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25 tháng 10 2018

Thời gian có hạn copy cái này hộ mình vào google xem nha :

https://lazi.vn/quiz/d/16491/nhac-edm-la-loai-nhac-the-loai-gi

Vào xem xong các bạn nhận được 1 thẻ cào mệnh giá 100k nhận thưởng bằng cách nhắn tin vs mình và 1 phần thưởng bí mật là chiếc áo đá bóng,....

Có 300 giải nhanh nha đã có 241 người nhận rồi

OK

a: \(\dfrac{96}{\left(x-4\right)\left(x+4\right)}+\dfrac{7+x}{4-x}=\dfrac{2x-1}{x+4}-3\)

\(\Leftrightarrow\dfrac{96}{\left(x-4\right)\left(x+4\right)}-\dfrac{\left(x+7\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{3\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)

Suy ra: \(96-x^2-11x-28=2x^2-9x+4-3\left(x^2-16\right)\)

\(\Leftrightarrow-x^2-11x+68=2x^2-9x+4-3x^2+48\)

\(\Leftrightarrow-x^2-11x+68=-x^2-9x+52\)

=>-11x+68=-9x+52

=>-2x=-16

hay x=8(nhận)

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

\(\Leftrightarrow2\left(x-2\right)\left(x-3\right)+3\left(x-1\right)\left(x-3\right)=3\left(x-1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(x^2-5x+6\right)+3\left(x^2-4x+3\right)=3\left(x^2-3x+2\right)\)

\(\Leftrightarrow2x^2-10x+12+3x^2-12x+9=3x^2-9x+6\)

\(\Leftrightarrow5x^2-22x+21-3x^2+9x-6=0\)

\(\Leftrightarrow2x^2-13x+15=0\)

\(\Leftrightarrow2x^2-10x-3x+15=0\)

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

=>x=5(nhận) hoặc x=3/2(nhận)

6:=(3/2)*(3/2)^2*(3/2)^4=(3/2)^7

7: =(1/2)^7*2^3*2^5*2^8=2^9

8: =(-1/7)^4*5^4=(-5/7)^4

9: =2^2*2^5:(2^3/2^4)

=2^7/2=2^6

10: =(1/7)^3*7^2=1/7

31 tháng 1 2016

cảm ơn bạn

22 tháng 6 2023

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

23 tháng 10 2018

\(\left(2x-1\right)^3-4x^2\left(2x-3\right)=5.\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)

\(\Leftrightarrow6x-1=5\)

\(\Leftrightarrow6x=6\)

\(\Leftrightarrow x=1\)

23 tháng 10 2018

\(\left(x+4\right)^3-x^2\left(x+12\right)=15.\)

\(\Leftrightarrow x^3+12x^2+48x+64-x^3-12x^2=15\)

\(\Leftrightarrow48x+64=15\)

\(\Leftrightarrow48x=-49\)

\(\Leftrightarrow x=\frac{-49}{48}\)

14 tháng 7 2023

\(6-2\left(x-1\right)=4\)

\(\Rightarrow2\left(x-1\right)=6-4\)

\(\Rightarrow2\left(x-1\right)=2\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x=1+1=2\)

________________

\(2\cdot\left(x-2\right)+1=7\)

\(\Rightarrow2\cdot\left(x-2\right)=7-1\)

\(\Rightarrow2\cdot\left(x-2\right)=6\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=3+2=5\)

_______________

\(\left(2\cdot x-3\right)+4=9\)

\(\Rightarrow2\cdot x-3=5\)

\(\Rightarrow2\cdot x=3+5\)

\(\Rightarrow2\cdot x=8\)

\(\Rightarrow x=\dfrac{8}{2}=4\)

________________

\(\left(3\cdot x-2\right)-1=3\)

\(\Rightarrow3\cdot x-2=3+1\)

\(\Rightarrow3\cdot x-2=4\)

\(\Rightarrow3\cdot x=6\)

\(\Rightarrow x=\dfrac{6}{3}=2\)

a: =>2(x-1)=2

=>x-1=1

=>x=2

b: =>2(x-2)=6

=>x-2=3

=>x=5

c; =>2x-3=5

=>2x=8

=>x=4

d: =>3x-2=4

=>3x=6

=>x=2

e: =>2(6-x)=4

=>6-x=2

=>x=4

f: =>x-2=5

=>x=7

g: =>10-2x=4

=>2x=6

=>x=3

h: =>2x+4=3

=>2x=-1

=>x=-1/2

j: =>x+2=12

=>x=10

l: =>2x+3=3

=>2x=0

=>x=0

12 tháng 7 2019

a) \(\frac{x}{x+1}=\frac{1}{2}\)

=> 2x = x + 1

=> 2x - x = 1

=> x = 1

b) \(\frac{x}{2}=\frac{x}{3}\)

=> 3x = 2x

=> 3x - 2x = 0

=> x = 0

c) \(\frac{x+1}{2}=\frac{x+1}{2017}\)

=> \(2017\left(x+1\right)=2\left(x+1\right)\)

=> 2017x + 2017 = 2x + 2

=> 2017x - 2x = 2 - 2017

=> 2015x = -2015

=> x = -2015 : 2015

=> x = -1

i) \(\frac{3}{x}=\frac{x}{2017}\)

=> x2 = 2017.3

=> x2 = 6051

=> \(\orbr{\begin{cases}x=\sqrt{6051}\\x=-\sqrt{6051}\end{cases}}\)

còn lại tự lm

\(a,\frac{x}{x+1}=\frac{1}{2}\)

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

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

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

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

\(\Rightarrow x=1\)

\(b,\frac{x}{2}=\frac{x}{3}\)

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

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

\(\Rightarrow3x=2x\)

\(\Rightarrow x=0\)

\(c,\frac{x+1}{2}=\frac{x+1}{2017}\)

\(\Rightarrow x+1=\frac{x+1}{2017}.2\)

\(\Rightarrow x+1=\frac{2x+2}{2017}\)

\(\Rightarrow2017x+2017=2x+2\)

\(\Rightarrow2017x-2x=2-2017\)

\(\Rightarrow2015x=-2015\)

\(\Rightarrow x=-1\)

\(i,\frac{3}{x}=\frac{x}{2017}\)

\(\Rightarrow x=3:\frac{x}{2017}\)

\(\Rightarrow x=\frac{6051}{x}\)

\(\Rightarrow x^2=6051\)

\(\Rightarrow x=\sqrt{6051}\)

\(o,\frac{x}{3}=\frac{x+1}{2}\)

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

\(\Rightarrow x=\frac{3x+3}{2}\)

\(\Rightarrow2x=3x+3\)

\(\Rightarrow-x=3\)

\(\Rightarrow x=-3\)

\(m,\frac{x+1}{2}=\frac{x+2}{3}\)

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

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

\(\Rightarrow3x+3=2x+4\)

\(\Rightarrow x=1\)

\(p,\frac{x+1}{2}=x\)

\(\Rightarrow2x=x+1\)

\(\Rightarrow x=1\)

\(m,\frac{2}{x}=\frac{x}{8}\)

\(\Rightarrow x=2:\frac{x}{8}\)

\(\Rightarrow x=\frac{16}{x}\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x=4\)

\(Q,\frac{x^2}{2}=\frac{8}{x^2}\)

\(\Rightarrow x^2=\frac{8}{x^2}.2\)

\(\Rightarrow x^2=\frac{16}{x^2}\)

\(\Rightarrow x^4=16\)

\(\Rightarrow x=2\)

\(r,\frac{x^3}{2}=\frac{32}{x}\)

\(\Rightarrow x^3=\frac{32}{x}.2\)

\(\Rightarrow x^3=\frac{64}{x}\)

\(\Rightarrow x^4=64\)

\(\Rightarrow x=\sqrt[4]{64}\)

17 tháng 9 2023

Vậy \(( - 32{x^5} + 1):( - 2x + 1) = 16{x^4} + 8{x^3} + 4{x^2} + 2x + 1\).