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Kaijo
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Ánh Tuyết
16 tháng 3 2020 lúc 8:42

a, \(\frac{4x+1}{2}-\frac{3x+2}{3}=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{12x+3-6x-4}{6}=\frac{6x-1}{6}\)

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

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

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

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

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๖²⁴ʱTú❄⁀ᶦᵈᵒᶫ
16 tháng 3 2020 lúc 8:44

\(\frac{4x+1}{2}-\frac{3x+2}{3}\)

\(=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}\)

tương tự đến hết nha a hay cj gì đps ! 

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✰๖ۣۜŠɦαɗøω✰
16 tháng 3 2020 lúc 8:57

a) \(\frac{4.x+1}{2}-\frac{3.x+2}{3}=\frac{3.\left(4.x+1\right)-2.\left(3.x+2\right)}{6}\)

                                                \(=\frac{12.x+3-6.x-4}{6}\)

                                                   \(=\frac{6.x-1}{6}\)

b)\(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Phạm Tú Uyên
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Akai Haruma
12 tháng 8 2020 lúc 9:21

Các ĐKXĐ: bạn tự tìm

a)

\(\frac{11x+10}{3x-3}+\frac{15x+13}{4-4x}=\frac{11x+10}{3(x-1)}-\frac{15x+13}{4(x-1)}=\frac{4(11x+10)-3(15x+13)}{12(x-1)}\)

\(=\frac{-x+1}{12(x-1)}=\frac{-(x-1)}{12(x-1)}=\frac{-1}{12}\)

b)

\(\frac{5x+3}{x^2-3x}+\frac{9-x}{9-3x}=\frac{5x+3}{x(x-3)}+\frac{x-9}{3x-9}=\frac{5x+3}{x(x-3)}+\frac{x-9}{3(x-3)}\)

\(=\frac{3(5x+3)}{3x(x-3)}+\frac{x(x-9)}{3x(x-3)}=\frac{x^2+6x+9}{3x(x-3)}=\frac{(x+3)^2}{3x(x-3)}\)

c)

\(\frac{4xy-1}{5x^2y}-\frac{2xy-1}{5x^2y}=\frac{(4xy-1)-(2xy-1)}{5x^2y}=\frac{2xy}{5x^2y}=\frac{2}{5x}\)

Akai Haruma
12 tháng 8 2020 lúc 9:24

d)

$\frac{x+8}{x^2-16}-\frac{2}{x^2+4x}=\frac{x+8}{(x-4)(x+4)}-\frac{2}{x(x+4)}$

$=\frac{x(x+8)}{x(x-4)(x+4)}-\frac{2(x-4)}{x(x+4)(x-4)}$

$=\frac{x^2+8x-2(x-4)}{x(x+4)(x-4)}=\frac{x^2+6x+8}{x(x+4)(x-4)}$

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

$\frac{x^2-49}{2x+1}.\frac{3}{7-x}=\frac{(x-7)(x+7)}{2x+1}.\frac{-3}{x-7}$

$=\frac{-3(x+7)}{2x+1}$

Akai Haruma
12 tháng 8 2020 lúc 9:30

f)

$\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{(2-3x)^3}$

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

$=\frac{x(x^2+1)}{(2-3x)^2}$
g)

$\frac{5xy}{2x-3}:\frac{15xy^3}{12-8x}=\frac{5xy}{2x-3}.\frac{12-8x}{15xy^3}$

$=\frac{5xy}{2x-3}.\frac{-4(2x-3)}{15xy^3}=\frac{-4}{3y^2}$

h)

$\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}=\frac{x(x+2)}{3(x-1)^2}:\frac{2(x+2)}{5(x-1)}$

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

$=\frac{5x}{6(x-1)}$

Hoàng Linh Nhi
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Kaijo
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๖²⁴ʱTú❄⁀ᶦᵈᵒᶫ
16 tháng 3 2020 lúc 9:06

\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)

\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)

\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)

\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)

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๖²⁴ʱTú❄⁀ᶦᵈᵒᶫ
16 tháng 3 2020 lúc 9:13

chỗ cuối tớ sai 

\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)

đây nha , e xin lỗi

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KAl(SO4)2·12H2O
16 tháng 3 2020 lúc 15:27

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

                                                          \(=\frac{3\left(x-1\right)+\left(2x-1\right)-2.2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x+1\right)}\)

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

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

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

b) \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)

                                                   \(=\frac{3x.10\left(x-y\right)-x.5\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{30x\left(x-y\right)+5x\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{5x\left[6\left(x-y\right)-\left(x+y\right)\right]}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{5x\left(5x-7y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{x\left(5x-7y\right)}{10\left(x-y\right)\left(x+y\right)}\)

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

                                                \(=\frac{5x^2-y-3x^2+2xy}{xy}\)

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

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

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

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

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

                                            \(=\frac{2}{2x}=\frac{1}{x}\) 

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Triệu Bảo Ngọc
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shi nit chi
6 tháng 11 2016 lúc 22:02

mk ko biết làm 

xin lỗi bn nhae

xin lỗi vì đã ko giúp được bn

chcus bn học gioi!

nhae@@@

hoang phuc
6 tháng 11 2016 lúc 22:06

mình không biết làm

tk nhé@@@@@@@@@@@@@@@@@@@@

LOL

hihi

Trương Minh Trọng
3 tháng 7 2017 lúc 16:47

a) ... \(=\frac{3\left(2x-1\right)+2x\left(3x+3\right)+2x^2+1}{2x\left(2x-1\right)}=\frac{6x-3+6x^2+6x+2x^2+1}{2x\left(2x-1\right)}\)

\(=\frac{8x^2+12x-2}{2x\left(2x-1\right)}=\frac{4x^2+6x-1}{x\left(2x-1\right)}\)(hình như hết đơn giản được rồi, kết quả tạm vậy bạn nhé!)

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

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

c) ... \(=\frac{4\left(x-2\right)+2\left(x+2\right)-5x+6}{\left(x-2\right)\left(x+2\right)}=\frac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x-2}\)

Long Nguyễn
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Phương An
1 tháng 12 2016 lúc 18:06

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

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

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

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

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

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

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

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

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

\(=\frac{1}{x}\)

Nguyễn Thanh Thảo
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Nguyễn Thanh Thảo
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Kim TaeHyung
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o0o I am a studious pers...
20 tháng 12 2016 lúc 15:03

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

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

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

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

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

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