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27 tháng 11 2020

a, \(\frac{3x}{2x+4};\frac{x+3}{x^2-4}\)

Ta có : \(2x+4=2\left(x+2\right)\)

\(x^2-4=\left(x-2\right)\left(x+2\right)\)

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

\(\frac{3x}{2x+4}=\frac{3x}{2\left(x+2\right)}=\frac{3x\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}=\frac{3x^2-6x}{2\left(x-2\right)\left(x+2\right)}\)

\(\frac{x+3}{x^2-4}=\frac{x+3}{\left(x-2\right)\left(x+2\right)}=\frac{2x+6}{\left(x-2\right)\left(x+2\right)}\)

27 tháng 11 2020

c, \(\frac{2x}{x^2-8x+16};\frac{x}{3x^2-12x}\)

Ta có : \(x^2-8x+16=\left(x-4\right)^2\)

\(3x^2-12x=3x\left(x-4\right)\)

MTC : \(3x\left(x-4\right)^2\)

\(\frac{2x}{x^2-8x+16}=\frac{2x}{\left(x-4\right)^2}=\frac{6x^2}{3x\left(x-4\right)^2}\)

\(\frac{x}{3x^2-12x}=\frac{x}{3x\left(x-4\right)}=\frac{x^2+4x}{3x\left(x-4\right)\left(x+4\right)}\)

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

\(\dfrac{2x-3}{2x^2-18}=\dfrac{2x-3}{2\left(x-3\right)\left(x+3\right)}=\dfrac{\left(2x-3\right)\cdot\left(2x-3\right)}{2\left(2x-3\right)\left(x-3\right)\left(x+3\right)}\)

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

\(\dfrac{2}{2x^2+3x-9}=\dfrac{2}{\left(x+3\right)\left(2x-3\right)}=\dfrac{2\cdot2\cdot\left(x-3\right)}{2\left(x-3\right)\cdot\left(x+3\right)\left(2x-3\right)}\)

\(=\dfrac{4x-12}{2\left(x-3\right)\left(x+3\right)\left(2x-3\right)}\)

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

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

28 tháng 7 2023

Mik cảm ơn ạ

26 tháng 11 2021

rzddddddfg

1 tháng 7 2017

MTC : \(150\left(x-2\right)\left(x-3\right)\)

\(\frac{5}{2x-4}=\frac{5}{2\left(x-2\right)}=\frac{5.3.\left(-25\right)\left(x-3\right)}{2.3.\left(-25\right)\left(x-2\right)\left(x-3\right)}=\frac{375\left(x-3\right)}{150\left(x-2\right)\left(x-3\right)}\)

\(\frac{z}{3x-9}=\frac{z}{3\left(x-3\right)}=\frac{z.2.\left(-25\right).\left(x-2\right)}{3.2.\left(-25\right)\left(x-3\right)\left(x-2\right)}=\frac{-50z\left(x-2\right)}{150\left(x-2\right)\left(x-3\right)}\)

\(\frac{7}{50-25x}=\frac{7}{-25\left(x-2\right)}=\frac{7.2.3.\left(x-3\right)}{-25.2.3\left(x-2\right)\left(x-3\right)}=\frac{42\left(x-3\right)}{150\left(x-2\right)\left(x-3\right)}\)

1 tháng 7 2017

Giúp mk đi maf

16 tháng 11 2016

phân thức 1 :3x(x^2-4)/(2x+4)(x^2-4)=? ( tính ra )
phan thức 2 ;(2x+4)(x+3)/(2x+4)(x^2-4)=?(tính ra )

kết quả tính dc là phrp1 qd của 2 phân thức đó

29 tháng 11 2018

\(2x+4=2\left(x+2\right)\)

\(x^2-4=x^2-2^2=\left(x+2\right)\left(x-2\right)\)

\(\Rightarrow MTC=2\left(x+2\right)\left(x-2\right)\)

\(\frac{3x}{2x+4}=\frac{3x}{2\left(x+2\right)}=\frac{3x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}\)

\(\frac{x-3}{x^2-4}=\frac{x-3}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x-3\right)}{2\left(x+2\right)\left(x-2\right)}\)

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

\(2x+2y=2\left(x+y\right)=\dfrac{2\left(x+y\right)\cdot x\left(y+3\right)\left(x+y\right)^2}{x\left(y+3\right)\left(x+y\right)^2}\)

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

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

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

\(\dfrac{1}{x^2+2xy+y^2}=\dfrac{1}{\left(x+y\right)^2}=\dfrac{6}{6\left(x+y\right)^2}\)

23 tháng 1

đề viết là 1/2y+2x mà bạn