\(^2\) + \(\dfrac{y}{4}\))\(^3\)
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5 tháng 9 2022

1) (3x2 + \(\dfrac{y}{4}\) )3 = (3x2)3 + 3.(3x2)2\(\dfrac{y}{4}\) + 3.3x2 . \(\dfrac{y^2}{16}\) + \(\dfrac{y^3}{64}\)

2) (4 - 26)3 = 43 - 3.42.26 + 3.4.262 - 263

3) (2a + 3b)2 = (2a)+ 2.2a.3b + (3b)2

4) (\(\dfrac{b}{2}\) - 5).(\(\dfrac{b}{2}\) + 5) = \(\left(\dfrac{b}{2}\right)^2\) - 52

1: \(\Leftrightarrow\left(x+2\right)\left(x-2\right)+3\left(x+1\right)=3+x^2-x-2\)

\(\Leftrightarrow x^2-x+1=x^2-4+3x+3=x^2+3x-1\)

=>-4x=-2

hay x=1/2

2: \(\Leftrightarrow\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)

\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

\(\Leftrightarrow2x^2+23x+61=2x^2+2x+11\)

=>21x=-50

hay x=-50/21

3: \(\Leftrightarrow6\left(x-8\right)+\left(x+2\right)\left(x-5\right)=-18-\left(x-5\right)\left(x-8\right)\)

\(\Leftrightarrow6x-48+x^2-3x-10+18+x^2-13x+40=0\)

\(\Leftrightarrow2x^2-10x=0\)

=>2x(x-5)=0

=>x=0(nhận) hoặc x=5(loại)

28 tháng 6 2017

Phép nhân các phân thức đại số

a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)

\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}\)

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

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

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

\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)

7 tháng 9 2018

B1:a)(3x-5)2-(3x+1)2=8

[(3x-5)+(3x+1)].[(3x-5)-(3x+1)]=8

(3x-5+3x+1)(3x-5-3x-1)=8

9x2-15x-9x2-3x-15x+25+15x+5+9x2-15x-9x2-3x+3x-5-3x-1=8

-36x+24=8

-36x=8-24=16

x=16:(-36)=\(\dfrac{-4}{9}\)

Bài 5: 

a: \(=\left(xy-u^2v^3\right)\left(xy+u^2v^3\right)\)

b: \(=\left(2xy^2-3xy^2+1\right)\left(2xy^2+3xy^2-1\right)\)

\(=\left(1-xy^2\right)\left(5xy^2-1\right)\)

Bài 6:

a: \(\left(a+b+c-d\right)\left(a+b-c+d\right)\)

\(=\left(a+b\right)^2+\left(c-d\right)^2\)

\(=a^2+2ab+b^2+c^2-2cd+d^2\)

b: \(\left(a+b-c-d\right)\left(a-b+c-d\right)\)

\(=\left(a-d\right)^2-\left(b-c\right)^2\)

\(=a^2-2ad+d^2-b^2+2bc-c^2\)

a: \(=\dfrac{5}{2}x-2x+\dfrac{7}{2}=\dfrac{1}{2}x+\dfrac{7}{2}\)

b: \(=\dfrac{-1}{4}x^4-3x^2+\dfrac{9}{4}x\)

c: \(=\dfrac{1}{5}x+\dfrac{1}{15}xy+\dfrac{7}{10}x^2\)

d: \(=-9x^3-1-12y+27xy\)

Bài 1: Thực hiện phép tính a, \(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}\)+\(\dfrac{2}{x^2+3}\)+\(\dfrac{1}{x+1}\) b, \(\dfrac{x+y}{2\left(x-y\right)}\)-\(\dfrac{x-y}{2\left(x+y\right)}\)+\(\dfrac{2y^2}{x^2-y^2}\) c, \(\dfrac{x-1}{x^3}\)-\(\dfrac{x+1}{x^3-x^2}\)+\(\dfrac{3}{x^3-2x^2+x}\) d, \(\dfrac{xy}{ab}\)+\(\dfrac{\left(x-a\right)\left(y-a\right)}{a\left(a-b\right)}\)-\(\dfrac{\left(x-b\right)\left(y-b\right)}{b\left(a-b\right)}\) e,...
Đọc tiếp

Bài 1: Thực hiện phép tính

a, \(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}\)+\(\dfrac{2}{x^2+3}\)+\(\dfrac{1}{x+1}\)

b, \(\dfrac{x+y}{2\left(x-y\right)}\)-\(\dfrac{x-y}{2\left(x+y\right)}\)+\(\dfrac{2y^2}{x^2-y^2}\)

c, \(\dfrac{x-1}{x^3}\)-\(\dfrac{x+1}{x^3-x^2}\)+\(\dfrac{3}{x^3-2x^2+x}\)

d, \(\dfrac{xy}{ab}\)+\(\dfrac{\left(x-a\right)\left(y-a\right)}{a\left(a-b\right)}\)-\(\dfrac{\left(x-b\right)\left(y-b\right)}{b\left(a-b\right)}\)

e, \(\dfrac{x^3}{x-1}\)-\(\dfrac{x^2}{x+1}\)-\(\dfrac{1}{x-1}\)+\(\dfrac{1}{x+1}\)

f, \(\dfrac{x^3+x^2-2x-20}{x^2-4}\)-\(\dfrac{5}{x+2}\)+\(\dfrac{3}{x-2}\)

g, \(\left\{\dfrac{x-y}{x+y}+\dfrac{x+y}{x-y}\right\}\).\(\left\{\dfrac{x^2+y^2}{2xy}\right\}\).\(\dfrac{xy}{x^2+y^2}\)

h, \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}\)+\(\dfrac{1}{\left(b-c\right)\left(c-a\right)}\)+\(\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)

i, \(\dfrac{\left[a^2-\left(b+c\right)^2\right]\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)

k, \(\left[\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left\{\dfrac{x^2}{y}-\dfrac{y^2}{x}\right\}\right]\):\(\dfrac{x-y}{x}\)

Bài 2: Rút gọn các phân thức:

a, \(\dfrac{25x^2-20x+4}{25x^2-4}\)

b, \(\dfrac{5x^2+10xy+5y^2}{3x^3+3y^3}\)

c, \(\dfrac{x^2-1}{x^3-x^2-x+1}\)

d, \(\dfrac{x^3+x^2-4x-4}{x^4-16}\)

e, \(\dfrac{4x^4-20x^3+13x^2+30x+9}{\left(4x^2-1\right)^2}\)

Bài 3: Rút gọn rồi tính giá trị các biểu thức:

a, \(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\) với a = 4, b = -5, c = 6

b, \(\dfrac{16x^2-40xy}{8x^2-24xy}\) với \(\dfrac{x}{y}\) = \(\dfrac{10}{3}\)

c, \(\dfrac{\dfrac{x^2+xy+y^2}{x+y}-\dfrac{x^2-xy+y^2}{x-y}}{x-y-\dfrac{x^2}{x+y}}\) với x = 9, y = 10

Bài 4: Tìm các giá trị nguyên của biến số x để biểu thức đã cho cũng có giá trị nguyên:

a, \(\dfrac{x^3-x^2+2}{x-1}\)

b, \(\dfrac{x^3-2x^2+4}{x-2}\)

c, \(\dfrac{2x^3+x^2+2x+2}{2x+1}\)

d, \(\dfrac{3x^3-7x^2+11x-1}{3x-1}\)

e, \(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\)

2
8 tháng 12 2017

Giúp mình nhé mọi người ! leuleu

8 tháng 12 2017

\(1.\)

\(a.\)

\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)

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

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

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

\(=x-1\)

\(b.\)

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

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

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

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

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

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

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

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

Tương tự các câu còn lại

1 tháng 5 2017

ai giải giúp mk vs đg cần gấp

19 tháng 6 2018

Giải:

1) \(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)

\(=\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)

\(=\dfrac{-1}{12}-\dfrac{55}{24}\)

\(=\dfrac{-19}{8}\)

2) \(-1,75-\left(\dfrac{-1}{9}-2\dfrac{1}{18}\right)\)

\(=-\dfrac{7}{4}+\dfrac{1}{9}+2\dfrac{1}{18}\)

\(=-\dfrac{7}{4}+\dfrac{1}{9}+\dfrac{37}{18}\)

\(=\dfrac{5}{12}\)

3) \(-\dfrac{5}{6}-\left(-\dfrac{3}{8}+\dfrac{1}{10}\right)\)

\(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\)

\(=-\dfrac{67}{120}\)

4) \(\dfrac{2}{5}+\left(-\dfrac{4}{3}\right)+\left(-\dfrac{1}{2}\right)\)

\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{1}{2}\)

\(=-\dfrac{43}{30}\)

5) \(\dfrac{3}{12}-\left(\dfrac{6}{15}-\dfrac{3}{10}\right)\)

\(=\dfrac{3}{12}-\dfrac{6}{15}+\dfrac{3}{10}\)

\(=\dfrac{3}{20}\)

6) \(\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}\)

\(=8\dfrac{5}{11}+3\dfrac{5}{8}-3\dfrac{5}{11}\)

\(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}\)

\(=8+\dfrac{5}{8}\)

\(=\dfrac{69}{8}\)

7) \(-\dfrac{1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)

\(=-\dfrac{1}{4}.13\dfrac{9}{11}-\dfrac{1}{4}.6\dfrac{2}{11}\)

\(=-\dfrac{1}{4}\left(13\dfrac{9}{11}+6\dfrac{2}{11}\right)\)

\(=-\dfrac{1}{4}\left(13+\dfrac{9}{11}+6+\dfrac{2}{11}\right)\)

\(=-\dfrac{1}{4}\left(13+6+1\right)\)

\(=-\dfrac{1}{4}.20=-5\)

8) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)

\(=\dfrac{4}{9}\left(-7\right)+6\dfrac{5}{9}\left(-7\right)\)

\(=-7\left(\dfrac{4}{9}+6\dfrac{5}{9}\right)\)

\(=-7\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)\)

\(=-7\left(6+1\right)\)

\(=-7.7=-49\)

Vậy ...

Bài 3: 

a: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=21\)

\(\Leftrightarrow x^3-27-x\left(x^2-16\right)=21\)

\(\Leftrightarrow x^3-27-x^3+16x=21\)

=>16x=48

hay x=3

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

\(\Leftrightarrow x^3+8-x^3-2x=4\)

=>-2x=4-8=-4

hay x=2

18 tháng 3 2018

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