Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2\left[x-\frac{1}{2}\right]+3\left[x-\frac{1}{3}\right]=-22\)
\(\Rightarrow2x-1+3x-1=-22\)
\(\Rightarrow2x+3x-1-1=-22\)
\(\Rightarrow5x=-22+1+1\)
\(\Rightarrow5x=-20\Leftrightarrow x=-4\)
a) \(A=1+2+2^2+...+2^{80}\)
\(2A=2+2^2+2^3+...+2^{81}\)
\(2A-A=2+2^2+2^3+...+2^{81}-1-2-2^2-...-2^{80}\)
\(A=2^{81}-1\)
Nên A + 1 là:
\(A+1=2^{81}-1+1=2^{81}\)
b) \(B=1+3+3^2+...+3^{99}\)
\(3B=3+3^2+3^3+...+3^{100}\)
\(3B-B=3+3^2+3^3+...+3^{100}-1-3-3^2-...-3^{99}\)
\(2B=3^{100}-1\)
Nên 2B + 1 là:
\(2B+1=3^{100}-1+1=3^{100}\)
2)
a) \(2^x\cdot\left(1+2+2^2+...+2^{2015}\right)+1=2^{2016}\)
Gọi:
\(A=1+2+2^2+...+2^{2015}\)
\(2A=2+2^2+2^3+...+2^{2016}\)
\(A=2^{2016}-1\)
Ta có:
\(2^x\cdot\left(2^{2016}-1\right)+1=2^{2016}\)
\(\Rightarrow2^x\cdot\left(2^{2016}-1\right)=2^{2016}-1\)
\(\Rightarrow2^x=\dfrac{2^{2016}-1}{2^{2016}-1}=1\)
\(\Rightarrow2^x=2^0\)
\(\Rightarrow x=0\)
b) \(8^x-1=1+2+2^2+...+2^{2015}\)
Gọi: \(B=1+2+2^2+...+2^{2015}\)
\(2B=2+2^2+2^3+...+2^{2016}\)
\(B=2^{2016}-1\)
Ta có:
\(8^x-1=2^{2016}-1\)
\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)
\(\Rightarrow2^{3x}-1=2^{2016}-1\)
\(\Rightarrow2^{3x}=2^{2016}\)
\(\Rightarrow3x=2016\)
\(\Rightarrow x=\dfrac{2016}{3}\)
\(\Rightarrow x=672\)
\(T=1\cdot2\cdot3+2\cdot3\cdot4+...+20\cdot21\cdot22\)
\(4T=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+...+20\cdot21\cdot22\cdot4\)
\(4T=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\cdot\left(5-1\right)+..+20\cdot21\cdot22\cdot\left(23-19\right)\)
\(4T=0\cdot1\cdot2\cdot3-1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+..+20\cdot21\cdot22\cdot23-19\cdot20\cdot21\cdot22\)
\(4T=21\cdot22\cdot23\cdot24\)
\(T=\frac{21\cdot22\cdot23\cdot24}{4}=21\cdot22\cdot23\cdot6=63756\\ k.cho.mk.nha\)
1.2.3+2.3.4+3.4.5+...+20.21.22
=1/4.(1.2.3.4+2.3.4.4+3.4.5.4+...+20.21.22.4)
=1/4.[1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+...+20.21.22.(23-19)]
=1/4.(1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+....-19.20.21.22+20.21.22.23)
=1/4.20.21.22.23
=53130
a: \(=\dfrac{3}{5}:\dfrac{7}{5}=\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{7}\)
b: \(=\dfrac{9}{17}\left(\dfrac{8}{5}-\dfrac{3}{5}\right)+\dfrac{8}{17}\)
=9/17+8/17=1
c: =>x-3/10=7/15*1/5=7/75
=>x=7/75+3/10=59/150
a/ 2/3.x+1/2=1/10
nên:2/3x=1/10-1/2
nên:2/3x=-2/5
suy ra :x=-2/5:2/3
vậy x=-3/5
a. 2/3x+1/2=1/10
2/3x=1/10-1/2
2/3x=-2/5
x=-2/5:2/3
x=-3/5
b.(7/2-2x).4/3=22/3
(7/2-2x).=22/3:4/3
7/2-2x=11/2
2x=7/2-11/2
2x=-2
x=-1
nho k cho minh voi nhe
\(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\) (sửa đề)
\(\Rightarrow3+2^{x-1}=24-\left[16-\left(4-1\right)\right]\)
\(\Rightarrow3+2^{x-1}=24-\left(16-3\right)\)
\(\Rightarrow3+2^{x-1}=24-13\)
\(\Rightarrow3+2^{x-1}=11\)
\(\Rightarrow2^{x-1}=11-3\)
\(\Rightarrow2^{x-1}=8\)
\(\Rightarrow2^{x-1}=2^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=3+1=4\)
Vậy \(x=4.\)
#\(Toru\)
a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)
\(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17x}\) = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)
\(\dfrac{72}{17x}\) = - \(\dfrac{1}{14}\)
17\(x\) = 72.(-14)
17\(x\) = - 1008
\(x\) = - 1008 : 17
\(x\) = - \(\dfrac{1008}{17}\)
Vậy \(x\) \(=-\dfrac{1008}{17}\)
b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))3 = - \(\dfrac{24}{27}\)
- \(\dfrac{32}{27}\) + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))3
(3\(x-\dfrac{7}{9}\))3 = - \(\dfrac{8}{27}\)
(3\(x-\dfrac{7}{9}\))3 = (- \(\dfrac{2}{3}\))3
3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)
3\(x\) = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
3\(x\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) : 3
\(x\) = \(\dfrac{1}{27}\)
Vậy \(x=\dfrac{1}{27}\)
3(x2 -1) -2 =22
=> 3(x2 -1)= 24
=> x2 -1 = 8
=> x2 =9
Trường hợp 1: x2 = 32
=> x= 3
Trường hợp 2: x2 =(-3)2
=> x= -3
Vậy x=3 hoặc x =-3