Bài 1: Phân thức đại số.

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

\(\dfrac{x^2-4x+3}{x^2-x}\\ =\dfrac{x^2-x-3x+3}{x\left(x-1\right)}\\ \\ =\dfrac{\left(x^2-x\right)-\left(3x-3\right)}{x\left(x-1\right)}\\ \\ =\dfrac{x\left(x-1\right)-3\left(x-1\right)}{x\left(x-1\right)}\\ \\ =\dfrac{\left(x-3\right)\left(x-1\right)}{x\left(x-1\right)}\\ \\ =\dfrac{x-3}{x}\text{ }\text{ }\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra : \(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x-3}{x}=\dfrac{x^2-4x+3}{x^2-x}\)

Vậy 3 phân thức \(\dfrac{x^2-2x-3}{x^2+x};\dfrac{x-3}{x};\dfrac{x^2-4x+3}{x^2-x}\) bằng nhau

Ta có:

(...) . (x-4) = x . ( x²- 16 )

= x(x - 4)(x + 4) = (x² + 4x)(x -4)

Vậy phải điền vào chỗ trống đa thức x(x + 4) hay x² + 4x.

a. \(x^2y^3.35xy=5.7x^3y^4\)

\(\Leftrightarrow35x^3y^4=35x^3y^4\Rightarrowđpcm\)

\(b.x^2\left(x+2\right).\left(x+2\right)=x\left(x+2\right)^2.x\)

\(\Leftrightarrow x^2\left(x+2\right)^2=x^2\left(x+2\right)^2\Rightarrowđpcm\)

\(c.\left(3-x\right)\left(9-x^2\right)=\left(3+x\right)\left(x^2-6x+9\right)\)

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

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

\(\Rightarrowđpcm\)

\(d.5\left(x^3-4x\right)=\left(10-5x\right)\left(-x^2-2x\right)\)

\(\Leftrightarrow5x^3-20x=5x^3-20x\Rightarrowđpcm\)

Bài 3: (SBT/24):

a. \(\dfrac{5x+3}{x-2}\)=\(\dfrac{5x^2+13x+6}{x^2-4}\)

(5x+3) . (x²-4) = 5x³-20x+3x³-12

(x-2) . (5x²+13x+6) = 5x³+13x²+6x-10x²-26x-12 = 5x³-20x+3x²-12

=> (5x+3) (x²-4) = (x-2) (5x²+13x+6)

Vậy \(\dfrac{5x+3}{x-2}\)=\(\dfrac{5x^2+13x+6}{x^2-4}\)(đẳng thức đúng)

b. \(\dfrac{x+1}{x+3}\)=\(\dfrac{x^2+3}{x^2+6x+9}\)

(x+1) . (x²+6x+9) = x³+6x²+9x+x²+6x+9 = x³+7x²+15x+9

(x+3) . (x²+3) = x³+3x+3x²+9

=> (x+1) (x²+6x+9) ≠ (x+3) (x²+3)

Vậy \(\dfrac{x+1}{x+3}\)≠\(\dfrac{x^2+3}{x^2+6x+9}\)(đẳng thức sai)

Chữa lại: \(\dfrac{x+1}{x+3}\)=\(\dfrac{x^2+3}{x^{2_{ }}+6x+9}\)

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

(x²-2) . (x+1) = x³+x²-2x-2

(x²-1) . (x+2) = x³+2x²-x-2

=> (x²-2) (x+1) ≠ (x²-1) (x+2)

Vậy \(\dfrac{x^2-2}{x^2-1}\)≠\(\dfrac{x+2}{x+1}\)(đẳng thức sai)

Chữa lại: \(\dfrac{x^2+x-2}{x^2-1}\)=\(\dfrac{x+2}{x+1}\)

d. \(\dfrac{2x^2-5x+3}{x^2+3x-4}\)=\(\dfrac{2x^2-x-3}{x^2+5x+4}\)

(2x²-5x+3) . (x²+5x+4) = 2x⁴+10x³+8x²-5x³-25x²-20x+3x²+15x+12

= 2x⁴+5x³-14x²-5x+12

(x²+3x-4) . (2x²-x-3) = 2x⁴-x³-3x²+6x³-3x²-9x-8x²+4x+12

= 2x⁴+5x³-14x²-5x+12

=> (2x²-5x+3) (x²+5x+4) = (x²+3x-4) (2x²-x-3)

Vậy \(\dfrac{2x^2-5x+3}{x^2+3x-4}\)=\(\dfrac{2x^2-x-3}{x^2+5x+4}\)

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

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

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

\(\Leftrightarrow P=\left(x-3\right)\left(x-1\right)=x^2-4x+3\)

Vậy chọn (B) là đúng

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

\(\Leftrightarrow\left(x^2-4\right)\left(x+2\right)P=\left(x-2\right)\left(x-1\right)Q\)

\(\Leftrightarrow\)\(\left(x+2\right)^2\left(x-2\right)P=\left(x-2\right)\left(x-1\right)Q\)

\(\Leftrightarrow\)\(\left(x+2\right)^2P=\left(x-1\right)Q\)

\(\Leftrightarrow P=x-1\)

\(Q=\left(x+2\right)^2=x^2+4x+4\)

b)\(\dfrac{\left(x+2\right)P}{x^2-1}=\dfrac{\left(x-2\right)Q}{x^2-2x+1}\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)P=\left(x+1\right)\left(x-1\right)\left(x-2\right)Q\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)P=\left(x+1\right)\left(x-2\right)Q\)

\(\Leftrightarrow P=\left(x+1\right)\left(x-2\right)=x^2-x-2\)

\(Q=\left(x-1\right)\left(x+2\right)=x^2+x-2\)

a)

\(\dfrac{P}{Q}=\dfrac{R}{S}\Rightarrow PS=QR\)

\(\Leftrightarrow PS+QS=QR+QS\)

\(\Leftrightarrow S\left(P+Q\right)=Q\left(R+S\right)\)

điều kiện Q,s khác 0 => chia hau vế cho QS

\(\Leftrightarrow\dfrac{S\left(P+Q\right)}{QS}=\dfrac{Q\left(R+S\right)}{QS}\Leftrightarrow\dfrac{\left(P+Q\right)}{Q}=\dfrac{\left(R+S\right)}{S}\) đpcm