a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:

\(A=\left[\frac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right]\div\frac{2x-1}{4\left(x-2\right)}\)

\(A=\frac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}=\frac{4\left(2x^2-x\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4x\left(2x-1\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4}{x+2}\)

b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)

=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)

c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)

d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6

a.ĐKXĐ \(x\ne0,x\ne1\),\(x\ne-1\)

B=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2-1}{x^3-x}.\frac{x^3+x}{\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x.\left(x^2+1\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2+1}{\left(x-1\right)^2}\)

=\(\frac{3-x^2}{\left(x-1\right)^2}\)

b.TH1 x=3\(\Rightarrow\)B=\(\frac{3-3^2}{2^2}=\frac{-3}{2}\)

TH2 x=-1\(\Rightarrow\)B=\(\frac{3-\left(-1\right)^2}{4}=\frac{1}{2}\)

c.B=-1\(\Leftrightarrow\frac{3-x^2}{\left(x-1\right)^2}=-1\)\(\Leftrightarrow x^2-3=x^2-2x+1\)\(\Leftrightarrow2x=4\Leftrightarrow x=2\)

d.B+2=\(\frac{3-x^2}{\left(x-1\right)^2}+2=\frac{x^2-4x+5}{\left(x-1\right)^2}=\frac{\left(x-2\right)^2+1}{\left(x-1\right)^2}\ge0\)với mọi x\(\Rightarrow B\)>-2

ĐKXĐ \(x\ne0;x\ne1;x\ne-1\)

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

\(A=\frac{\left(x-3\right)}{\left(1-x^2\right)}:\frac{\left(1-2x\right)}{\left(x^2-1\right)}\)

\(A=\frac{\left(3-x\right)}{\left(x^2-1\right)}:\frac{\left(1-2x\right)}{\left(x^2-1\right)}\)

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

\(a,ĐKXĐ:x\ne\pm1;x\ne\frac{1}{2}\)

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

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

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

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

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

\(b,Vớix\ne\pm1;x\ne\frac{1}{2}\)ta có \(A=\frac{2x+4}{1-2x}=\frac{-1\left(1-2x\right)+5}{1-2x}=-1+\frac{5}{1-2x}\)

Với x thuộc Z để A nguyên thì \(5⋮1-2x\Rightarrow1-2x\inƯ\left\{5\right\}=\left\{\pm1;\pm5\right\}\)

Với 1-2x=1 => x= 0(TMĐKXĐ)

với 1-2x=-1 => x=1(loại)

với 1-2x=5 => x=-2(tmđkxđ)

với 1-2x=-5 => x=3(tmđkxđ)

Vậy với \(x\in\left\{0;-2;-3\right\}\)thì A nguyên

a) A xác định \(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x+1\ne0\\1-2x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\\2x\ne1\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne-1\\x\ne\frac{1}{2}\end{cases}\Leftrightarrow}x\ne\pm1;x\ne\frac{1}{2}}\)

\(A=\left(\frac{1}{x-1}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\left(x\ne\pm1;x\ne\frac{1}{2}\right)\)

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

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

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

a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\)

\(A=\frac{3}{x+4}-\frac{x\left(x-1\right)}{x+4}\times\frac{2x-5}{x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)

\(=\frac{3\left(x+4\right)}{\left(x+4\right)^2}-\frac{x\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)x\left(x-2\right)\left(x+4\right)}-\frac{17}{\left(x+4\right)^2}\)

\(=\frac{3x+12}{\left(x+4\right)^2}-\frac{\left(x-1\right)\left(2x-5\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{17}{\left(x+4\right)^2}\)

\(=\frac{\left(3x+12\right)\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}-\frac{2x^2-7x+5}{\left(x+4\right)^2\left(x-2\right)}-\frac{17\left(x-2\right)}{\left(x+4\right)^2\left(x-2\right)}\)

\(=\frac{3x^2+6x-24-2x^2+7x-5-17x+34}{\left(x+4\right)^2\left(x-2\right)}\)

\(=\frac{x^2-4x+5}{\left(x+4\right)^2\left(x-2\right)}=\frac{x^2-4x+5}{x^3+6x^2-32}\)

b) \(18A=1\)

<=> \(18\times\frac{x^2-4x+5}{x^3+6x^2-32}=1\)( ĐK : \(\hept{\begin{cases}x\ne0\\x\ne2\\x\ne-4\end{cases}}\))

<=> \(\frac{x^2-4x+5}{x^3+6x^2-32}=\frac{1}{18}\)

<=> 18( x² - 4x + 5 ) = x³ + 6x² - 32

<=> 18x² - 72x + 90 = x³ + 6x² - 32

<=> x³ + 6x² - 32 - 18x² + 72x - 90 = 0

<=> x³ - 12x² + 72x - 122 = 0

Rồi đến đây chịu á :) 

Ý lộn == là \(\frac{x^2-2x}{x+4}\)ạ ==

1. Cho bt P= (1/√x+2 + 1/√x-2 ) . √x-2/√x với x>0, x khác 4

a) rút gọn P

b) tìm x để P>1/3

c) tìm các giá trị thực của x để Q=9/2P có giá trị nguyên

2. Cho 2 biểu thức

A= 1-√x / 1+√ x và B= ( 15-√x/ x-25 + 2/√x+5) : √x+1/√ x-5 với x lớn hơn hoặc bằng 0, x khác 25

a) tính giá trị của A khi x= 6-2√5

b) rút gọn B

c) tìm a để pt A-B=a có nghiệm

chúc bạn học tốt

Bài 1 :

\(a,P=\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}=\left[\frac{x}{\left(x+6\right)\left(x-6\right)}-\frac{x-6}{x\left(x+6\right)}\right]:\frac{2x-6}{x\left(x+6\right)}\)

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

\(=\frac{6}{x-6}\)

\(b,\)Với \(x\ne-6;x\ne6;x\ne0;x\ne3\)  Thì

\(P=1\Rightarrow\frac{6}{X-6}=1\Rightarrow6=x-6\Rightarrow x=12\)(Thỏa mãn \(ĐKXĐ\))

\(c,\)Ta có :

\(P< 0\Rightarrow\frac{6}{X-6}< 0\Rightarrow X-6< 0\Rightarrow X< 6\)

Do :  \(x\ne-6;x\ne6;x\ne0;x\ne3\)  ,Nên với \(x< 6\)và  \(x\ne-6;x\ne0;x\ne3\)  thì \(P< 0\)

Bài 1 : 

a ) Ta có : 

\(P=\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}\)

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

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

\(=\frac{x^2-x^2+12x-36}{x-6}.\frac{1}{2\left(x-3\right)}\)

\(=\frac{12\left(x-3\right)}{x-6}.\frac{1}{2\left(x-3\right)}\)

\(=\frac{6}{x-6}\)

b ) \(P=1\Leftrightarrow\frac{6}{x-6}=1\Leftrightarrow x-6=6\Leftrightarrow x=12\left(tm\right)\)

c ) \(p< 0\Leftrightarrow\frac{6}{x-6}< 0\Leftrightarrow x-6< 0\Rightarrow x< 6\)

Bài làm:

a) \(đkxd:x\ne2;x\ne-2;x\ne0;x\ne3\)

Ta có: \(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

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

\(A=\left[\frac{x^2+4x+4+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\right]:\frac{x-3}{x\left(2-x\right)}\)

\(A=\frac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)

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

\(A=\frac{4x^2}{x-3}\)

b) Ta có: \(4x^2>0\left(\forall x\ne0\right)\)

=> Để A>0 thì \(x-3>0\)

\(\Rightarrow x>3\)

Vậy với \(x>3\)thì A>0

c) Ta có: \(\left|x-7\right|=4\)\(\Rightarrow\orbr{\begin{cases}x-7=4\\x-7=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=11\\x=3\end{cases}}\)

Mà theo điều kiện xác định, \(x\ne3\)

\(\Rightarrow x=11\)

Khi đó, \(A=\frac{4.11^2}{11-3}=\frac{121}{2}\)

Vậy \(A=\frac{121}{2}\)

Học tốt!!!!