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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: \(2\left(x-1\right)\left(x+1\right)\left(x-5\right)\)
\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}\\ =\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}=\dfrac{2\left(x+1\right)\left(3x-6\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)
\(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x-5\right)\left(5x-5\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)
Bài 1 :
a, \(\left(a-2\right)^2-b^2=\left(a-2-b\right)\left(a-2+b\right)\)
b, \(2a^3-54b^3=2\left(a^3-27b^3\right)=2\left(a-3b\right)\left(a^2+3ab+9b\right)\)
Bài 2 : tự kết luận nhé, ngại mà lười :(
a, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
\(\Leftrightarrow\frac{4x-3}{5}-\frac{5x-4}{3}=\frac{6x-2}{7}+3\)
\(\Leftrightarrow\frac{12x-9-25x+20}{15}=\frac{6x-2+21}{7}\)
\(\Leftrightarrow\frac{-13x-29}{15}=\frac{6x+19}{7}\Rightarrow-91x-203=90x+285\)
\(\Leftrightarrow181x=-488\Leftrightarrow x=-\frac{488}{181}\)
b, \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4x+8+9\left(2x-1\right)}{12}-\frac{10x-6}{12}=\frac{12x+5}{12}\)
\(\Rightarrow4x+8+18x-9-10x+6=12x+5\)
\(\Leftrightarrow12x+5=12x+5\Leftrightarrow0x=0\)
Vậy phương trình có vô số nghiệm
c, \(\left|2x-3\right|=4\)
Với \(x\ge\frac{3}{2}\)pt có dạng : \(2x-3=4\Leftrightarrow x=\frac{7}{2}\)
Với \(x< \frac{3}{2}\)pt có dạng : \(2x-3=-4\Leftrightarrow x=-\frac{1}{2}\)
d, \(\left|3x-1\right|-x=2\Leftrightarrow\left|3x-1\right|=x+2\)
Với \(x\ge\frac{1}{3}\)pt có dạng : \(3x-1=x+2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Với \(x< \frac{1}{3}\)pt có dạng : \(3x-1=-x-2\Leftrightarrow4x=-1\Leftrightarrow x=-\frac{1}{4}\)
\(a,\dfrac{x^3-2^3}{x^2-4}=\dfrac{\left(x-2\right)\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-2x+4}{x+2}\\ b,\dfrac{2}{2x-4}=\dfrac{2}{2\left(x-2\right)}=\dfrac{1}{x-2}\\ \dfrac{3}{3x-6}=\dfrac{3}{3\left(x-2\right)}=\dfrac{1}{x-2}\)
