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Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
a) x15= x.
=> x15- x= 0.
=> x( x14- 1)= 0.
=> \(\orbr{\begin{cases}x=0.\\x^{14}-1=0.\end{cases}}\)
=> \(\orbr{\begin{cases}x=0.\\x^{14}=1.\end{cases}}\)
=> \(\orbr{\begin{cases}x=0.\\x=1.\end{cases}}\)
Vậy x\(\in\) { 0; 1}
b) 16x< 128.
Nếu x= 0 thì 16x= 160= 0( chọn)
Nếu x= 1 thì 16x= 161= 16( chọn)
Nếu x= 2 thì 16x= 162= 256( loại)
Vậy x\(\in\) { 0; 1}
c) 5x. 5x+ 1. 5x+ 2\(\le\) 1000...00: 218( 18 chữ số 0)
=> 5x+ x+ 1+ x+ 2\(\le\) 1018: 218.
=> 53x+ 3\(\le\) 518.
=> 3x+ 3\(\le\) 18.
=> 3x\(\le\) 15.
=> x\(\le\) 5.
=> x\(\in\){ 0; 1; 2; 3; 4; 5}
Vậy x\(\in\){ 0; 1; 2; 3; 4; 5}
d) 2x.( 22)2=( 23)2.
=> 2x. 24= 26.
=> 2x= 26: 24.
=> 2x= 22.
=> x= 2.
Vậy x= 2.
e)( x5)10= x.
=> x50- x= 0.
=> x( x49- 1)= 0.
=> \(\orbr{\begin{cases}x=0.\\x^{49}-1=0.\end{cases}}\)
=> \(\orbr{\begin{cases}x=0.\\x^{49}=1.\end{cases}}\)
=> \(\orbr{\begin{cases}x=0.\\x=1.\end{cases}}\)
Vậy x\(\in\) { 0; 1}
\(x^{15}=x\)
\(\Rightarrow x^{15}-x=0\)
\(\Rightarrow x\left(x^{14}-1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x^{14}-1=0\Rightarrow x=\pm1\end{cases}}\)
b)
\(5^x.5^{x+1}.5^{x+2}=5^x.5^x.5.5^x.5^2=5^{x+x+x+1+3}=5^{3x+3}\le10^{18}:2^{118}\)
\(=>5^{3x+3}\le5^{18}=>3x+3\le18=>x\le5=>x\in\left\{0;1;2;3;4;5\right\}\)
**** bn, câu a tự lm nhé
Bài 1:tính nhanh:
a)46.93 + 87.93 - 33.93 = (46+87-33) . 93
=100. 93
= 9300
b)14+16+18 + .......... +98+100= [(100+14):2+1].114
=28.114
=3192
c)2009 - {264:[42.2+(12-2)2 ]}=2009 - {264:[42.2+102]}
=2009 - {264:[16.2+100]}
=2009 - {264:[32+100]}
=2009 - {264:132}
=2009 - 2
=2007
d) (24.23.54.7-27.252.6) : [26.53]=(16.8.625.7-128.625.6) : [26.53]
=(560000-480000) : [26.53]
=80000 : [64.125]
=80000 : 8000
=10
Bài 2: Tìm số tự nhiên x, biết:
a) (2x+8) : 5= 27 : 24
(2x+8) : 5=128:16
(2x+8) : 5=8
(2x+8) =8.5
(2x+8) =40
2x =40-8
2x =32
x =16
b)8.(x+25)-155=181
8.(x+25)=181+155
8.(x+25)=336
(x+25)=336:8
(x+25)=42
x=42-25
x=17
c)5x =125
x.x.x.x.x=125
x=3
d)720 : [41 - (2x - 5)] = 23.5
720 : [41 - (2x - 5)] =8.5
720 : [41 - (2x - 5)] =40
[41 - (2x - 5)] =720:40
[41 - (2x - 5)] =18
(2x - 5)=41-18
(2x - 5)=23
2x=23+5
2x=28
x=14
Chúc bạn làm bài tốt!!
a) (2x+8) : 5 = 2^3 = 8
2x+8 = 8 . 5
2x + 8 = 40
2x = 40 -8 = 32
x = 32:2
x = 16
b) = 8 . ( x + 25 ) = 181 +155
8 . (x + 25) = 336
x + 25 = 366 : 8
x + 25 = 42
x = 17
a, 34\(x\) + 4 = 81\(x+3\)
3\(^{4x+4}\) = (34)\(x+3\)
3\(^{4x+4}\) = 3\(4x+12\)
4\(x\) + 4 = 4\(x\) + 12
0 = 8 (vô lý)
Vậy \(x\) \(\in\) \(\varnothing\)
b, (4\(x\) - 1)3 = 27.125
(4\(x\) - 1)3 = 33.53
(4\(x\) - 1)3 = 153
4\(x\) - 1 = 15
4\(x\) = 16
\(x\) = 4