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Khách
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27 tháng 11 2021
\(9,\dfrac{2}{x^2-2x}=\dfrac{6}{3x\left(x-2\right)};\dfrac{x}{3x-6}=\dfrac{x^2}{3x\left(x-2\right)}\\ 10,\dfrac{x}{x-5}=\dfrac{x}{x-5};x+1=\dfrac{\left(x+1\right)\left(x-5\right)}{x-5}\\ 11,-3=\dfrac{-3\left(x^2+x+5\right)}{x^2+x+5}\\ 12,\dfrac{x}{2x-8}=\dfrac{x^2}{2x\left(x-4\right)};\dfrac{x+1}{4x-x^2}=\dfrac{-2\left(x+1\right)}{2x\left(x-4\right)}\)
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 : \(\left(x-1\right)\left(x^2+x+1\right)\)
\(\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}\)
29 tháng 7 2021
cho mình hỏi là giữa khác phân số với nhua là phải có dấu như là công, trừ, nhân hay chia chứ?
NT
0
22 tháng 2 2022
a: \(\dfrac{-7}{x^2-4}=\dfrac{-7}{\left(x-2\right)\left(x+2\right)}=\dfrac{-14}{2\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{11}{2x+4}=\dfrac{11}{2\left(x+2\right)}=\dfrac{11\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}\)
b: \(\dfrac{2}{9x^2-1}=\dfrac{2}{\left(3x-1\right)\left(3x+1\right)}\)
\(\dfrac{4x}{1-3x}=\dfrac{-4x}{3x-1}=\dfrac{-4x\left(3x+1\right)}{\left(3x-1\right)\left(3x+1\right)}\)
c: \(\dfrac{3}{x+2}=\dfrac{6\left(x^2-2x+4\right)}{2\left(x+2\right)\left(x^2-2x+4\right)}\)
\(\dfrac{x+1}{x^3+8}=\dfrac{2x+2}{2\left(x+1\right)\left(x^2-2x+4\right)}\)
\(\dfrac{x+2}{2\left(x+2\right)}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{2\left(x+2\right)\left(x^2-2x+4\right)}\)
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:
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) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
AH
Akai Haruma
Giáo viên
13 tháng 12 2021
Lời giải:
$\frac{4x^2-3x+8}{x^3-1}$
$\frac{2x}{x^2+x+1}=\frac{2x(x-1)}{(x-1)(x^2+x+1)}=\frac{2x^2-2x}{x^3-1}$
$\frac{6}{1-x}=\frac{-6(x^2+x+1)}{(x-1)(x^2+x+1)}=\frac{-6x^2-6x-6}{x^3-1}$