Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
D= 1x1x2x2x3x3x4x4x...x100x100
D= (1x2x3x4x...x100) x (1x2x3x4x...x100)
D=(1x2x3x4x...x100)2
Ta có \(5^x=125\)
\(\Rightarrow5^x=5^3\Rightarrow x=3\)
Vậy x = 3
b,\(3^{2x}=81\)
\(\Rightarrow3^{2x}=3^4\Rightarrow2x=4\Rightarrow x=2\)
Vậy x = 2
c,\(5^{2x-3}-2.5^2=5^2.3\)
\(\Rightarrow5^{2x-3}=5^2.3+5^2.2\Rightarrow5^{2x-3}=5^2.\left(2+3\right)\Rightarrow5^{2x-3}=5^3\)
\(\Rightarrow2x-3=3\Rightarrow2x=6\Rightarrow x=3\)
Vậy x = 3
53 - ( x - 32 ) = 53 - 43
<=> x - 32 = 43
<=> x - 9 = 64
<=> x = 73
S = 1 + 2 + 22 + 23 +24 + 25 +...+ 260 + 261 + 262 + 263
= ( 1 + 22) +( 2 + 23) + (24 + 26) + ( 25 + 27) +...+ (260 + 262) + ( 261 + 263)
=( 1 + 22) + 2 ( 1 + 22) + 24 (1 + 22) + 25 (1 +22)+...+ 260 ( 1 + 22) + 261( 1 + 22)
= ( 1 + 22)( 1 + 2 +24 + 25 +...+ 260)
= 5 ( 1 + 2 +24 + 25 +...+ 260)
Vậy S chia hết cho 5 vì có một thừa số là 5.
a) 52x+1=125
=>52x+1= 53
=>2x+1=3
=>2x = 2
=>x =1
c)73<x <=93
=>343<x<=729
=> x = { 344;345;...;729}
d) (2x+3)4=81
=>(2x+3)4=34
=>2x+3 =3
=>2x =0
=> x = 0
nhớ tk nha
a)52x+1=125
52x+1=53
⏩ 2x+1=3
2x=4
x=4:2
x=2
b) (2x)2(2x) 3 =25.25
(2x) 2 (2x) 3 =210
(2x) 5=1024
(2x) 5 =45
2x=4
x=4 :2=2
d)(2x +3)4=81
(2x +3)4=34
2x+3=3
2x=0
x=0
Câu c bạn tự giải nhé
Nhớ tk giùm mình đấy
1) 10 - [ 20 - ( 6 - 8) ]
= 10 - [ 20 - (-2) ]
= 10 - 22
= -12
2)
5 - l x - 2 l = 3
\(=>\orbr{\begin{cases}5-x-2=3\\5-x-2=-3\end{cases}}\)
\(=>\orbr{\begin{cases}5-x=3+2\\5-x=\left(-3\right)+2\end{cases}}\)
\(=>\orbr{\begin{cases}5-x=5\\5-x=-1\end{cases}}\)
\(=>\orbr{\begin{cases}x=5+5=10\\x=\left(-1\right)+5=4\end{cases}}\)
\(=>\orbr{\begin{cases}x=10\\x=4\end{cases}}\)
b) \(\left(x-1\right)^2+2=11\)
\(\left(x-1\right)^2=11-2=9\)
\(=>x-1=3\)
\(=>x=3+1\)
\(=>x=4\)
\(x^2+2x+4⋮x+1\)
\(\Leftrightarrow\left(x^2+x\right)+\left(x+1\right)+3⋮x+1\)
\(\Leftrightarrow x\left(x+1\right)+\left(x+1\right)+3⋮x+1\)
\(\Leftrightarrow3⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;2;-4\right\}\)
Ta có: \(x^2+2x+4\)
\(=\left(x^2+x\right)+\left(x+1\right)+3\)
\(=x\left(x+1\right)+\left(x+1\right)+3\)
\(=\left(x+1\right)\left(x+1\right)+3\)
Để \(x^2+2x+4\) chia hết cho x + 1 thì 3 phải chia hết cho x + 1
\(\Rightarrow\left(x+1\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x\in\left\{-4;-2;0;2\right\}\)
a, 4^x +4^x .4 =20
4^x(1+4) =20
4^x .5 =20
4^x =20:5
4^x = 4
x = 1
b,7^x .7^2=7^90
7^x = 7^90:7^2
7^x =7^88
x = 88
c, 5^x +5^x.5 = 750
5^x(1+5) =750
5^x . 6 =750
5^x = 750:6
5^x =125
5^x = 5^3
x =3
-_-? ?????????????????????????????????????????????????????????????????????????????????????????????????????????????????
⇔5x.25-5x=3.103
⇔5x(25-1)=3.1000
⇔5x.24=3000
⇔5x=125
⇔5x=53
⇒x=3
Vậy x=3
5x+2-5x=3.103
5x.52-5x=3000
5x.52-5x.50=3000
5x.(52.50)=3000
5x.52=3000
5x=210
mà 5...đề sai à bạn?