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1. Tìm x:
a) 135 - 3 x ( x- 1) = 3^4 + 6^2
x=-2, x=3
b) 3 x ( x + 7 ) = 5^2 + 5
x=-căn bậc hai(89)/2-7/2, x=căn bậc hai(89)/2-7/2
2. So sánh 2^20 và 3^15
2 ^20 < 3 ^15
B1. 2x + 3 + 22 = 72
=> 2x + 3 + 4 = 72
=> 2x + 3 = 72 - 4
=> 2x + 3 = 68
=> ko có gtri x
B2 : Ta có : A = 1 + 2 + 22 + 23 + 24 + 25 + 26 + ... + 22001 + 22002
= (1 + 2) + (22 + 23 + 24) + (25 + 26 + 27) + ... + (22000 + 22001 + 22002)
= 3 + 22.(1 + 2 + 22) + 25.(1 + 2 + 22 ) + ... + 22000 . (1 + 2 + 22)
= 3 + 22.7 + 25.7 + ... + 22000 . 7
= 3 + (22 + 25 + .... + 22000) . 7
=> Số dư của 7 là 3
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) 2x = 64
=> 2x = 26
=> x = 6
b) 5x = 7x
=> 7x - 5x = 0
=> 5x(2x - 1) = 0
=> \(\orbr{\begin{cases}5^x=0\\2^x-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x\in\varnothing\\2^x=1\end{cases}}\)\(\Rightarrow2^x=2^0\Rightarrow x=0\)
c) 5x . 53 = 125
=> 5x + 3 = 53
=> x + 3 = 3
=> x = 0
d) 3x - 17 = 64
=> 3x = 64 + 17
=> 3x = 81
=> 3x = 34
=> x = 4
e) (3x - 5)3 = 52 . 24 + 600
=> (3x - 5)3 = 25.16 + 600
=> (3x - 5)3 = 400+ 600
=> (3x - 5)3 = 1000
=> (3x - 5)3 = 103
=> 3x - 5 = 10
=> 3x = 15
=> x = 15 : 3
=> x = 5
g) (5x - 15)3 = (5x - 15)7
=> (5x - 15)7 - (5x - 15)3 = 0
=> (5x - 15)3. [(5x - 15)4 - 1] = 0
=> \(\orbr{\begin{cases}\left(5x-15\right)^3=0\\\left(5x-15\right)^4-1=0\end{cases}\Rightarrow\orbr{\begin{cases}\left(5x-15\right)=0\\\left(5x-15\right)^4=1\end{cases}\Rightarrow}\orbr{\begin{cases}5x-15=0\\5x-15=\pm1\end{cases}}}\)
Nếu 5x - 15 =0
=> 5x = 15
=> x = 3
Nếu 5x - 15 = 1
=> 5x = 16
=> x = 16 : 5
=> x = 16/5
Nếu 5x - 16 = -1
=> 5x = 14
=> x = 14 : 5
=> x = 14/5
Vậy \(x\in\left\{3;\frac{16}{5};\frac{14}{5}\right\}\)
a) 2x = 64
Vì 26 = 64 nên x = 6
Vậy x = 6
b) 5x = 7x
Vì 50 = 1 và 70 = 1
=> x = 0
Vậy, x = 0
c) 5x . 53 = 625
Ta có 625 = 54
nên 5x . 53 = 54
5x+3 = 54
=> x = 1
Vậy x = 1
d) 3x - 17 = 64
3x = 64 + 17 = 81 = 34
=> x = 4
Vậy x = 4
e) ( 3x - 5 ) 3 = 52 . 24 + 600
( 3x - 5 ) 3 = 25 . 16 + 600 = 1000 = 103
=> 3x - 5 = 10
3x = 10 + 5 = 15
x = 15 : 3 = 5
Vậy x = 5
g) ( 5x - 15 ) 3 = ( 5x - 15 ) 7
=> (5x - 15 ) 3 : ( 5x - 15 ) 7 = 1
( 5x - 15 ) 3 - 7 = 1
( 5x - 15 ) -4 = 1 = 1-4 = -1-4
=> 5x - 15 = 1 hoặc 5x - 15 = -1
5x = 1 + 15 hoặc 5x = -1 + 15
5x = 16 hoặc 5x = 14
\(x=\frac{16}{5}\) hoặc \(x=\frac{14}{5}\)
Vậy, \(x\in\left\{\frac{16}{5};\frac{14}{5}\right\}\)
Cbht
Bài 1:
a) \(x^{10}=1^x\Rightarrow\orbr{\begin{cases}x=1\\x=10\end{cases}}\)
b) \(x^{10}=x\Rightarrow x=1\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\left(2x-15\right)^5.\left(2x-15\right)^3=\left(2x-15\right)^3\)
\(\left(2x-15\right)^2=1\Rightarrow x=8\)
Bài 2:
\(a;2^{16}=2^{13}\cdot2^3=2^{13}\cdot8>7\cdot2^{13}\)
\(b;49^8\cdot27^5=7^{16}\cdot3^{15}=21^{15}\cdot7>21^5\)
C;Ta có:\(199^{20}< 200^{20}=2^{20}\cdot10^{40}=2^{15}\cdot10^{40}\cdot2^5\)
\(2003^{15}>2000^{15}=2^{15}\cdot10^{45}=2^{15}\cdot10^{40}\cdot10^5\)
Vì 25<105 nên 19920<200315
\(d;3^{39}< 3^{40}=9^{20}< 11^{20}< 11^{21}\)
\(a)x^{15}=x\)
\(\Rightarrow x^{15}-x=0\)
\(\Leftrightarrow x\left(x^{14}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
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_
\(a)x^{15}=x\)
\(\Rightarrow x=1\)
\(b)2^x-15=17\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\Rightarrow x=5\)
\(c)\left(2x+1\right)^3=125\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow x=2\)
Bài làm
a. ( | x | + 3 ) . 15 - 5 = 70
( | x | + 3 ) . 15 = 70 + 5
( | x | + 3 ) . 15 = 75
| x | + 3 = 75 : 15
| x | + 3 = 5
| x | = 5 - 3
suy ra | x | = 2 hoặc | x | = -2
vì |2| = 2 ; |-2| = 2
b.86 : [ 2 . ( 2x - 1 )2 - 7 ] + 42 = 2 . 32
86 : [ 2 . ( 2x - 1 )2 - 7 ] + 16 = 2 . 9
86 : [ 2 . ( 2x - 1 )2 - 7 ] + 16 = 18
86 : [ 2 . ( 2x - 1 )2 - 7 ] = 18 - 16
86 : [ 2 . ( 2x - 1 )2 - 7 ] = 2
[ 2 . ( 2x - 1 )2 - 7 ] = 86 : 2
2 . ( 2x - 1 )2 - 7 = 43
2 . ( 2x - 1 ) 2 = 43 + 7
2 . ( 2x - 1 ) 2 = 50
( 2x - 1 ) 2 = 50 : 2
( 2x - 1 ) 2 = 25
( 2x - 1 ) 2 = 52
suy ra ( 2.3 - 1 ) 2 = 52 = 25