Gọi phân thức cần tìm là \(A\)

Ta có:

\(\dfrac{1}{x}.\dfrac{x}{x+1}.\dfrac{x+1}{x+2}.\dfrac{x+2}{x+3}.\dfrac{x+3}{x+4}.\dfrac{x+4}{x+5}.\dfrac{x+5}{x+6}.\dfrac{x+6}{x+7}.\dfrac{x+7}{x+8}.\dfrac{x+8}{x+9}.\dfrac{x+9}{x+10}\)

\(=\dfrac{x\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)\left(x+8\right)\left(x+9\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)\left(x+8\right)\left(x+9\right)\left(x+10\right)}\)\(=\dfrac{x}{x+10}\)

Suy ra:

\(\dfrac{1}{x}.\dfrac{x}{x+1}.\dfrac{x+1}{x+2}.\dfrac{x+2}{x+3}.\dfrac{x+3}{x+4}.\dfrac{x+4}{x+5}.\dfrac{x+5}{x+6}.\dfrac{x+6}{x+7}.\dfrac{x+7}{x+8}.\dfrac{x+8}{x+9}.\dfrac{x+9}{x+10}.A=1\)

\(\Leftrightarrow\dfrac{x}{x+10}.A=1\)

\(\Leftrightarrow A=\dfrac{x+10}{x}\)

Vậy phân thức cần điền vào chỗ trống là \(\dfrac{x+10}{x}\)

\(\dfrac{x}{x+1}:\dfrac{x+2}{x+3}:\dfrac{x+3}{x+4}:\dfrac{x+4}{x+5}:\dfrac{x+5}{x+6}=\dfrac{x}{x+6}\)

tính chất quan trọng phần thức với

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\Rightarrow c=\dfrac{ad}{b}\)áp vào

\(\dfrac{x^5-1}{x^2-1}=\dfrac{A}{x+1}\Rightarrow A=\dfrac{\left(x^5-1\right)\left(x+1\right)}{x^2-1}\) {x khác +-1}

\(A=\dfrac{\left(x^5-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left[\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)\right]\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\left(x^4+x^3+x^2+x+1\right)\)

Vậy đa thức cần điền là

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

Vế phải chứng tỏ đã chia mẫu của vế trái cho x - 1 ( vì x² – 1 = (x - 1)(x + 1)

Vậy phải chia tử của vế trái x⁵ – 1 cho x - 1

Vậy phải điền vào chỗ trống : x⁴ + x³ + x² + x + 1

Ta có:

Để  thì 

Vậy phân thức cần điền là 

a, = x²

b, = 2x-2y hoặc 2(x-y)

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

a)\(\dfrac{x+5}{3x-2}=\dfrac{x\left(x+5\right)}{x\left(3x-2\right)}\) b)\(\dfrac{2x-1}{4}=\dfrac{\left(2x-1\right)\left(2x+1\right)}{8x+4}\) c)\(\dfrac{2x\left(x-2\right)}{x^2-4x+4}=\dfrac{2x}{x-2}\) d) \(\dfrac{5x^2+10x}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x}{x-2}\)