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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. 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: 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: \(2\left(x-1\right)\left(x+1\right)\left(x-5\right)\) \(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}\\
=\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}=\dfrac{2\left(x+1\right)\left(3x-6\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\) \(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x-5\right)\left(5x-5\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\) a) \(\dfrac{3x}{2x+4}\) và \(\dfrac{x+3}{x^2-4}\) Phân tích các mẫu thức thành nhân tử : \(2x+4 = 2(x+2)\) \(x^2 - 4 = (x-2)(x+2)\) MTC : \(2(x+2)(x-2)\) Nhân tử phụ của mẫu thức : \(2x + 4\) là \((x - 2)\) \(x^2 - 4\) là \(2\) QĐ: \(\dfrac{3x}{2x+4}=\dfrac{3x}{2\left(x+2\right)}=\dfrac{3x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}\) \(\dfrac{x+3}{x^2-4}=\dfrac{x+3}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x+3\right)}{2\left(x+2\right)\left(x-2\right)}\) b) \(\dfrac{x+5}{x^2+4x+4}\) và \(\dfrac{x}{3x+6}\) Phân tích các mẫu thức thành nhân tử : \(x^2+4x+4 = (x+2)^2\) \(3x + 6\) \(= 3(x+2)\) MTC : \(3(x+2)^2\) Nhân tử phụ của mẫu thức : \(x^2 + 4x +4 \) là \(3\) \(3x + 6\) là \((x+2)\) QĐ : \(\dfrac{x+5}{x^2+4x+4}=\dfrac{\left(x+5\right)}{\left(x+2\right)^2}=\dfrac{3\left(x+5\right)}{3\left(x+2\right)^2}\) \(\dfrac{x}{3x+6}=\dfrac{x}{3\left(x+2\right)}=\dfrac{x\left(x+2\right)}{3\left(x+2\right)^2}\) a) \(\frac{3x+6}{x^2-4}=\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3}{x-2}\)( ĐKXĐ : x ≠ ±2 ) \(\frac{2x+6}{x^3+3x^2-9x-27}=\frac{2\left(x+3\right)}{x^2\left(x+3\right)-9\left(x+3\right)}=\frac{2\left(x+3\right)}{\left(x+3\right)\left(x^2-9\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)( ĐKXĐ : x ≠ ±3 ) MTC : ( x - 2 )( x - 3 )( x + 3 ) => \(\hept{\begin{cases}\frac{3}{x-2}=\frac{3\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3\left(x^2-9\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3x-27}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\\\frac{2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{4x-4}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\end{cases}}\) b) \(\frac{x^2-4x+4}{2x^2-3x+1}=\frac{\left(x-2\right)^2}{2x^2-2x-x+1}=\frac{\left(x-2\right)^2}{2x\left(x-1\right)-\left(x-1\right)}=\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}\)( ĐKXĐ : \(\hept{\begin{cases}x\ne1\\x\ne\frac{1}{2}\end{cases}}\)) \(\frac{x+4}{2x-2}=\frac{x+4}{2\left(x-1\right)}\)( ĐKXĐ : x ≠ 1 ) MTC : \(2\left(x-1\right)\left(2x-1\right)\) => \(\hept{\begin{cases}\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}=\frac{2\left(x^2-4x+4\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2-8x+8}{2\left(x-1\right)\left(2x-1\right)}\\\frac{x+4}{2\left(x-1\right)}=\frac{\left(x+4\right)\left(2x-1\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2+7x-4}{2\left(x-1\right)\left(2x-1\right)}\end{cases}}\) c) \(\frac{6a}{a-b}\)( ĐKXĐ : a ≠ b ) ; \(\frac{2b}{b-a}=\frac{-2b}{a-b}\)( ĐKXĐ : a ≠ b) ; \(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\)( ĐKXĐ : a ≠ ±b ) MTC : \(\left(a-b\right)\left(a+b\right)\) => \(\frac{6a}{a-b}=\frac{6a\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{6a^2+6ab}{\left(a-b\right)\left(a+b\right)}\) \(\frac{-2b}{a-b}=\frac{-2b\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{-2ab-2b^2}{\left(a-b\right)\left(a+b\right)}\) \(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\) d) \(\frac{x}{x^2+11x+30}=\frac{x}{x^2+5x+6x+30}=\frac{x}{x\left(x+5\right)+6\left(x+5\right)}=\frac{x}{\left(x+5\right)\left(x+6\right)}\)( ĐKXĐ : x ≠ -5 ; x ≠ -6 ) \(\frac{5}{x^2+9x+20}=\frac{5}{x^2+4x+5x+20}=\frac{5}{x\left(x+4\right)+5\left(x+4\right)}=\frac{5}{\left(x+4\right)\left(x+5\right)}\)( ĐKXĐ : x ≠ -4 ; x ≠ -5 ) MTC : \(\left(x+4\right)\left(x+5\right)\left(x+6\right)\) => \(\hept{\begin{cases}\frac{x}{\left(x+5\right)\left(x+6\right)}=\frac{x\left(x+4\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{x^2+4x}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\\\frac{5}{\left(x+4\right)\left(x+5\right)}=\frac{5\left(x+6\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{5x+30}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\end{cases}}\) Sai chỗ nào bạn bỏ qua nhé a) Tìm MTC: 2x + 6 = 2(x + 3) x2 – 9 = (x – 3)(x + 3) MTC = 2(x – 3)(x + 3) = 2(x2 – 9) Nhân tử phụ: 2(x – 3)(x + 3) : 2(x + 3) = x – 3 2(x – 3)(x + 3) : (x2 – 9) = 2 Qui đồng: b) Tìm MTC: x2 – 8x + 16 = (x – 4)2 3x2 – 12x = 3x(x – 4) MTC = 3x(x – 4)2 Nhân tử phụ: 3x(x – 4)2 : (x – 4)2 = 3x 3x(x – 4)2 : 3x(x – 4) = x – 4 Qui đồng:
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CM
Cao Minh Tâm
15 tháng 7 2017
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