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8 tháng 3 2020

a) \(\frac{8xy}{3x-1}:\frac{12xy^3}{5-15x}\)

\(=\frac{8xy}{3x-1}.\frac{5-15x}{12xy^3}\)

\(=\frac{2}{3x-1}.\frac{5\left(1-3x\right)}{3y^2}\)

\(=\frac{10}{3y^2}\)

b) \(\frac{2x+1}{x-2}:\left(-\frac{2x-1}{x-2}\right)\)

\(=\frac{2x+1}{x-2}.\frac{x-2}{1-2x}=\frac{2x+1}{1-2x}\)

8 tháng 3 2020

=.=, làm nhanh lẫn

a) \(=\frac{8xy}{3x-1}.\frac{5\left(3x-1\right)}{-12xy^3}\)

\(=\frac{-10}{3y^2}\)

11 tháng 12 2019

a) \(\frac{3x-2}{2xy}-\frac{7x-4}{2xy}\)

\(=\frac{3x-2}{2xy}+\frac{-\left(7x-4\right)}{2xy}\)

\(=\frac{3x-2-7x+4}{2xy}\)

\(=\frac{-4x+2}{2xy}\)

\(=\frac{2.\left(-2x+1\right)}{2xy}.\)

\(=\frac{-2x+1}{xy}.\)

b) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}\)

Ta có:

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

\(2x+8=2.\left(x+4\right)\)

\(MTC:2x.\left(x+4\right)\)

\(\frac{6}{x^2+4x}+\frac{3}{2x+8}\)

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

\(=\frac{6.2}{2x.\left(x+4\right)}+\frac{3x}{2x.\left(x+4\right)}\)

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

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

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

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

\(=\frac{3}{2x}.\)

Chúc bạn học tốt!

11 tháng 12 2019

Không có gì. hihi

14 tháng 2 2020

a.=\(\frac{7x+2}{3xy^2}.\frac{x^2y}{14x+4}\)

=\(\frac{7x+2}{3y}.\frac{x^2y}{2\left(7x+2\right)}\)

=\(\frac{1}{3y}.\frac{x}{2}\)

=\(\frac{x}{6y}\)

b.=\(\frac{8xy}{3x-1}.\frac{5-15x}{12xy^3}\)

=\(\frac{2}{3x-1}.\frac{-15x+5}{3y^2}\)

=\(\frac{2}{3x-1}.\frac{-5\left(3x-1\right)}{3y^2}\)

=\(\frac{-10}{3y^2}\)

c.=\(\frac{3\left(x^3+1\right)}{x-1}.\frac{1}{x^2-x+1}\)

=\(\frac{3\left(x+1\right).\left(x^2-x+1\right)}{x-1}.\frac{1}{x^2-x+1}\)

=\(\frac{3x+3}{x-1}\)

d.=\(\frac{4\left(x+3\right)}{.\left(3x-1\right)}.\frac{1-3x}{x^2+3x}\)

=\(\frac{4\left(x+3\right)}{x.\left(3x-1\right)}.\frac{-\left(3x-1\right)}{x\left(x+3\right)}\)

=\(\frac{-4}{x^2}\)

e.=\(\frac{2\left(2x+3y\right)}{x-1}.\frac{1-x^3}{4x^2+12xy+9y^2}\)

=\(2.\frac{-\left(1+x+x^2\right)}{2x+3y}\)

=\(-\frac{2x^2+2x+2}{2x+3y}\)

14 tháng 2 2020

Phần C thiếu x3 , chỗ (x-1)

14 tháng 8 2020

a)\(ĐKXĐ:x\ne0;-1\)

Ta có:\(\frac{x^3+1}{x}.\left(\frac{1}{x+1}+\frac{x-1}{x^2-x+1}\right)=\frac{x^3+1}{x}.\frac{\left(x^2-x+1\right)+\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{x^3+1}{x}.\frac{x^2-x+1+\left(x^2-1\right)}{x^3+1}=\frac{2x^2-x}{x}=\frac{2x\left(x-1\right)}{x}=2\left(x-1\right)\)

30 tháng 11 2016

a. 2x

b.\({3x}\over x^2-1\)

17 tháng 3 2020

\(\frac{15x-10}{x^2+3}=0\)

\(\Leftrightarrow\frac{5\left(3x-2\right)}{x^2+3}=0\)

\(\Leftrightarrow5\left(3x-2\right)=0\)

\(\Leftrightarrow3x-2=0\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\frac{2}{3}\)

...

17 tháng 3 2020

What's wrong???

21 tháng 3 2020

a) \(\frac{15x-10}{x^2+3}=0\)

<=> 15x - 10 = 0

<=> 5(3x - 2) = 0

<=> 3x - 2 = 0

<=> 3x = 2

<=> x = 2/3

b) ĐKXĐ: \(x\ne1;x\ne-3\)

<=>\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}-\frac{8}{x^2+2x-3}=0\)

<=> \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}-\frac{8}{\left(x-1\right)\left(x+3\right)}=0\)

<=> (3x - 1)(x + 3) - (2x + 5)(x - 1) - 8 = (x - 1)(x + 3)

<=> 3x2 + 9x - x - 3 - 2x2 + 2x - 5x + 5 - 8 = 0

<=> x2 + 5x - 6 = 0

<=> (x - 1)(x + 6) = 0

<=> x - 1 = 0 hoặc x + 6 = 0

<=> x = 1 (ktm) hoặc x = -6 (tm)

=> x = -6

3 tháng 8 2018

|x-2|<3

|x+4|<5

|3+x|>2

really your name is crazy

and do you crazy

22 tháng 11 2019

a)\(\frac{2x^3-x^2-2x-1}{x^3+3x^2-x-3}=\frac{2x^3-\left(x^2+2x+1\right)}{x^2\left(x+3\right)-\left(x+3\right)}=\frac{2x^3-\left(x+1\right)^2}{\left(x+3\right)\left(x^2-1\right)}=\frac{2x^3-\left(x+1\right)^2}{\left(x+3\right)\left(x-1\right)\left(x+1\right)}=\frac{2x^3-\left(x+1\right)}{\left(x+3\right)\left(x-1\right)}\)b)

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

c)

\(\frac{9x^2y^2+3x^2}{12xy^5+4xy^3}=\frac{3x^2\left(3y^2+1\right)}{4xy^3\left(3y^2+1\right)}=\frac{3x^2}{4xy^3}\)

@@@@ỦNG HỘ NHOA@@@@

23 tháng 11 2019

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