\(P=\left(\frac{2x}{2x^2-5x+2}-\frac{5}{2x-3}\right):\left(3+\frac{2}{1-x}\right) \)(dk x khac 3/2 ; x khac 1)

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

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

\(P=\frac{-\left(3x-5\right)}{\left(2x-3\right)\left(x-1\right)}\cdot\frac{x-1}{3x-5}\)

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

b) TC: \(|2x-1|=3\)

TH1) \(|2x-1|=2x-1\)khi \(x\ge\frac{1}{2}\)

2x-1=3 suy ra x=2 ( thoa dk)

TH2) \(|2x-1|=-2x+1\)khi \(x< \frac{1}{2}\)

-2x+1=3 suy ra x=-1 ( thoa dk)

khi x= 2 thi P=-1 

khi x= -1 thi P=1/5

c) de P thuoc Z thi \(-\frac{1}{2x-3}\)thuoc Z 

suy ra \(\frac{1}{3-2x}\)thuoc Z
suy ra 3-2x thuoc \(Ư\left(1\right)\in\left\{\pm1\right\}\)

khi 3-2x=1 thi x= 1 (ko thoa dk x khac 1)

khi 3-2x=-1 thi x=2(thoa dk)

vay x=2 thi P thuoc Z

d) giai tg tu cau c

\(\dfrac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\dfrac{2\sqrt{x}-2}{\sqrt{x}+2}+\dfrac{39\sqrt{x}+12}{5x+9\sqrt{x}-2}\\ =\dfrac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\dfrac{2\sqrt{x}-2}{\sqrt{x}+2}+\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\\ =\dfrac{\left(-7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(5\sqrt{x}-1\right)}+\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\)

\(=\dfrac{-7x-14\sqrt{x}+7\sqrt{x}+14+10x-2\sqrt{x}-10\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\\ =\dfrac{3x+20\sqrt{x}+28}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}+2\right)\cdot\left(3\sqrt{x}+14\right)}{\left(5\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}+14}{5\sqrt{x}-1}\)

b) để Q = -3 thì \(\dfrac{3\sqrt{x}+14}{5\sqrt{x}-1}=-3\)

quy đồng và khử mẫu ta được:

\(3\sqrt{x}+14=-3\cdot\left(5\sqrt{x}-1\right)\\ 3\sqrt{x}+14=-15\sqrt{x}+3\\ 18\sqrt{x}=-11\\ \sqrt{x}=-\dfrac{11}{18}\)

vậy không có giá trị x nào để Q = -3

Đặt \(A=\frac{x^2+2x-1}{x-1}\)

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

      Vậy để A nguyên thì x thỏa mãn mõi số nguyên

chịu chưa học lớp 6

(đkxđ x khác 1)

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

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

Để biểu thức nguyên=>\(\frac{2}{x-1}\in Z\)<=>\(2⋮x-1=>x-1\inƯ=\){1,2,-1,-2}

=>x\(\in\){2,3,0,-1}

Bạn cần câu nào?

làm đc câu nào hay câu đây, càng nhiều càng tốt

cảm ơn nha

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

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

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

\(\Rightarrow A=\frac{2x^2+1}{x^2+1}:\frac{x-1}{x^2+1}\)

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