrt{x}-3}=7\Leftrightarrow2\sqrt{x}-3=\frac{9}{7}\)

\(\Leftrightarrow2\sqrt{x}=\frac{30}{7}\Leftrightarrow\sqrt{x}=\frac{15}{7}\Leftrightarrow x=\frac{225}{49}\)

d) P nguyên \(\Leftrightarrow9⋮2\sqrt{x}-3\)

\(\Leftrightarrow2\sqrt{x}-3\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)

Lập bảng:

Vậy \(x\in\left\{1;4;9;0;36\right\}\)

\(\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}=\frac{x^2+2x}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}=\)

\(=\frac{x\left(x^2+2x\right)+2\left(x+5\right)\left(x-5\right)+50-5x}{2x\left(x+5\right)}=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}=\)

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

a/ Để biểu thức xác đinh => 2x(x+5) khác 0 => x khác 0 và x khác -5

b/ Gọi biểu thức là A. Rút gọn A ta được: 

\(A=\frac{x\left(x-1\right)\left(x+5\right)}{2x\left(x+5\right)}=\frac{x-1}{2}\left(x\ne0;x\ne-5\right)\)

A=1 => x-1=2 => x=3

c/ A=-1/2 <=> x-1=-1 => x=0

d/ A=-3 <=> x-1=-6  => x=-5

Ta có \(A=[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)]:\frac{x-1}{x^3}\)

\(\Leftrightarrow A=\left[\frac{2}{\left(x+1\right)^3}.\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}.\frac{x^2+1}{x^2}\right].\frac{x^3}{x-1}\)

\(\Leftrightarrow A=\left[\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\right].\frac{x^3}{x+1}=\frac{x}{x+1}\)

Để \(A=\frac{x}{x+1}< 1\Leftrightarrow\frac{1}{x+1}>0\Leftrightarrow x>-1\)

Để \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)

d)  \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)

\(\Leftrightarrow x-2< 0\)  ( vì \(-1< 0\))

\(\Leftrightarrow x< 2\)

\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)

  \(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)

\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)

\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)

\(A=\frac{-1}{x-2}\)

theo câu a) \(A=\frac{-1}{x-2}\)  với ĐKXĐ: \(x\ne\pm2\)

b) \(\left|2x-1\right|=3\)

\(\Rightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)   \(\Rightarrow x=-1\)  ( vì \(x=2\)  ko TM ĐKXĐ )

+) khi \(x=-1\)thì \(A=\frac{-1}{-1-2}=\frac{-1}{-3}=\frac{1}{3}\)

vậy khi \(x=-1\)  thì \(A=\frac{1}{3}\)