Bài 1: Phân thức đại số

Ta có:

$$\begin{gathered} x^2{y}^5.35xy=35x^3{y}^4 \\ 5.7x^3{y}^4=35x^3{y}^4 \\ Suy ra: x^2{y}^3.35xy=5.7x^3{y}^4 \end{gathered}$$

Vậy

Ta có:

$$\begin{gathered} x^2\left(x+2\right)\left(x+2\right)=x^2{\left(x+2\right)}^2 \\ x{\left(x+2\right)}^2.x=x^2{\left(x+2\right)}^2 \\ Suy ra:\hspace{0.17em}{x}^2\left(x+2\right)\left(x+2\right)=x^2{\left(x+2\right)}^2 \end{gathered}$$

vậy

Ta có:

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

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

( vì ( x- 3) = - (3- x) nên ${\left(x-3\right)}^2=\left[-{\left(3-x\right)}^2\right]={\left(3-x\right)}^2$ )

Từ (1) và (2) suy ra: $\left(x-3\right).\left(9-x^2\right)=\left(3+x\right)\left(x^2-6x+9\right)$

Do đó: 

Ta có: $\left(x^3-4x\right).5=5x^3-20x$

$\left(10-5x\right)\left(-x^2-2x\right)=-10x^2-20x+5x^3+10x^2=5x^3-20x$

Suy ra: $\left(x^3-4x\right).5=\left(10-5x\right)\left(-x^2-2x\right)$

Vậy

⇒ $A\left(4x^2-1\right)=\left(2x-1\right)\left(6x^2+3x\right)$

      ⇒ A(2x – 1)(2x + 1) = (2x – 1).3x(2x + 1) ⇒ A = 3x

Vậy 

 ⇒ $\left(4x^2-3x-7\right)\left(2x+3\right)=A\left(4x-7\right)$

      ⇒ $\left(4x^2+4x-7x-7\right)\left(2x+3\right)=A\left(4x-7\right)$

      ⇒ [4x(x + 1) – 7(x + 1)](2x+ 3) = A(4x - 7)

      ⇒ (x + 1)(4x – 7)(2x + 3) = A(4x – 7)

      ⇒ A = (x + 1)(2x + 3) = $2x^3+3x+2x+3=2x^2+5x+3$

Vậy 

 ⇒ $\left(4x^2-7x+3\right)\left(x^2+2x+1\right)=A\left(x^2-1\right)$

      ⇒ $\left(4x^2-4x-3x+3\right){\left(x+1\right)}^2=A\left(x+1\right)\left(x-1\right)$

      ⇒ $\left[4x\left(x-1\right)-3\left(x-1\right)\right].{\left(x+1\right)}^2=A.\left(x+1\right)\left(x-1\right)$

      ⇒ $\left(x-1\right)\left(4x-3\right){\left(x+1\right)}^2=A\left(x+1\right)\left(x-1\right)$

      ⇒ A = $\left(4x-3\right)\left(x+1\right)=4x^2+4x-3x-3=4x^2+x-3$

Vậy 

 ⇒ $\left(x^2-2x\right).A=\left(2x^2-3x-2\right)\left(x^2+2x\right)$

      ⇒ $x\left(x-2\right).A=\left(2x^2-4x+x-2\right).x\left(x+2\right)$

      ⇒ $x\left(x-2\right).A=\left[2x\left(x-2\right)+\left(x-2\right)\right].x\left(x+2\right)$

      ⇒ x(x – 2).A = (x – 2)(2x + 1).x.(x + 2)

      ⇒ A = (2x + 1)(x + 2) = $2x^2+4x+x+2=2x^2+5x+2$

Vậy 

$\left(5x+3\right)\left(x^2-4\right)=5x^3-20x+3x^2-12$ (1)

$\left(x-2\right)\left(5x^2+13x+6\right)=5x^3+13x^2+6x-10x^2-26x-12$

      = $5x^3-20x+3x^2-12$ (2)

Từ (1) và (2) suy ra: $\left(5x+3\right)\left(x^2-4\right)$ = $\left(x-2\right)\left(5x^2+13x+6\right)$

Vậy đẳng thức đúng.

$$\begin{gathered} \left(x+1\right)\left(x^2+6x+9\right)=x^3+6x^2+9x+x^2+6x+9=x^3+7x^2+15x+9 \\ \left(x+3\right)\left(x^2+3\right)=x^3+3x+3x^2+9 \end{gathered}$$

Ta có: $\left(x+1\right)\left(x^2+6x+9\right)\ne \left(x+3\right)\left(x^2+3\right)$

Vậy đẳng thức sai.

$$\begin{gathered} \left(x^2-2\right)\left(x+1\right)=x^3+x^2-2x-2 \\ \left(x^2-1\right)\left(x+2\right)=x^3+2x^2-x-2 \end{gathered}$$

Ta có: $\left(x^2-2\right)\left(x+1\right)\ne \left(x^2-1\right)\left(x+2\right)$

Vậy đẳng thức sai.

$$\begin{gathered} \left(2x^2-5x+3\right)\left(x^2+5x+4\right) \\ =2x^4+10x^3+8x^2-5x^3-25x^2-20x+3x^2+15x+12 \\ =2x^4+5x^3-14x^2-5x+12 \end{gathered}$$

$$\begin{gathered} \left(x^2+3x-4\right)\left(2x^2-x-3\right) \\ =2x^4-x^3-3x^2+6x^3-3x^2-9x-8x^2+4x+12 \\ =2x^4+5x^3-4x^2-5x+12 \end{gathered}$$

Ta có: $\left(2x^2-5x+3\right)\left(x^2+5x+4\right)$ = $\left(x^2+3x-4\right)\left(2x^2-x-3\right)$

Vậy đẳng thức đúng.

Để  thì: $\left(x-3\right)\left(x^3-1\right)=\left(x^2+x+1\right).P$

Hay $\left(x-3\right)\left(x-1\right)\left(x^2+x+1\right)=\left(x^2+x+1\right).P$

Suy ra: P = $P=\left(x-3\right)\left(x-1\right)=x^2-x-3x+3=x^2-4x+3$

Chọn B. P = $x^2-4x+3$

$\frac{\left(x+2\right)P}{x-2}=\frac{\left(x-1\right)Q}{x^2-4}$

⇒ $\left(x+2\right).P.\left(x^2-4\right)=\left(x-2\right)\left(x-1\right).Q$

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

Chọn P = (x – 1) thì Q = ${\left(x+2\right)}^2$

$\frac{\left(x+2\right).P}{x^2-1}=\frac{\left(x-2\right).Q}{x^2-2x+1}$

⇒ $\left(x+2\right).P.\left(x^2-2x+1\right)=\left(x^2-1\right)\left(x-2\right).Q$

Hay $\left(x+2\right){\left(x-1\right)}^2.P=\left(x-1\right)\left(x+1\right)\left(x-2\right).Q$

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

Nếu  ⇒ PS = QR (1). Vì  là phân thức

⇒ Q, S khác không. Cộng vào hai vế của đẳng thức (1) với QS

PS + QS = QR + QS ⇒ (P + Q).S = Q.(R + S)

 ⇒ PS = QR (1) và P ≠ Q, R ≠ S

Trừ từng vế đẳng thức (1) với PR: PS – PR = QR – PR

⇒ P(S – R) = R(Q – P)