\(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

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

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

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

b) \(Voix>0\Rightarrow P\ne\varnothing\)(mk ko chac)

c) \(P\inℤ\Leftrightarrow x+3⋮x-3\Leftrightarrow x-3\in\left\{-1;-2;-3;-6;1;2;3;6\right\}\) 

sau do tinh

cau nay la toan lp 8 nha

P= O/ nha

\(ĐKXĐ:x\ne\pm3\)

\(P=\left(\frac{x^2-3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

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

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

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

\(\Leftrightarrow P=\frac{1}{x+3}:\frac{x-3}{x^2+9}\)

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

mk chịu

sửa đề: \(P=\frac{\frac{x^2+3}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}}{\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}}\)

a) Ta có: \(P=\frac{\frac{x^2+3}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}}{\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}}\)

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

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

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

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

b) Với x>0 thì P sẽ không nhận giá trị khi x=3(vì khi đó mẫu sẽ bằng 0 thì phân thức vô nghĩa)

c) Để \(P=\frac{x+3}{x-3}\) có giá trị nguyên

thì \(x+3⋮x-3\)

hay \(x-3+6⋮x-3\)

\(\Leftrightarrow6⋮x-3\)

\(\Leftrightarrow x-3\inƯ\left(6\right)\)

\(\Leftrightarrow x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

\(\Leftrightarrow x\in\left\{4;2;5;1;6;0;9;-3\right\}\)(tmđk \(x\ne3\))

Vậy: \(x\in\left\{4;2;5;1;6;0;9;-3\right\}\)

ĐKXĐ: \(x\ne\pm3\)

\(P=\left[\dfrac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)

\(=\left[\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}+\dfrac{3}{x^2+9}\right]:\left[\dfrac{1}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)

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

Ý 2 mình k hiểu ý bạn lắm

\(P=\dfrac{x+3}{x-3}=\dfrac{x-3+6}{x-3}=1+\dfrac{6}{x-3}\in Z\)

\(\Leftrightarrow\left(x-3\right)\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

Kết hợp vs ĐKXĐ \(\Rightarrow x\in\left\{0;1;2;4;5;6;9\right\}\)

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 : \(x\ne0;x\ne3\)

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

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

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