Bài 2.Tìm x
a,(x-2018) .3=0
b,25(x-8)=0
c,25+(15+x)=75
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
30 tháng 7 2021
a) \(\left(2x-1\right)^2-25=0\)
⇔ \(\left(2x-1\right)^2-5^2=0\)
⇔ \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)
⇒ \(2x-1-5=0\) hoặc \(2x-1+5=0\)
⇔ \(x=3\) hoặc \(x=-2\)
TC
30 tháng 7 2021
Bài 1: Tìm x
a) (2x-1) ² - 25 = 0
<=> (2x-1)2 = 25
<=> 2x-1 = 5 hay 2x-1 =-5
<=> 2x= 6 hay 2x=-4
<=> x=3 hay x= -2
Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0
<=> (x-1)(3x+1)=0
<=> x-1=0 hay 3x+1=0
<=> x=1 hay 3x=-1
<=> x=1 hay x=\(\dfrac{-1}{3}\)
Vậy S={1;\(\dfrac{-1}{3}\)}
c) 2(x+3) - x ² - 3x = 0
<=> 2(x+3)- x(x+3)=0
<=> (x+3)(2-x)=0
<=> x+3=0 hay 2-x=0
<=> x=-3 hay x=2
Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0
<=> x(x-2)+3(x-2)=0
<=> (x-2)(x+3)=0
<=> x-2=0 hay x+3=0
<=> x=2 hay x=-3
Vậy S={2;-3}
e) 4x ² - 4x +1 = 0
<=> (2x-1)2=0
<=> 2x-1=0
<=> 2x=1
<=> x=\(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
f) x +5x2 = 0
<=> x(1+5x)=0
<=>x=0 hay 1+5x=0
<=> x=0 hay 5x=-1
<=> x=0 hay x= \(\dfrac{-1}{5}\)
Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0
<=> x2-x+3x-3=0
<=> x(x-1)+3(x-1)=0
<=> (x-1)(x+3)=0
<=> x-1=0 hay x+3=0
<=> x=1 hay x=-3
Vậy S={1;-3}
LT
1
T
22 tháng 12 2023
a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)
\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)
b) \(\left(x+3\right)^2-5x-15=0\)
\(\Rightarrow\left(x+3\right)^2-5\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+3-5\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
c) \(2x^5-4x^3+2x=0\)
\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)
\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)
\(\Rightarrow2x\left(x^2-1\right)^2=0\)
\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
\(\text{#}Toru\)
31 tháng 7 2021
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
31 tháng 7 2021
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
Y
2
5 tháng 2 2021
\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)
29 tháng 7 2021
`a)4x(x-2)+x-2=0`
`<=>(x-2)(4x+1)=0`
`<=>[(x-2=0),(4x+1=0):}`
`<=>[(x=2),(x=-1/4):}`
Vậy `S={2;-1/4}.`
`b)(3x-1)^3-9=0`
`<=>(3x-1-3)(3x-1+3)=0`
`<=>(3x-4)(3x+2)=0`
`<=>[(3x-4=0),(3x+2=0):}`
`<=>[(x=4/3),(x=-2/3):}`
Vậy `S={4/3;-2/3}.`
`c)x^3-8+(x-2)(x+1)=0`
`<=>(x-2)(x^2+2x+4)+(x-2)(x+1)=0`
`<=>(x-2)(x^2+3x+5)=0`
Mà `x^2+3x+5=(x+3/2)^2+11/4>=11/4>0`
`<=>x-2=0`
`<=>x=2`
Vậy `S={2}`
29 tháng 7 2021
a) Ta có: \(4x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)
b)Ta có: \(\left(3x-1\right)^2-9=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
c) Ta có: \(x^3-8+\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4+x+1\right)=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
6 tháng 8 2021
a, \(4x\left(x-2\right)+x-2=0\Leftrightarrow\left(4x+1\right)\left(x-2\right)=0\Leftrightarrow x=-\dfrac{1}{4};x=2\)
b, \(\left(3x-1\right)^2-9=0\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\Leftrightarrow x=\dfrac{4}{3};x=-\dfrac{2}{3}\)
c, \(x^3-8+\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+5\ne0\right)=0\Leftrightarrow x=2\)
6 tháng 8 2021
a) Ta có: \(4x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)
b) Ta có: \(\left(3x-1\right)^2-9=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
7 tháng 10 2021
a: Ta có: \(2x\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)
KV
16 tháng 10 2023
a) \(6x^2-72x=0\)
\(6x\left(x-12\right)=0\)
\(6x=0\) hoặc \(x-72=0\)
*) \(6x=0\)
\(x=0\)
*) \(x-12=0\)
\(x=12\)
Vậy \(x=0;x=12\)
b) \(-2x^4+16x=0\)
\(-2x\left(x^3-8\right)=0\)
\(-2x=0\) hoặc \(x^3-8=0\)
*) \(-2x=0\)
\(x=0\)
*) \(x^3-8=0\)
\(x^3=8\)
\(x=2\)
Vậy \(x=0;x=2\)
c) \(x\left(x-5\right)-\left(x-3\right)^2=0\)
\(x^2-5x-x^2+6x-9=0\)
\(x-9=0\)
\(x=9\)
d) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(x^3-6x^2+12x-8-x^3+8=0\)
\(-6x^2+12x=0\)
\(-6x\left(x-2\right)=0\)
\(-6x=0\) hoặc \(x-2=0\)
*) \(-6x=0\)
\(x=0\)
*) \(x-2=0\)
\(x=2\)
Vậy \(x=0;x=2\)
a) \(\left(x-2018\right)\cdot3=0\)
\(\Leftrightarrow x-2018=0\)
\(\Leftrightarrow x=2018\)
b) \(25\left(x-8\right)=0\)
\(\Leftrightarrow x-8=0\)
\(\Leftrightarrow x=8\)
c) \(25+\left(15+x\right)=75\)
\(\Leftrightarrow15+x=50\)
\(\Leftrightarrow x=35\)
a) x=2018
b)x=8
c)x=35