a)

\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}=\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}\)

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

MTC: \(2\left(x-1\right)\left(x+1\right)\left(x-5\right)\)

\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}\\ =\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}=\dfrac{2\left(x+1\right)\left(3x-6\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)

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

M.n jup mk câu b,c vs ạk

Bạn siêng thật !!!

Bài giải

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

\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=-\dfrac{8}{x\left(x-2\right)}=-\dfrac{8.\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)

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

\(\dfrac{x^4}{x^2-1}\) giữ nguyên.

c) \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3.y}{\left(x-y\right)^3.y}=\dfrac{x^3y}{y\left(x-y\right)^3}\)

\(\dfrac{x}{y^2-xy}=\dfrac{x}{y.\left(y-x\right)}=-\dfrac{x}{y.\left(x-y\right)}=-\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right).\left(x-y\right)^2}=\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right)^3}\)

a) Tìm MTC: x3 – 1 = (x – 1)(x2 + x + 1)

Nên MTC = (x – 1)(x2 + x + 1)

Nhân tử phụ:

(x3 – 1) : (x3 – 1) = 1

(x – 1)(x2 + x + 1) : (x2 + x + 1) = x – 1

(x – 1)(x2+ x + 1) : 1 = (x – 1)(x2 + x + 1)

Qui đồng:

b) Tìm MTC: x + 2

2x – 4 = 2(x – 2)

6 – 3x = 3(2 – x)

MTC = 6(x – 2)(x + 2)

Nhân tử phụ:

6(x – 2)(x + 2) : (x + 2) = 6(x – 2)

6(x – 2)(x + 2) : 2(x – 2) = 3(x + 2)

6(x – 2)(x + 2) : -3(x – 2) = -2(x + 2)

Qui đồng:

Bạn giỏi quá !!!

1)

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

MTC: \(\left(x-1\right)\left(x^2+x+1\right)\)

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

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

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

?1 . Có . Mẫu thức chung : 12x²y³z đơn giản hơn

?2 . \(\dfrac{3}{x^2-5x}=\dfrac{3}{x\left(x-5\right)}=\dfrac{6}{2x\left(x-5\right)}\)

\(\dfrac{5}{2x-10}=\dfrac{5}{2\left(x-5\right)}=\dfrac{5x}{2x\left(x-5\right)}\)

?3 . \(\dfrac{3}{x^2-5x}=\dfrac{3}{x\left(x-5\right)}=\dfrac{6}{2x\left(x-5\right)}\)

\(\dfrac{-5}{10-2x}=\dfrac{5}{2x-10}=\dfrac{5}{2\left(x-5\right)}=\dfrac{5x}{2x\left(x-5\right)}\)

a)

\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{-\left(x-y\right)}{2x^2-8y^2}=\dfrac{-\left(x-y\right)}{2\left(x^2-4y^2\right)}=\dfrac{-\left(x-y\right)}{2\left(x-2y\right)\left(x+2y\right)}\)

MTC: \(10x\left(x-2y\right)\left(x+2y\right)\)

\(\dfrac{7}{5x}=\dfrac{7.2\left(x-2y\right)\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{14\left(x-2y\right)\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)

\(\dfrac{4}{x-2y}=\dfrac{4.10x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{40x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)

\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{-\left(x-y\right)}{2x^2-8y^2}=\dfrac{-\left(x-y\right)}{2\left(x^2-4y^2\right)}=\dfrac{-\left(x-y\right)}{2\left(x-2y\right)\left(x+2y\right)}=\dfrac{-\left(x-y\right).5x}{10x\left(x-2y\right)\left(x+2y\right)}\)

\(\)

b)

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

MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)

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

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

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