ĐKXĐ: x<>0; x<>1/2; x<>3; x<>-3

a: \(P=\left(\frac{6}{x^2-9}-\frac{5}{3-x}+\frac{1}{x+3}\right):\frac{2x-1}{x^2-3x}\)

\(=\left(\frac{6}{\left(x-3\right)\left(x+3\right)}+\frac{5}{x-3}+\frac{1}{x+3}\right):\frac{2x-1}{x\left(x-3\right)}\)

\(=\frac{6+5\left(x+3\right)+x-3}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x\left(x-3\right)}{2x-1}\)

\(=\frac{x+3+5x+15}{x+3}\cdot\frac{x}{2x-1}=\frac{6x+18}{x+3}\cdot\frac{x}{2x-1}=\frac{6x}{2x-1}\)

b: \(P=\frac13\)

=>\(\frac{6x}{2x-1}=\frac13\)

=>18x=2x-1

=>16x=-1

=>\(x=-\frac{1}{16}\) (nhận)

c: Để P nguyên thì 6x⋮2x-1

=>6x-3+3⋮2x-1

=>3⋮2x-1

=>2x-1∈{1;-1;3;-3}

=>2x∈{2;0;4;-2}

=>x∈{1;0;2;-1}

Kết hợp ĐKXĐ, ta được: x∈{1;2;-1}

a,Ta có \(\frac{x+5}{x+3}=\frac{x+3+2}{x+3}\)\(=1+\frac{2}{x+3}\)

Để \(\frac{x+5}{x+3}< 1\)\(\Leftrightarrow1+\frac{2}{x+3}< 0\)

=>\(\frac{2}{x+3}< 0\)

=>x+3<0

=>x<-3

b, Ta có \(\frac{x+3}{x+4}=\frac{x+4-1}{x+4}=1-\frac{1}{x+4}\)

Để \(\frac{x+3}{x+4}>1\Leftrightarrow\)\(\frac{-1}{x+4}>0\)=>x+4>0=>x>-4

thank you vinamilk

a. x+5/x+3<1

<=> (x+5/x+3) -1 <0

<=>(x+5-x-3)/x+3 <0

<=>2/x+3 <0

<=> x+3 <0 ( do 2 > 0 )

<=> x< -3 

câu b tương tự