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a) Ta có: ( 2 x + 1 ) 3 = 3 3 nên 2x + 1 = 3. Do đó x = 1.
b) Ta có: ( 2 x - 1 ) 3 = 5 3 nên 2x - 1 = 5. Do đó x = 3.
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: ( 2x+1) . (y - 3) = 12
=> 2x + 1 = 12 => 2x = 11 => x = \(\frac{11}{2}\)
y - 3 = 12 => y = 12 + 3 = 15
vậy x = \(\frac{11}{2}\) và y = 15
t i c k nhé!! 865856856878769696896856858788567567479647
(2x+1).(y-3)=12
\(\Rightarrow2x+1;y-3\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
Mà 2x + 1 lẻ.
Ta có bảng :
2x+1 | x | y-3 | y |
-3 | -2 | -4 | -1 |
-1 | -1 | -12 | -9 |
1 | 0 | 12 | 15 |
3 | 1 | 4 | 7 |
![](https://rs.olm.vn/images/avt/0.png?1311)
32x34x3n=310
3n=310-2-4
3n=34
bài 2:
9<3.x<235
A={4;5;6;...;78}
bài 2 nếu sai thì bảo mk nhé
Bài3:
so sánh 230 và 320
bài 3 chịu ahihi
![](https://rs.olm.vn/images/avt/0.png?1311)
(x+1)(x+2)+(x+3)+...+(x+100)=5750
(x+x+...+x)(1+2+3+...+100)=5750
100x + 5050 =5750
100x =700
x =7
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750
( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750
x . 100 + ( 100 + 1 ). 100 : 2 =5750
x .100 + 5050 = 5750
x .100 = 5750 - 5050
x . 100 = 700
x = 700 : 100
x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
(2x+1)3=125
=> 2x + 1 = 41,666667
=> 2x = 42,66667
=> x = 21,33333
\(\left(2x+1\right)3=125\)
\(2x+1=125.3\)
\(2x+1=375\)
\(2x=375-1\)
\(2x=374\)
\(x=374:2\)
\(x=187\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\frac{1}{3}+\frac{1}{6}+......+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(A=\frac{1}{6}+\frac{1}{12}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{2002}.\frac{1}{2}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{4004}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2000}{4004}\)
\(A=\frac{1}{2}-\frac{1}{x+1}=\frac{2000}{4004}\)
\(A=\frac{1}{x+1}=\frac{1}{2}-\frac{2000}{4004}\)
\(A=\frac{1}{x+1}=\frac{1}{2002}\)
\(x+1=2002\)
nên \(x=2002-1=2001\)
Vậy x = 2001
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`@` `\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+1)^3=5^3
=>th1: 2x+1=5 =>. 2x=4 =>x=2
th2: 2x+1=-5 => 2x=-6 =>x=-3
vậy x=2 hoặc x=-3
( 2x + 1 )3 = 125
( 2x + 1 )3 = 53
=> 2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2
Vậy x = 2
=))