Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a)x^{15}=x\)
\(\Rightarrow x^{15}-x=0\)
\(\Leftrightarrow x\left(x^{14}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy....
\(b)2^x-15=17\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy...
\(c)\left(2x+1\right)^3=125\)
\(\Leftrightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Leftrightarrow2x=4\Rightarrow x=2\)
Vậy...
_Y nguyệt_
\(a)x^{15}=x\)
\(\Rightarrow x=1\)
\(b)2^x-15=17\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\Rightarrow x=5\)
\(c)\left(2x+1\right)^3=125\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow x=2\)
a. ( x - 1 )2 = 25
<=> \(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)
<=>\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
b. 2x + 2 - 2x = 96
<=> 2x.4-2x=96
<=> 2x.3=96
<=> 2x=32=25
<=> x=5
c. 2x . 7 = 224
<=> 2x=32=25
<=> x=5
d. ( 7x - 11 )3 = 25 . 52 = 200 (xem lại đề)
e. 9 < 3x < 81
<=> 32<3x<34
<=> 2<x<4
\(A=1+5^2+5^4+...+5^{200}\)
\(5^2A=5^2+5^4+5^6+...+5^{202}\)
\(25A-A=\left(5^2+5^4+5^6+...+5^{202}\right)-\left(1+5^2+5^4+...+5^{200}\right)\)
\(24A=5^{202}-1\)
\(A=\frac{5^{202}-1}{24}\)
Câu B có sai đề ko vậy
a)
\(2A=2+2^2+2^3+...+2^{101}\)
\(2A-A=\left(2+2^2+2^3+....+2^{101}\right)-\left(1+2+2^2+...+2^{100}\right)\)
\(A=2^{101}-1\)
b)
Tách ra thành 2 tổng :\(D=3+3^3+...+3^{99}\) và \(E=3^2+3^4+...+3^{100}\)
\(3^2D=3^3+3^5+...+3^{101}\)
\(9D-D=\left(3^3+3^5+...+3^{101}\right)-\left(3+3^3+...+3^{99}\right)\)
\(8D=3^{101}-3\Leftrightarrow D=\frac{3^{101}-3}{8}\)
Tương tự \(E=\frac{3^{102}-3^2}{8}\)
Ta có \(D-E=B\)
Do đó \(\frac{3^{101}-3-3^{102}+3^2}{8}\)
Tương tự phần a, b tính được \(C=\frac{5^{202}-1}{24}\)
c,\(C=1+5^2+5^4+5^6+...+5^{200}\)
\(\Rightarrow25C=5^2+5^4+5^6+5^8+...+5^{202}\)
\(\Rightarrow25C-C=24C=\left(5^2+5^4+...+5^{202}\right)-\left(1+5^2+...+5^{200}\right)\)
\(=5^{202}-1\)
\(\Rightarrow C=\frac{5^{202}-1}{24}\)
A = 1 + 2 + 22 + ... + 2100
=> 2A = 2 + 22 + 23 + ... + 2100 + 2101
=> 2A - A = ( 2 + 22 + 23 + ... + 2100 + 2101 ) - ( 1 + 2 + 22 + ... + 2100 )
=> A = 2101 - 1
b) 3^2 . [(5^2 - 3 ) : 11 ] - 2^4 + 2.10^3
= 9 . [(25 - 3 ) : 11 ] - 16 + 2.1000
= 9 . [22 : 11 ] - 16 + 2000
= 9 . 2 - 16 + 2000
= 18 - 16 + 2000
= 2 + 2000
= 2002
(72005 + 72004) : 72004
= 72005 : 72004 + 72004 : 72004
= 72005 - 2004 + 1
= 71 + 1
= 7 + 1
= 8
a) ( 3^5 . 3^7 ) : 3^10 + 5.2^4 - 7^3 : 7
= 3^10 : 3^10 + 80 - 7^2
= 1 + 80 - 49
= 32
1)\(79-5\left(11-x\right)=34\)
\(\Rightarrow79-55+5x=34\)
\(\Rightarrow24+5x=34\)
\(\Rightarrow5x=-10\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
2)\(32+2\left(7-x\right)=40\)
\(\Rightarrow32+14-2x=40\)
\(\Rightarrow46-2x=40\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
3)\(\left(166-2x\right).8^9=2.8^{11}\)
\(\Rightarrow\left(83-x\right).2.8^9=2.8^{11}\)
\(\Rightarrow83-x=8^3\)
\(\Rightarrow83-x=512\)
\(\Rightarrow x=-429\)
Vậy \(x=-429\)
4)\(5^2.x-2^3.x=51\)
\(\Rightarrow x\left(5^2-2^3\right)=51\)
\(\Rightarrow x\left(25-8\right)=51\)
\(\Rightarrow17x=51\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
5)\(3^x+4.3^x=5.3^7\)
\(\Rightarrow3^x\left(1+4\right)=5.3^7\)
\(\Rightarrow5.3^x=5.3^7\)
\(\Rightarrow3^x=3^7\)
\(\Rightarrow x=7\)
Vậy \(x=7\)
6)\(7.2^x-2^x=6.32\)
\(\Rightarrow2^x\left(7-1\right)=6.2^5\)
\(\Rightarrow6.2^x=6.2^5\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
7)\(15^{3-x}=225\)
\(\Rightarrow15^{3-x}=15^2\)
\(\Rightarrow3-x=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
8)\(4.5^x-3=97\)
\(\Rightarrow4.5^x=100\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
9)\(171-3.2^x=123\)
\(\Rightarrow3.2^x=48\)
\(\Rightarrow2^x=16\)
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
10)\(180-4.x^5=32\)
\(\Rightarrow4.x^5=148\)
\(\Rightarrow x^5=37\)//Đề có lỗi không ???
a) => 7 . x - 11 = -193
=> 7 . x = - 193 + 11
=> 7 . x = - 182
=> x = - 182 : 7
=> x = -26
a)(7x-11)=25.52.200
(7x-11)=32.25.200
(7x-11)=160000
7x=160000+11
7x=160011
x=160011:7
x=160011/7
b)16.4x=325
24.4x=(25)5
24.4x=225
4x=225:24
4x=221
(22)x=221
22.x=221
=>2.x=21
x=21:2
x=10,5
Đề sai