Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=5+52 +53 +..........+575
5A=52+53+54+...+576
5A-A=576-5
a=\(\dfrac{5^{76}-5}{4}\)
Ta có: ( x + 2)( x - 5) = -12
=> \(x+2\inƯ\left(-12\right);x-5\inƯ\left(-12\right)\)
mà Ư (-12) = \(\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x+2\in\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\\x-5\in\left\{"....."\right\}\end{matrix}\right.\)
Xét các t/h:
\(\dfrac{2n-1}{n+1}=\dfrac{2\left(n+1\right)-3}{n+1}\)
Để \(\dfrac{2\left(n+1\right)-3}{n+1}\in Z\Rightarrow3⋮n+1\)
\(\Rightarrow n+1\inƯ\left(3\right)=\left\{-1;-3;1;3\right\}\)
\(n+1=-1\Rightarrow n=-2\)
\(n+1=-3\Rightarrow n=-4\)
\(n+1=1\Rightarrow n=0\)
\(n+1=3\Rightarrow n=2\)
(sữa đề tìm \(x\) nguyên )
\(2^x+3+2^x=144\Leftrightarrow2^x+2^x=141\)
ta có : \(2^x+2^x\) là số chẳn
mà \(141\) là số lẽ \(\Rightarrow\) phương trình vô nghiệm
\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\)
\(\Leftrightarrow6\left(x-7\right)=7\left(y-6\right)\)
\(6x-42=7y-42\)
\(6x=7y\Leftrightarrow x=\dfrac{7}{6}y\)
\(x=-4:\left(7-6\right).7=-28\)
\(y=-28-4=-24\)
b tương tự
Giải:b)
\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\) nên \(6\left(x-7\right)=7\left(y-6\right)\)
Do đó \(6x-42=7y-42\) nên \(6x=7y\)
Suy ra \(6x-6y=y\) hay \(6\left(x-y\right)=y\)
Nên 6.(-4) = y
Vậy y = -24, x = \(\dfrac{7.\left(-24\right)}{6}\)= -28
c)
\(\dfrac{x+3}{y+5}=\dfrac{3}{5}\) nên \(5\left(x+3\right)=3\left(y+5\right)\)
Do đó \(5x+15=3y+15\) nên \(5x=3y\)
Suy ra \(5x+5y=3y+5y\)
\(5\left(x+y\right)=8y\)
\(5.16=8y\)
Nên \(y=\dfrac{5.16}{8}=\dfrac{80}{8}=10\)
Vậy y = 10, x = 16 - 10 =6
Đề sai, tớ sửa lại
Ta có :
\(A=2+2^2+..............+2^{60}\)
\(\Leftrightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...........+\left(2^{59}+2^{60}\right)\)
\(\Leftrightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+.........+2^{59}\left(1+2\right)\)
\(\Leftrightarrow A=2.3+2^3.3+...........+2^{59}.3\)
\(\Leftrightarrow A=3\left(2+2^2+..........+2^{59}\right)\)
\(\Leftrightarrow A⋮3\rightarrowđpcm\)
Lại có :
\(A=2+2^2+2^3+............+2^{60}\)
\(\Leftrightarrow A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+..........+\left(2^{58}+2^{59}+2^{60}\right)\)
\(\Leftrightarrow A=2\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+..........+2^{59}\left(1+2+2^2\right)\)
\(\Leftrightarrow A=2.7+2^4.7+............+2^{58}.7\)
\(\Leftrightarrow A=7\left(2+2^3+..........+2^{58}\right)\)
\(\Leftrightarrow A⋮7\rightarrowđpcm\)
Ta tiếp tục có :
\(A=2+2^2+2^3+............+2^{60}\)
\(\Leftrightarrow A=\left(2+2^2+2^3+2^4\right)+..............+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(\Leftrightarrow A=2\left(1+2+2^2+2^3\right)+.............+2^{57}\left(1+2+2^2+2^3\right)\)
\(\Leftrightarrow A=2.15+............+2^{57}.15\)
\(\Leftrightarrow A=15\left(2+.........+2^{57}\right)\)
\(\Leftrightarrow A⋮15\rightarrowđpcm\)
a) 2.(3x - 8)=64:23
vậy : 2.(3x - 8 )=64 : 8
2.(3x - 8) = 8
(3x - 8)= 8:2
(3x - 8)=4
3x = 8+4=12
x=12 : 3
x = 4
b)2+4+6+....+2x=210
vì mỗi số cách nhau 2 đơn vị =>
2+4+6+8+10+12+14+16+18+20+22+24+26+28
vậy 2x=28
x=28:2=14
c)1+3+5+...+(2x-1)=225
Vì mỗi số cách nhau 2 đơn vị=>
1+3+5+7+9+11+13+15+17+19+21+23+25+27+29
vậy (2x - 1)=29
2x=29+1=30
x=30:2=15
like nha
a) 2 . (3x - 8) = 64 : 23
2 . (3x - 8) = 64 : 8
2. (3x - 8) = 8
3x - 8 = 8 : 2
3x - 8 = 4
3x = 4 + 8
3x = 12
x = 12 : 3
x = 4
b) 2 + 4 + 6 + ... + 2x = 210
(2 + 2x) . [(2x - 2) : 2 + 1] : 2 = 210
[(2 + 2x) : 2]. (x - 1 + 1) : 2 = 210
(1 + x) . x : 2 = 210
x . (x + 1) : 2 = 210
x . (x + 1) = 210 . 2
x . (x + 1) = 420
Ta có: 420 = 42 . 10 = 21 . 2 . 10 = 21 . 20
=> x = 20
c) 1 + 3 + 5 +...+ (2x - 1) = 225
(2x - 1 + 1) . [(2x - 1 - 1) : 2 + 1] : 2 = 225
2x . [(2x - 2) : 2 + 1) : 2 = 225
x . (x - 1 + 1) = 225
x . x = 225
Ta có: 225 = 5 . 45 = 5 . 5 . 9 = 5 . 5 . 3 . 3 = (5 . 3) . (5 . 3) = 15 . 15
=> x = 15
\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+....\dfrac{2}{2016^2}\)
Ta thấy: \(\dfrac{2}{3^2}< \dfrac{2}{2.3}\)
\(\dfrac{2}{4^2}< \dfrac{2}{3.4}\)
...\(\dfrac{2}{2016^2}< \dfrac{2}{2015.2016}\)
Đặt:A=\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+...+\dfrac{2}{2016^2}\)
=>\(A< \dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{2015.2016}\)
=>\(A< \dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}-\dfrac{2}{5}+...+\dfrac{2}{2015}-\dfrac{2}{2016}\)
=>A<\(\dfrac{2}{2}-\dfrac{2}{2016}\)
=>A<\(\dfrac{1007}{1008}\) mà \(\dfrac{1007}{1008}\) < 1
=>A<1
Vậy \(\dfrac{2}{3^2}+\dfrac{2}{4^2}+\dfrac{2}{5^2}+...+\dfrac{2}{2016^2}\)<1 (\(đpcm\))
\(\dfrac{2}{3^2}+\dfrac{2}{4^2}+...+\dfrac{2}{2016^2}=2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)\)
Ta có: \(\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{2016^2}< \dfrac{1}{2015.2016}\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}\right)\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(\Rightarrow2\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2016^2}\right)< 2\left(\dfrac{1}{2}-\dfrac{1}{2017}\right)=1-\dfrac{2}{2017}< 1\)
=> đpcm
Ta có :
\(A=5+5^2+..........+5^{75}\)
\(\Leftrightarrow5A=5^2+5^3+.........+5^{75}+5^{76}\)
\(\Leftrightarrow5A-A=\left(5^2+5^3+.........+5^{76}\right)-\left(5+5^2+..........+5^{75}\right)\)
\(\Leftrightarrow4A=5^{76}-5\)
\(\Leftrightarrow4A+5=5^{76}\)
Mà \(4A+5=5^n+3\)
\(\Leftrightarrow5^{76}=5^n+3\)
hình như sai đề, hoặc là \(n\in\varnothing\)