tìm x
(3x-1)3 = 125
(10+x) : 25 = 30 . 22
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\(a,5x-16=40-1\)
<=>\(5x=40-1+16\)
<=>\(5x=55\)
<=>\(x=11\)
\(b,x-10=-25\)
<=>\(x=\left(-25\right)-10\)
<=>\(x=-35\)
\(c,-12+x=-30\)
<=>\(x=\left(-30\right)+12\)
<=>\(x=-18\)
a) 5x-16=40-1
5x-16=39
5x=39+16
5x=55
x=55:5
x=11
b)x-10=(-25)
x=(-25)+10
x=(-15)
c) -12+x= -30
x= -30-(-12)
x= -30+12
x= -18
d) 2x +12 = -40+ 6
2x+12=(-34)
2x=(-34)-12
2x=(-46)
x=(-46):2
x=(-23)
e)125:(3x-13)=25
3x-13=125:25
3x-13=5
3x=5+13
3x=18
x=18:3
x=6
f)541+(218-x)=735
218-x=735-541
218-x=194
x=218-194
x= 24
g) 3(2x+1)-19=14
3(2x+1)=14+19
3(2x+1)=33
2x+1=33:3
2x+1=11
2x=11-1
2x=10
x=10:2
x=5
h)175-5(x+3)=85
5(x+3)=175-85
5(x+3)=90
x+3=90:5
x+3=18
x=18-3
x=15
i)8x+(-3)=39
8x=39-(-3)
8x=42
x=42:8
x=42/8
L)2x+4x=36+(-6)
6x=30
x=30:6
x=5
a ) 126 - 2x = 12 - ( -28 )
126 - 2x = 40
2x = 126 - 40
2x = 86
x = 86 : 2
x = 43
b ) 25 - 5( x + 1 ) = 30 - 125
25 - ( 5x + 5 ) = -95
25 - 5x - 5 = -95
20 - 5x = -95
5x = 20 - ( - 95 )
5x = 115
x = 115 : 5
x = 23
c ) I 3x - 1 I - 25 = 73
I 3x - 1 I = 73 + 25
I 3x - 1 I = 98
=> 3x - 1 = 98 hoặc 3x - 1 = -98
=> 3x = 98 + 1 hoặc 3x = -98 + 1
=> 3x = 99 hoặc 3x = -97
=> x = 99 : 3 hoặc x = -97 : 3
=> x = 33 hoặc x = -93/3
d ) 2( x + 2 ) - 6 = 120 - 4x
2x + 4 - 6 = 120 - 4x
2x - 2 = 120 - 4x
2x + 4x = 120 + 2
6x = 122
x = 122 : 6
x = 61/3
a) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
b) \(3^x+25=26\times2^2+2\times3^0\)
\(3^x+25=26\times4+2\times1\)
\(3^x+25=106\)
\(3^x=106-25\)
\(3^x=81\)
\(3^x=3^4\)
\(x=4\)
(2x+1)3 = 125
a)<=> (2x+1)3 = 53
<=> 2x+1 = 5
<=> 2x = 4
<=> x = 2
3^x+25=26 . 2^2 + 2. 3^0
b)3^x+25=104 +2
3^x+25=106
3^x=106+25
3^x=81=3^4
=> x=4
Bài 2:
\(a,45+170+25+30\)
\(=\left(45+25\right)+\left(170+30\right)\)
\(=60+200=260\)
Bài 3:
\(a,\left(x-6\right).5=150\)
\(x-6=150:5\)
\(x-6=30\)
\(x=30+6\)
\(x=36\)
\(b,2^5.\left(3x-2\right)=2^3.2^6\)
\(2^5.\left(3x-2\right)=2^{3+6}\)
\(2^5.\left(3x-2\right)=2^9\)
\(3x-2=2^9:2^5\)
\(3x-2=2^4=16\)
\(3x=16+2\)
\(3x=22\)
\(x=22:3\)
\(x\approx7,3\)
\(c,100-7.\left(x-5\right)=51\)
\(7.\left(x-5\right)=100-51\)
\(7.\left(x-5\right)=49\)
\(x-5=49:7\)
\(x-5=7\)
\(x=7+5\)
\(x=12\)
Phần d) bạn thiếu dữ liệu ạ.
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}.`
125-5(3x-1)=25
5(3x-1)=125-25
5(3x-1)=100
3x-1=100:5
3x-1=20
3x =20+1
3x = 21
x=21:3
x= 7
Vậy x = 7
125-5 3x = 25+1
125-5 3x = 26
120 3x = 26
3x = 120 . 26
3x = 3,120 : 3
x = 1,040
Làm đúng ko ?
( 3x - 1 )3 = 125
( 3x - 1 )3 = 53
=> 3x - 1 = 5
3x = 6
x = 2
( 10 + x ) : 25 = 30 . 22
( 10 + x ) : 25 = 4
10 + x = 100
x = 90