2X . 4 =128
X15- x
( 2x + 1 )3 = 125
( x + 5)2 (x + 1 ) =0
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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
1) 3x+4 = 243
3x+4 = 35
=> x + 4 = 5
x = 5 - 4
x = 1
Vậy x = 1
2) (x-3)2 = 25
(x-3)2 = 52
=> x - 3 = 5
x = 5 + 3
x = 8
Vậy x = 8
3) (2x-1)3 = 125
(2x-1)3 = 53
=> 2x - 1 =5
2x = 5 + 1
2x = 6
x = 6:2
x = 3
Vậy x=3
5) 15 - x = 7 - (-2)
15 - x = 7 + 2
15 - x = 9
x = 15 - 9
x = 6
6) / x - 3 / = 7 - (-2)
/ x - 3 / = 7 + 2
/ x - 3 / = 9
TH1: x - 3 = 9
x = 9 + 3
x = 12
TH2: x - 3 = -9
x = -9 + 3
x = -6
Vậy x = 12,-6
7) /x-5/ = /-7/
x - 5 = -7
x = -7 + 5
x = -2
1,=\(x^2-3x-2x^2+6x=-x^2+3x\)
2,=\(3x^2-x-5+15x=3x^2+14x-5\)
3,=\(5x+15-6x^2-6x=-6x^2-x+15\)
4,=\(4x^2+12x-x-3=4x^2+11x-3\)
5: =>(x+5)^3=0
=>x+5=0
=>x=-5
6: =>(2x-3)^2=0
=>2x-3=0
=>x=3/2
7: =>(x-6)(x-10)=0
=>x=10 hoặc x=6
8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)
=>(x-4)^3=0
=>x-4=0
=>x=4
Cho mình câu chả lời sớm nhất vào 4h30 chiều ngày 28/9/2021
1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
5: =>4x^2-1/9=0
=>(2x-1/3)(2x+1/3)=0
=>x=1/6 hoặc x=-1/6
6: =>x-1=2
=>x=3
7:=>(2x-1)^3=-27
=>2x-1=-3
=>2x=-2
=>x=-1
8: =>1/8(x-1)^3=-125
=>(x-1)^3=-1000
=>x-1=-10
=>x=-9
3: =>(5x-5)^2-4=0
=>(5x-7)(5x-3)=0
=>x=3/5 hoặc x=7/5
4: =>(5x-1)^2=0
=>5x-1=0
=>x=1/5
1: =>(3x-1)(2x-1)=0
=>x=1/3 hoặc x=1/2
2: =>x^2(2x-3)-4(2x-3)=0
=>(2x-3)(x^2-4)=0
=>(2x-3)(x-2)(x+2)=0
=>x=3/2;x=2;x=-2
`@` `\text {Answer}`
`\downarrow`
`1,`
\(2x\left(3x-1\right)+1-3x=0\)
`<=> 2x(3x - 1) - 3x + 1 = 0`
`<=> 2x(3x - 1) - (3x - 1) = 0`
`<=> (2x - 1)(3x-1) = 0`
`<=>`\(\left[{}\begin{matrix}2x-1=0\\3x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy, `S = {1/2; 1/3}`
`2,`
\(x^2\left(2x-3\right)+12-8x=0\)
`<=> x^2(2x - 3) - 8x + 12 =0`
`<=> x^2(2x - 3) - (8x - 12) = 0`
`<=> x^2(2x - 3) - 4(2x - 3) = 0`
`<=> (x^2 - 4)(2x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy, `S = {+-2; 3/2}`
`3,`
\(25\left(x-1\right)^2-4=0\)
`<=> 25(x-1)(x-1) - 4 = 0`
`<=> 25(x^2 - 2x + 1) - 4 = 0`
`<=> 25x^2 - 50x + 25 - 4 = 0`
`<=> 25x^2 - 15x - 35x + 21 = 0`
`<=> (25x^2 - 15x) - (35x - 21) = 0`
`<=> 5x(5x - 3) - 7(5x - 3) = 0`
`<=> (5x - 7)(5x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy, `S = {7/5; 3/5}`
`4,`
\(25x^2-10x+1=0\)
`<=> 25x^2 - 5x - 5x + 1 = 0`
`<=> (25x^2 - 5x) - (5x - 1) = 0`
`<=> 5x(5x - 1) - (5x - 1) = 0`
`<=> (5x - 1)(5x-1)=0`
`<=> (5x-1)^2 = 0`
`<=> 5x - 1 = 0`
`<=> 5x = 1`
`<=> x = 1/5`
Vậy,` S = {1/5}.`
2x.4=128
=>2x=128:4
=>2x=32
Mà: 25=32
=>2x=25
=>x=5
2x.4=128
=>2x=128:4
=>2x=32
Mà: 25=32
=>2x=25
=>x=5