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PN
25 tháng 11 2015
\(a.\) Ta có:
\(MTC:\) \(\left(x+1\right)\left(x+2\right)\)
Do đó
\(\frac{3x}{x+1}=\frac{3x\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}\)
\(\frac{x+4}{x+2}=\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x+2\right)}\)
\(b.\) Ta có:
\(x^2+x=x\left(x+1\right)\)
\(x^2-1=\left(x-1\right)\left(x+1\right)\)
nên \(MTC:\) \(x\left(x-1\right)\left(x+1\right)\)
Do đó:
\(\frac{5}{x^2+x}=\frac{5}{x\left(x+1\right)}=\frac{5\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(\frac{6}{x^2-1}=\frac{6}{\left(x-1\right)\left(x+1\right)}=\frac{6x}{x\left(x-1\right)\left(x+1\right)}\)
\(c.\) Ta có:
\(x^2-5x+4=x^2-x-4x+4=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
\(2x^2-8x=2x\left(x-4\right)\)
nên \(MTC:\) \(2x\left(x-1\right)\left(x-4\right)\)
Do đó:
\(\frac{4}{x^2-5x+4}=\frac{4}{\left(x-1\right)\left(x-4\right)}=\frac{8x}{2x\left(x-1\right)\left(x-4\right)}\)
\(\frac{x+1}{2x^2-8x}=\frac{x+1}{2x\left(x-4\right)}=\frac{\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x-4\right)}\)
27 tháng 11 2020
Làm nốt d :P
\(\frac{x+3}{2x^2-15x-8};\frac{3}{x^2-8x}\)
Ta có : \(2x^2-15x-8=\left(2x+1\right)\left(x-8\right)\)
\(x^2-8x=x\left(x-8\right)\)
MTC : \(x\left(x-8\right)\left(2x+1\right)\)
\(\frac{x+3}{2x^2-15x-8}=\frac{x+3}{\left(2x+1\right)\left(x-8\right)}=\frac{x^2+3x}{x\left(x-8\right)\left(2x+1\right)}\)
\(\frac{3}{x^2-8x}=\frac{3}{x\left(x-8\right)}=\frac{6x+3}{x\left(x-8\right)\left(2x+1\right)}\)
H9
HT.Phong (9A5)
CTVHS
29 tháng 9 2023
a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
\(\dfrac{3}{4-2x}=\dfrac{-3}{2\left(x-2\right)}=\dfrac{-3\left(x^2+2x+4\right)}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
b) \(\dfrac{x}{x^2-1}=\dfrac{x}{\left(x+1\right)\left(x-1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\)
\(\dfrac{1}{x^2+2x+1}=\dfrac{1}{\left(x+1\right)^2}=\dfrac{x-1}{\left(x+1\right)^2\left(x-1\right)}\)
c) \(\dfrac{1}{x+2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)
d) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
LY
3
23 tháng 11 2020
MTC : ( x - 1 )( x2 + x + 1 )
Ta có : \(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{6x^2+6x+6}{\left(x-1\right)\left(x^2+x+1\right)}\)
25 tháng 11 2020
Hnay mới học thì hnay trả lời nhá :P
\(\frac{4x^2-3x+5}{x^3-1};\frac{2x}{x^2+x+1}\)
Ta có : \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(x^2+x+1=x^2+x+1\)
MTC : \(\left(x-1\right)\left(x^2+x+1\right)\)
\(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
TN
0
27 tháng 11 2021
\(9,\dfrac{2}{x^2-2x}=\dfrac{6}{3x\left(x-2\right)};\dfrac{x}{3x-6}=\dfrac{x^2}{3x\left(x-2\right)}\\ 10,\dfrac{x}{x-5}=\dfrac{x}{x-5};x+1=\dfrac{\left(x+1\right)\left(x-5\right)}{x-5}\\ 11,-3=\dfrac{-3\left(x^2+x+5\right)}{x^2+x+5}\\ 12,\dfrac{x}{2x-8}=\dfrac{x^2}{2x\left(x-4\right)};\dfrac{x+1}{4x-x^2}=\dfrac{-2\left(x+1\right)}{2x\left(x-4\right)}\)
14 tháng 11 2017
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:
11 tháng 10 2023
a: \(\dfrac{3}{x-1}=\dfrac{3\cdot9}{9\cdot\left(x-1\right)}=\dfrac{27}{9\left(x-1\right)}\)
\(\dfrac{4}{3x-3}=\dfrac{12}{9x-9}=\dfrac{12}{9\left(x-1\right)}\)
\(\dfrac{10}{9-9x}=\dfrac{-10}{9x-9}=-\dfrac{10}{9\left(x-1\right)}\)
b: \(\dfrac{3}{2\left(x-3\right)}=\dfrac{3x-9}{2\left(x-3\right)^2}\)
\(\dfrac{3x-2}{x^2-6x+9}=\dfrac{6x-4}{2\left(x-3\right)^2}\)
c: \(\dfrac{3}{x^2+2x+1}=\dfrac{3}{\left(x+1\right)^2}=\dfrac{3x}{x\left(x+1\right)^2}\)
\(-\dfrac{2}{x^2+x}=\dfrac{-2}{x\left(x+1\right)}=\dfrac{-2\left(x+1\right)}{x\left(x+1\right)^2}\)
\(\frac{10}{x+2};\frac{5}{2x-4};\frac{1}{6-3x}\)
Ta có : \(x+2=x+2\)
\(2x-4=2\left(x-2\right)\)
\(6-3x=3\left(2-x\right)=-3\left(x-2\right)\)
MTC : \(-6\left(x-2\right)\left(x+2\right)\)
\(\frac{10}{x+2}=\frac{10.\left(-6\right)\left(x-2\right)}{-6\left(x-2\right)\left(x+2\right)}=\frac{-60x+120}{-6\left(x-2\right)\left(x+2\right)}\)
\(\frac{5}{2x-4}=\frac{5}{2\left(x-2\right)}=\frac{5.\left(-3\right)\left(x+2\right)}{-6\left(x-2\right)\left(x+2\right)}=\frac{-15x-30x}{-6\left(x-2\right)\left(x+2\right)}\)
\(\frac{1}{6-3x}=\frac{1}{-3\left(x-2\right)}=\frac{2\left(x+2\right)}{-6\left(x-2\right)\left(x+2\right)}=\frac{2x+4}{-6\left(x-2\right)\left(x+2\right)}\)