Bài 5:
a: ĐKXĐ: x<>3; x<>-3; x<>-1
\(M=\left(\frac{2x}{x+3}+\frac{x}{x-3}-\frac{3\left(x+3\right)}{x^2-9}\right):\left(\frac{2x-2}{x-3}-1\right)\)
\(=\frac{2x\left(x-3\right)+x\left(x+3\right)-3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{2x-2-x+3}{x-3}\)
\(=\frac{2x\left(x-3\right)+\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x-3}{x+1}\)
\(=\frac{\left(x-3\right)\left(2x+x+3\right)}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x-3}{x+1}=\frac{3x+3}{\left(x+3\right)}\cdot\frac{x-3}{x+1}=\frac{3\left(x-3\right)}{x+3}=\frac{3x-9}{x+3}\)
b: \(M=\frac23\)
=>\(\frac{3x-9}{x+3}=\frac23\)
=>3(3x-9)=2(x+3)
=>9x-27=2x+6
=>7x=33
=>x=33/7(nhận)
c: M<3
=>M-3<0
=>\(\frac{3x-9-3x-9}{x+3}\) <0
=>\(\frac{-18}{x+3}\) <0
=>x+3>0
=>x>-3
Kết hợp ĐKXĐ, ta được: x>-3 và x∉{3;-1}
Bài 4:
a: ĐKXĐ: x<>0; x<>2; x<>-2
\(Q=\left(\frac{x}{x-2}-\frac{4}{x^2-2x}\right)\cdot\left(\frac{1}{x+2}+\frac{4}{x^2-4}\right)\)
\(=\frac{x^2-4}{x\left(x-2\right)}\cdot\frac{x-2+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)}{\left.\left(x-2\right)\left(x+2\right)\right.}\cdot\frac{x+2}{x\left(x-2\right)}=\frac{x+2}{x\left(x-2\right)}\)
b: Khi x=-1 thì \(Q=\frac{-1+2}{-1\cdot\left(-1-2\right)}=\frac{1}{-1\cdot\left(-3\right)}=\frac13\)
Vì x=-2 không thỏa mãn ĐKXĐ
nên khi x=-2 thì Q không có giá trị
c: Đặt A=x*Q
\(=\frac{x\left(x+2\right)}{x\left(x-2\right)}=\frac{x+2}{x-2}\)
Để A là số nguyên thì x+2⋮x-2
=>x-2+4⋮x-2
=>4⋮x-2
=>x-2∈{1;-1;2;-2;4;-4}
=>x∈{3;1;4;0;6;-2}
Kết hợp ĐKXĐ, ta được: x∈{3;1;4;6}
