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Kaijo
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Uchiha
15 tháng 3 2020 lúc 21:39

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

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

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

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Uchiha
15 tháng 3 2020 lúc 21:42

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

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Uchiha
15 tháng 3 2020 lúc 21:46

3,1/x(y-x)-1/y(y-x)=y/xy(y-x)-x/xy(y-x)=(y-x)/xy(x-y)=1/xy

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Sakura서 정
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Nguyễn Lê Phước Thịnh
27 tháng 3 2020 lúc 10:37

a) Ta có: \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)

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

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

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

\(=\frac{x^2+5x+6}{2x\left(x+3\right)}\)

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

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

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

b) Ta có: \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)

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

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

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

c) Ta có: \(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)

\(=\frac{5\left(x+2\right)\cdot2\cdot\left(2-x\right)}{4\cdot\left(x-2\right)\cdot\left(x+2\right)}\)

\(=\frac{5\cdot2\cdot\left(2-x\right)}{-4\left(2-x\right)}=\frac{5\cdot2}{-4}=\frac{-5}{2}\)

d) Ta có: \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)

\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3x}{x\left(x+4\right)\cdot2\left(2-x\right)}\)

\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3}{2\left(x+4\right)\cdot\left(2-x\right)}=\frac{3\left(1-4x^2\right)}{2\left(-x^2-2x+8\right)}\)

\(=\frac{3-12x^2}{-2x^2-4x+16}\)

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Jeong Soo In
27 tháng 3 2020 lúc 10:35

a) \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)

\(=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne-3;x\ne0\right)\)

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

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

b) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne0;x\ne-3\right)\)

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

c) \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}\) \(\left(ĐKXĐ:x\ne\pm2\right)\)

\(=\frac{-5\left(x-2\right)}{2\left(x-2\right)}=\frac{-5}{2}\)

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lê nhật duẫn
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Nguyễn Ngọc Lộc
28 tháng 3 2020 lúc 19:25

ĐKXĐ bạn tự tìm nha : )

k, Ta có : \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}\)

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

j, Ta có : \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{x+y}{y-x}:\frac{x\left(x+y\right)}{3\left(x^2-y^2\right)}=\frac{x+y}{y-x}.\frac{3\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}\)

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

i, Ta có : \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right)}{-\left(a-b\right)}:\frac{a+b}{2\left(a^2-b^2\right)}=\frac{a\left(a+b\right)}{-\left(a-b\right)}.\frac{2\left(a-b\right)\left(a+b\right)}{a+b}\)

\(=\frac{2a\left(a+b\right)\left(a-b\right)}{-\left(a-b\right)}=-2a\left(a+b\right)\)

h, = k,

f, Ta có : \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{-3}{x-6}=\frac{-3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(x-6\right)}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)

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💋Amanda💋
28 tháng 3 2020 lúc 19:27
https://i.imgur.com/1LeIfCN.jpg
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Bùi Lan Anh
28 tháng 3 2020 lúc 19:29

a. \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right).2\left(2-x\right)}{4\left(x-2\right)\left(x+2\right)}=\frac{-5}{2}\)

b. \(\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{12x.15y^4}{5y^3.8x^3}=\frac{3.3y}{2x^2}=\frac{9y}{2x^2}\)

c. \(\frac{4y^2}{11x^4}.\left(\frac{-3x^2}{8y}\right)=\frac{4y^2.\left(-3x^2\right)}{11x^4.8y}=\frac{-3y}{22x^2}\)

d. \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}=\frac{\left(x-2\right)\left(x+2\right)\left(x+4\right)}{3\left(x+4\right).2\left(x-2\right)}=\frac{x+2}{6}\)

f. \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x+6\right)\left(x-6\right).3}{\left(2x+10\right)\left(6-x\right)}=\frac{-3x-18}{2x+10}\)

g. \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6}=\frac{\left(x^2-9y^2\right).3xy}{x^2y^2.\left(2x-6\right)}=\frac{3x^2-27y^2}{2x^2y-6xy}\)

h. \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right).3x}{x\left(x+4\right).2\left(1-2x\right)}=\frac{3+6x}{2x+8}\)

i. \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right).2\left(a-b\right)\left(a+b\right)}{\left(b-a\right)\left(a+b\right)}=-2a^2-2ab\)

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

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Kaijo
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Nguyễn Ngọc Linh
15 tháng 3 2020 lúc 20:52

1, \(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)\(=\frac{4y.y}{11x^2.x^2}.\frac{-3x^2}{2.4y}\)\(=\frac{y}{11x^2}.\frac{-3}{2}=\frac{-3y}{22x^2}\)

2, \(\frac{4x^2}{5y^2}:\frac{6x}{5y}:\frac{2x}{3y}\)\(=\frac{4x^2}{5y^2}.\frac{5y}{6x}.\frac{3y}{2x}\)\(=\frac{2x.2x}{5y.y}.\frac{5y}{3.2x}.\frac{3y}{2x}\)\(=\frac{2x}{y}.\frac{1}{3}.\frac{3y}{2x}\)

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

3, \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}\)\(=\frac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\frac{x+4}{2\left(x-2\right)}\)\(=\frac{\left(x+2\right)}{3}.\frac{1}{2}=\frac{x+2}{6}\)

4, \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\left(-\frac{2\left(x-2\right)}{x+2}\right)=\frac{5}{4}.\frac{-2}{1}=-\frac{5}{2}\)

5, \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{3}{-\left(x-6\right)}=\frac{x+6}{2\left(x+5\right)}.\frac{-3}{1}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)

6, \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}=\frac{\left(x-3y\right)\left(x+3y\right)}{\left(xy\right)^2}.\frac{3xy}{2\left(x-3y\right)}=\frac{x+3y}{xy}.\frac{3}{2}=\frac{3\left(x+3y\right)}{2xy}\)

7, \(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}=\frac{3\left(x-y\right)\left(x+y\right)}{5xy}.\frac{5xy.3x}{-2\left(x-y\right)}=\frac{3\left(x+y\right)}{1}.\frac{3x}{-2}=\frac{-9x\left(x+y\right)}{2}\)

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Nguyễn Ngọc Linh
15 tháng 3 2020 lúc 20:52

Làm rõ lâu.

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BoY
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Edogawa Conan
20 tháng 7 2020 lúc 10:26

\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}=-\frac{3\left(x+6\right)}{2x+10}=-\frac{3x+18}{2x+10}\)

\(\frac{x^2-4}{x^2-9}\cdot\frac{3x+9}{x+2}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{3\left(x+3\right)}{x+2}=\frac{3\left(x-2\right)}{x-3}\)

\(\frac{x^3-8}{5x+20}\cdot\frac{x^2+4x}{x^2+2x+4}=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}\cdot\frac{x\left(x+4\right)}{x^2+2x+4}=\frac{x\left(x-2\right)}{5}\)

\(\frac{4x+12}{\left(x+4\right)^2}:\frac{3x+9}{x+4}=\frac{4\left(x+3\right)}{\left(x+4\right)^2}\cdot\frac{x+4}{3\left(x+3\right)}=\frac{4}{3\left(x+4\right)}\)

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Kaijo
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Bùi Lan Anh
10 tháng 3 2020 lúc 20:10

Phép nhân và phép chia các đa thức

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nguyễn hoàng khải phong
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Ngô Vũ Quỳnh Dao
15 tháng 12 2017 lúc 10:08

a) \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)

b) \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}=\frac{3+6x}{2x+8}\)

Kaijo
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lê thị hương giang
15 tháng 3 2020 lúc 20:02

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

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

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

\(4,\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)

\(5,\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2x\left(x+4\right)}\)

\(6,\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{9y}{2x^2}\)

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Dô Dụng
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Võ Đông Anh Tuấn
28 tháng 8 2016 lúc 10:06

\(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{\left(5x+10\right).\left(4-2x\right)}{\left(4x-8\right).\left(x+2\right)}\)

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

                      \(=\frac{5\left(2-x\right)}{2\left(x-2\right)}=-\frac{5}{2}\)

Ngô Tấn Đạt
28 tháng 8 2016 lúc 16:29

\(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{\left(5x+10\right)\left(4-2x\right)}{\left(4x-8\right)\left(x+2\right)}\\ =\frac{5\left(x+2\right).2\left(2-x\right)}{4\left(x-2\right)\left(x+2\right)}\\ =\frac{5\left(2-x\right)}{2\left(x-2\right)}=-\frac{5}{2}\)