a: \(\Leftrightarrow\dfrac{5x^2-13x+6}{A}=\dfrac{5x-3}{2x+5}\)

\(\Leftrightarrow\dfrac{5x^2-10x-3x+6}{A}=\dfrac{5x-3}{2x+5}\)

\(\Leftrightarrow A=\dfrac{\left(2x+5\right)\left(x-2\right)\left(5x-3\right)}{\left(5x-3\right)}=\left(2x+5\right)\left(x-2\right)\)

b: \(\Leftrightarrow\dfrac{x\left(x+4\right)}{A}=\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(2x-1\right)}=\dfrac{x}{\left(2x-1\right)}\)

=>A=(x+4)(2x-1)

Bài 1:

a) Ta có: \(VT=\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}\)

\(=\frac{-\left(u^2-3u+2\right)}{\left(u+2\right)\left(u-1\right)}\)

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

\(=\frac{-\left[u\left(u-1\right)-2\left(u-1\right)\right]}{\left(u+2\right)\left(u-1\right)}\)

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

\(=\frac{2-u}{u+2}\)(1)

Ta có: \(VP=\frac{u^2-4u+4}{4-u^2}\)

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

\(=\frac{-\left(u-2\right)}{u+2}\)

\(=\frac{2-u}{u+2}\)(2)

Từ (1) và (2) suy ra \(\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}=\frac{u^2-4u+4}{4-u^2}\)

b) Ta có: \(VT=\frac{v^3+27}{v^2-3v+9}\)

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

\(=v+3=VP\)(đpcm)

Bài 2:

a) Ta có: \(\frac{3x^2-2x-5}{M}=\frac{3x-5}{2x-3}\)

\(\Leftrightarrow\frac{3x^2-5x+3x-5}{M}=\frac{3x-5}{2x-3}\)

\(\Leftrightarrow\frac{x\left(3x-5\right)+\left(3x-5\right)}{M}=\frac{3x-5}{2x-3}\)

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

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

\(\Leftrightarrow M=\left(x+1\right)\left(2x-3\right)\)

\(\Leftrightarrow M=2x^2-3x+2x-3\)

hay \(M=2x^2-x-3\)

Vậy: \(M=2x^2-x-3\)

b) Ta có: \(\frac{2x^2+3x-2}{x^2-4}=\frac{M}{x^2-4x+4}\)

\(\Leftrightarrow\frac{2x^2+4x-x-2}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)

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

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

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

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

\(\Leftrightarrow M=\left(2x-1\right)\left(x-2\right)\)

\(\Leftrightarrow M=2x^2-4x-x+2\)

hay \(M=2x^2-5x+2\)

Vậy: \(M=2x^2-5x+2\)

Bài 3:

a) Ta có: \(\frac{x+1}{N}=\frac{x^2-2x+4}{x^3+8}\)

\(\Leftrightarrow\frac{x+1}{N}=\frac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Leftrightarrow\frac{x+1}{N}=\frac{1}{x+2}\)

\(\Leftrightarrow N=\left(x+1\right)\left(x+2\right)\)

hay \(N=x^2+3x+2\)

Vậy: \(N=x^2+3x+2\)

n) Ta có: \(\frac{\left(x-3\right)\cdot N}{3+x}=\frac{2x^3-8x^2-6x+36}{2+x}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{2x^3+4x^2-12x^2-24x+18x+36}{x+2}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{\left(x+3\right)}=\frac{2x^2\left(x+2\right)-12x\left(x+2\right)+18\left(x+2\right)}{x+2}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{\left(x+2\right)\left(2x^2-12x+18\right)}{x+2}\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-12x+18\)

\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-6x-6x+18=2x\left(x-3\right)-6\left(x-3\right)=2\cdot\left(x-3\right)^2\)

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

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

\(\Leftrightarrow N=\frac{2\left(x-3\right)}{x+3}\)

hay \(N=\frac{2x-6}{x+3}\)

Vậy: \(N=\frac{2x-6}{x+3}\)

a) Kết quả N = (x + 1)(x + 2);

b) Kết quả N = 2(x + 3)(x - 3).

x + 2 . P x 2 - 1 = x - 2 . Q x 2 - 2 x + 1

⇒ x + 2 . P . x 2 - 2 x + 1 = x 2 - 1 x - 2 . Q

Hay  x + 2 x - 1 2 . P = x - 1 x + 1 x - 2 . Q

Chọn P = (x – 2)(x + 1) =  x 2 - x - 2  thì Q = (x + 2)(x – 1) = x 2 + x - 2

x + 2 P x - 2 = x - 1 Q x 2 - 4

⇒ x + 2 . P . x 2 - 4 = x - 2 x - 1 . Q

Hay (x + 2)(x – 2)(x + 2).P = (x – 2)(x – 1).Q

Chọn P = (x – 1) thì Q = x + 2 2

Câu 2: A

Câu 1: B