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

Câu 1 :

a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)

b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)

\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)

\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)

\(\Leftrightarrow2x^2+8x+6=0\)

\(\Leftrightarrow x^2+4x+4-1=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)

Vậy : \(x=-3\) thì P = 1.

\(a,x\ne2;x\ne-2;x\ne0\)

\(b,A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)

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

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

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

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

\(c,\)Để A > 0 thi \(\frac{1}{2-x}>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\)

a)

DK:tồn tại P \(\hept{\begin{cases}x\ne0\\x\ne-+6\\x\ne3\end{cases}}\)

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

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

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

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

b)6/(x-6)=1=> x-6=6=> x=12

c)x-6<0=> x<6

dieu kien xac  dinh cua bieu thuc tren la x khac -+6,x khac 3