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`#3107`
b)
`2.3^x = 162`
`\Rightarrow 3^x = 162 \div 2`
`\Rightarrow 3^x = 81`
`\Rightarrow 3^x = 3^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
c)
`(2x - 15)^5 = (2 - 15)^3`
\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`
\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)
`d)`
\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!
`e)`
\(7\cdot4^{x-1}+4^{x-1}=23?\)
\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)
Bạn xem lại đề!
`f)`
\(2\cdot2^{2x}+4^3\cdot4^x=1056\)
\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
_____
\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)
\(\Rightarrow\left(x\div3+17\right)\div10=2\)
\(\Rightarrow x\div3+17=20\)
\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)
Vậy, `x = 9.`
1: =>3^x=81
=>x=4
2: =>2^x=8
=>x=3
3: =>x^3=2^3
=>x=2
4: =>x^20-x=0
=>x(x^19-1)=0
=>x=0 hoặc x=1
5: =>2^x=32
=>x=5
6: =>(2x+1)^3=9^3
=>2x+1=9
=>2x=8
=>x=4
7: =>x^3=115
=>\(x=\sqrt[3]{115}\)
8: =>(2x-15)^5-(2x-15)^3=0
=>(2x-15)^3*[(2x-15)^2-1]=0
=>2x-15=0 hoặc (2x-15)^2-1=0
=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1
=>x=15/2 hoặc x=8 hoặc x=7
1. Tìm số tự nhiên x biết:
1) \(3^x.3=243\)
\(3^x=243:3\)
\(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
_____
2) \(7.2^x=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
_____
3) \(x^3=8\)
\(x^3=2^3\)
\(\Rightarrow x=3\)
_____
4) \(x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x=1\)
5) \(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
_____
6) \(\left(2x+1\right)^3=9.81\)
\(\left(2x+1\right)^3=729=9^3\)
\(\rightarrow2x+1=9\)
\(2x=9-1\)
\(2x=8\)
\(x=8:2\)
\(\Rightarrow x=4\)
_____
7) \(x^6:x^3=125\)
\(x^3=125\)
\(x^3=5^3\)
\(\Rightarrow x=5\)
_____
8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)
_____
9) \(3^{x+2}-5.3^x=36\)
\(3^x.\left(3^2-5\right)=36\)
\(3^x.\left(9-5\right)=36\)
\(3^x.4=36\)
\(3^x=36:4\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
_____
10) \(7.4^{x-1}+4^{x+1}=23\)
\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)
\(4^{x-1}.\left(7+4^2\right)=23\)
\(4^{x-1}.\left(7+16\right)=23\)
\(4^{x-1}.23=23\)
\(4^{x-1}=23:23\)
\(4^{x-1}=1\)
\(4^{x-1}=4^1\)
\(\rightarrow x-1=0\)
\(x=0+1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt
a) Ta có: ( x + 1 ) 4 = ( 2 x ) 4 nên x +1 = 2x. Do đó x = 1.
b) Ta có: ( 2 x - l ) 5 = x 5 nên 2x - l = x. Do đó x = l.
a) Để x + 5 chia hết cho x + 2
hay (x + 2) + 3 chia hết x + 2
vì x+ 2 chia hết cho x+2 nên 3 sẽ chia hết cho x + 2
hay x + 2 thuộc Ư(3)= {-1, 1, 3, -3}
x + 2 | -1 | 1 | 3 | -3 |
x | -3 | -1 | 1 | -5 |
Vậy x= -3, -1, 1, -5
b, \(2x+3⋮x+1\)
\(2\left(x+1\right)+1⋮x+1\)
\(1⋮x+1\)hay \(x+1\inƯ\left(1\right)=\left\{\pm1\right\}\)
x + 1 | 1 | -1 |
x | 0 | -2 |
d, \(3x+13⋮2x+6\)
\(6x+26⋮2x+6\)
\(3\left(2x+6\right)+8⋮2x+6\)
\(8⋮2x+6\)hay \(2x+6\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
2x + 6 | 1 | -1 | 2 | -2 | 4 | -4 | 8 | -8 |
2x | -5 | -7 | -4 | -8 | -2 | -10 | 2 | -14 |
x | -5/2 | -7/2 | -2 | -4 | -1 | -5 | 1 | -7 |
9x-1.5y=75x
=>32(x-1).5y=75x
=>32x-2.5y=75x
=>3x.3x-2.5y=75x
=>3x-2.5y=25x=52x
52x không chia hết cho 3
=>3x-2.5y không chia hết cho 3
=>3x-2=1
=>x-2=0
=>x=2
=>32-2.5y=52.2
=>30.5y=54
=>5y=54
=>y=4
vậy x=2;y=4
`@` `\text {Ans}`
`\downarrow`
`2^x * 4 = 128`
`=> 2^x = 128 * 4`
`=> 2^x = 512`
`=> 2^x = 2^9`
`=> x = 9`
Vậy, `x = 9`
`x^15 = x`
`=> x^15 - x = 0`
`=> x(x^14 - 1) = 0`
`=>` TH1: `x = 0`
`TH2: x^14 - 1 = 0`
`=> x^14 = 1`
`=> x = 1`
Vậy, `x \in {0; 1}`
`(2x+1)^3 = 125`
`=> (2x+1)^3 = 5^3`
`=> 2x + 1 = 5`
`=> 2x = 5 - 1`
`=> 2x =4`
`=> x = 4 \div 2`
`=> x = 2`
Vậy,` x = 2.`
`(x - 5)^4 = (x-5)^6`
`=> (x-5)^4 - (x-5)^6 = 0`
`=> (x-5)^4 * [ 1 - (x-5)^2] = 0`
`=> - (x-6)(x-5)^4(x-4) = 0`
`TH1: (x - 5)^4 = 0`
`=> x - 5 = 0`
`=> x = 0 +5`
`=> x = 5`
`TH2: x - 6=0`
`=> x=6`
`TH3: x-4=0`
`=> x = 4`
Vậy, `x \in {4; 5; 6}`
a: =>2^x=32
=>x=5
b: =>x^15-x=0
=>x(x^14-1)=0
=>x=0; x=1;x=-1
c: =>2x+1=5
=>2x=4
=>x=2
d: =>(x-5)^4[(x-5)^2-1]=0
=>(x-5)(x-4)(x-6)=0
=>x=5;x=4;x=6
2x - 17 \(⋮\)x - 1
=> 2( x - 1 ) - 15 \(⋮\)x - 1
=> - 15 \(⋮\)x -1
=> x - 1 \(\inƯ\left(-15\right)=\left\{\pm1;\pm3;\pm5;\pm15\right\}\)
Lập bảng
x-1 | -15 | -5 | -3 | -1 | 1 | 3 | 5 | 15 |
x |
Tự tìm
a) 720:[41-(2x-5)=8x5
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 =28:2
x =14
a) (2x - 17 )5 = (2x -17 )5
=> Với mọi x \(\in\)N thì : ( 2x - 17 )5 = (2x -17 )5
Vậy :...
b)
=> Với x = 0 hoặc x = 1 thì 5x = 3x
Vậy :...
\(5^{\left(2x+1+x+1\right)}=5^{17}\)
\(\Rightarrow\)2x + x + 2 = 17
( 2 + 1)x = 17 - 2
3x = 15
x = 15 : 3
x = 5