Câu 3 : 

\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\)  ĐKXđ : \(x\ne\pm1\)

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

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

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

\(A=\frac{10}{x+1}\)

\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)

ĐKXđ : \(x\ne0;x\ne3\)

\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)

\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)

\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)

a) ĐKXĐ : \(x\ne0,x\ne\frac{3}{2},x\ne-\frac{3}{2}\)

Ta có : \(M=\frac{\left(2x^3+3x^2\right)\left(2x+1\right)}{4x^3-9x}\)

\(=\frac{x^2\left(2x+3\right)\left(2x+1\right)}{x\left(4x^2-9\right)}\)

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

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

Vậy : \(M=\frac{x\left(2x+1\right)}{2x-3}\) với \(x\ne0,x\ne\frac{3}{2},x\ne-\frac{3}{2}\)

b) Để \(M=0\Leftrightarrow\frac{x\left(2x+1\right)}{2x-3}=0\)

\(\Rightarrow x\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(loại\right)\\x=-\frac{1}{2}\left(tm\right)\end{cases}}\)

Vậy : \(x=-\frac{1}{2}\) để M=0.

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm\frac{3}{2}\end{cases}}\)

a) \(M=\frac{\left(2x^3+3x^2\right)\left(2x+1\right)}{4x^3-9x}\)

\(\Leftrightarrow M=\frac{x^2\left(2x+3\right)\left(2x+1\right)}{x\left(4x^2-9\right)}\)

\(\Leftrightarrow M=\frac{x\left(2x+3\right)\left(2x+1\right)}{\left(2x+3\right)\left(2x-3\right)}\)

\(\Leftrightarrow M=\frac{x\left(2x+1\right)}{2x-3}\)

b) Để M =0

\(\Leftrightarrow\frac{x\left(2x+1\right)}{2x-3}=0\)

\(\Leftrightarrow x\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(KTM\right)\\x=\frac{-1}{2}\left(TM\right)\end{cases}}}\)

Vậy ..........

c) Ta có :

\(M=\frac{x\left(2x+1\right)}{2x-3}=x+2+\frac{6}{2x-3}\)

Để M có giá trị nguyên

\(\Leftrightarrow2x-3\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)( Không lấy âm vì n thuộc N )

Ta có bảng sau :

Vậy..........

a,ĐK:  \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

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

\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)

c, Với x = 4 thỏa mãn ĐKXĐ thì

\(A=\frac{-3}{4-3}=-3\)

d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)

\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)

Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)

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

a) Đk: x > 0 và x khác +-1

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

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

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

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

b) Ta có: A = \(\frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}=-\left(\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\frac{1}{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> 1/x - 1/2 = 0 <=> x = 2 (tm)

Vậy MaxA = 1/4 <=> x = 2

\(a,P=\left[\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right]\cdot\dfrac{2x}{1-x}\left(x\ne1;x\ne-1;x\ne0\right)\\ P=\left(\dfrac{1}{3x}-\dfrac{1}{3x}-1\right)\cdot\dfrac{2x}{1-x}\\ P=-1\cdot\dfrac{2x}{1-x}=\dfrac{2x}{x-1}\\ b,P=2+\dfrac{2}{x-1}\in Z\\ \Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{2;3\right\}\left(x\ne-1;x\ne0\right)\\ c,P\le1\Leftrightarrow\dfrac{2x}{x-1}-1\le0\\ \Leftrightarrow\dfrac{x+1}{x-1}\le0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\le0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\ge0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-1\le x< 1\)

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

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

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

a) đặt mẫu chứng là x-2