Bạn ơi mik ra \(\dfrac{x^3+45x-54}{12\left(x-3\right)\left(x+3\right)}\) có đúng không bạn?

Mình rút chx hết bạn bạn gửi cách làm bạn qua mình tham khảo đc k ạ?

d) Ta có: \(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)

\(=\dfrac{5\sqrt{x}-6-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-9+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

\(=\dfrac{3\sqrt{x}}{x-3}\)

f) Ta có: \(\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)

\(=\dfrac{3+\sqrt{1-x^2}}{\sqrt{1+x}}:\dfrac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}\)

\(=\dfrac{\sqrt{1-x^2}}{\sqrt{1+x}}=\sqrt{1-x}\)

Ta có: \(P=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\right)\)

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

\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+3}\right):\left(\dfrac{-\left(x-9\right)+x-4\sqrt{x}+4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}:\dfrac{-x+9+2x-4\sqrt{x}-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x-4\sqrt{x}+4}\)

\(=\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2}=\dfrac{3}{\sqrt{x}-2}\)

\(\left(\dfrac{x+2\sqrt{x}+7}{x-9}+\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\) (ĐK: \(x\ne1,x\ge0,x\ne9\))

\(=\left[\dfrac{x+2\sqrt{x}+7}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-1}\right)\)

\(=\dfrac{x+2\sqrt{x}+7-x-2\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\left[\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\right]\)

\(=\dfrac{10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{-4}\)

\(=-\dfrac{5\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)}\)

\(=-\dfrac{5\sqrt{x}-5}{2\sqrt{x}-6}\)

a: Sửa đề: \(M=\frac{x}{x+3}+\frac{3-x}{x+3}\cdot\frac{x^2+3x+9}{x^2-9}\)

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

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

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

b: \(A=\left(\frac{3x}{1-3x}-\frac{2x}{3x+1}\right):\frac{6x^2+10x}{1-6x+9x^2}\)

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

\(=\frac{-9x^2-3x-6x^2+2x}{3x+1}\cdot\frac{3x-1}{2x\left(3x+5\right)}=\frac{-15x^2-x}{3x+1}\cdot\frac{3x-1}{2x\left(3x+5\right)}\)

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

Đề sai rồi bạn

\(B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\cdot\dfrac{\sqrt{x}-2+3}{3}\)

\(=\dfrac{-3\left(\sqrt{x}+3\right)}{x-9}\cdot\dfrac{\sqrt{x}+1}{3}=\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}\)

\(b,P=\left[\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-1\right]:\dfrac{9-x^2+\left(x-3\right)\left(x+3\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\left(x\ne\pm3;x\ne2\right)\\ P=\left(\dfrac{x}{x+3}-1\right)\cdot\dfrac{\left(x-2\right)\left(x+3\right)}{9-x^2+x^2-9-\left(x-2\right)^2}\\ P=\dfrac{x-x-3}{x+3}\cdot\dfrac{\left(x-2\right)\left(x+3\right)}{-\left(x-2\right)^2}\\ P=\dfrac{-3}{-\left(x-2\right)}=\dfrac{3}{x-2}\)

Với \(x^3-4x=0\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\\x=-2\end{matrix}\right.\)

Với \(x=0\Leftrightarrow P=\dfrac{3}{0-2}=-\dfrac{3}{2}\)

Với \(x=-2\Leftrightarrow P=\dfrac{3}{-2-2}=-\dfrac{3}{4}\)