tìm x:
x^2-2x-120=0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) <=>5x-x=57+7
<=> 4x=64
<=> x=16
b) <=>4x=28+12
<=> x=40/4
<=>10
c) th1 x=0
th2 x-3=0
<=>x=3
d) th1 2x-4=0
<=> x=2
th2 x+1=0
<=> x=-1
câu e mình ko biết làm hihi
a, 5x - 7 = x + 57
5x - x = 57 + 7
4x = 64
x = 64 : 4
x = 16
b, x + 3x - 12 = 28
4x = 28 + 12
4x = 40
x = 10
c, x(x-3) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)
d, (2x-4)(x+1)=0
\(\Rightarrow\orbr{\begin{cases}2x-4=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
e, 1+2+...+x=120
=> \(\frac{x\left(x+1\right)}{2}=120\)
=> x(x+1) = 120.2
=>x(x+1) = 240
=> x(x+1) = 15.16
=> x=16
126-2.(x-1)=20 120+3.(x-3)=180
2.(x-1)=126-20 3.(x-3)=180-120
2.(x-1)=106 3.(x-3)=60
x-1=106:2 x-3=60:3
x-1=53 x-3=20
x=53+1 x=20+3
x=54 x=23
a. ( x + 3 )( x - 3 ) = 16
⇔x2-9=16
⇔x2-16-9=0
⇔x2-25=0
⇔(x-5)(x+5)=0
⇔\(\left[{}\begin{matrix}x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=5\end{matrix}\right.\)
\(\left(x.7-7\right)\left(x.12+24\right)=0\)
=> \(\orbr{\begin{cases}x.7-7=0\\x.12+24=0\end{cases}}\)
=> \(\orbr{\begin{cases}7x=7\\12x=-24\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
\(170-\left(4+5x\right)=101\)
=> \(4+5x=170-101\)
=> \(4+5x=69\)
=> \(5x=69-4\)
=> \(5x=65\)
=> \(x=65:5=13\)
(x.7-7).(x.12+24)=0
* x.7-7=0 * x.12+24=0
x.7=7 x.12=24
x=7:7 x=24:12
x=1 x=2
vạy x=1 hoặc x=2
170-(4+5x)=101
4+5x=170-101
4+5x=69
5x=69-4
5x=65
x=65:5
x=13
vậy x=13
250:(2x-4)=145-120
250:(2x+4)=25
2x+4=250:25
2x+4=10
2x=10+4
2x=14
x=14:2
x=7
vậy x=7
6.x+3.x-4.x=105
x.(6+3-4)=105
x.(9-4)=105
x.5=105
x=105:5
x=21
vậy x=21
a: (x-35)-120=0
=>x-35=120
=>x=120+35=155
b: 59(x-29)=0
=>x-29=0
=>x=29
c: 2x+3x=45:3
=>5x=15
=>x=3
`@` `\text {Ans}`
`\downarrow`
`(x - 35) - 120 = 0`
`x - 35 = 0 + 120`
`x - 35 = 120`
`x = 120+ 35`
`x = 155`
Vậy, `x = 155`
____
`(x - 29)*59 = 0`
`x - 29 = 0 \div 59`
`x - 29 = 0`
`x = 0 + 29`
`x = 29`
Vậy, `x= 29`
____
`2x + 3x = 45 \div 3`
`(2 + 3)x = 15`
`5x = 15`
`x = 15 \div 5`
`x = 3`
Vậy, `x = 3.`
\(x^2-2x-120=0\)
=>\(x^2-12x+10x-120=0\)
=>x(x-12)+10(x-12)=0
=>(x-12)(x+10)=0
=>\(\left[\begin{array}{l}x-12=0\\ x+10=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=12\\ x=-10\end{array}\right.\)
\(x^2-2x-120=0\)
\(x^2-2x+1-121=0\)
\(\left(x-1\right)^2-11^2\) =0
\(\left(x-1+11\right)\left(x-1-11\right)=0\)
\(\left(x+10\right)\left(x-12\right)=0\)
\(\left[\begin{array}{l}x+10=0\Rightarrow x=-10\\ x-12=0\Rightarrow x=12\end{array}\right.\)
Vậy x = -10; x = 12