K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
1 tháng 8 2018
4x2 - 12x -7 = 0
<=> 4x2 -14x +2x -7 = 0
<=> 2x(2x-7) + (2x-7) = 0
<=> (2x-7)(2x+1) = 0
<=> 2x-7 = 0 hoặc 2x+1 = 0
<=> 2x = 7 hoặc 2x = -1
<=> x= \(\frac{7}{2}\)hoặc x= \(\frac{-1}{2}\)
Vậy tập nghiệm của phương trình là S={\(\frac{7}{2}\);\(\frac{-1}{2}\)}
NH
23 tháng 9 2018
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,=\(4x^2+12x-x-3=4x^2+11x-3\)
4 tháng 9 2022
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
10 tháng 1 2017
Bài 1:
\(x^4-4x^3+12x-9=0\)
\(\Rightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)
\(\Rightarrow x^2\left(x^2-4x+3\right)-3\left(x^2-4x+3\right)=0\)
\(\Rightarrow\left(x^2-4x+3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left(x^2-3x-x+3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[x\left(x-3\right)-\left(x-3\right)\right]\left(x^2-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-1\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-3=0\\x-1=0\\x^2-3=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=3\\x=1\\x=\pm\sqrt{3}\end{matrix}\right.\)
Bài 2:
\(x^4-4x^3+3x^2+4x-4=0\)
\(\Rightarrow x^4-4x^3+4x^2-x^2+4x-4=0\)
\(\Rightarrow x^2\left(x^2-4x+4\right)-\left(x^2-4x+4\right)=0\)
\(\Rightarrow\left(x^2-4x+4\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x-2\right)^2\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-2=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
LN
3
23 tháng 10 2017
a)
\(12x^2-3x=6\\ \Leftrightarrow x^2-\dfrac{1}{4}x=\dfrac{1}{2}\\ \Leftrightarrow x^2-2.\dfrac{1}{8}x+\left(\dfrac{1}{8}\right)^2=\dfrac{1}{2}+\left(\dfrac{1}{8}\right)^2=\dfrac{33}{64}\\ \Leftrightarrow\left(x-\dfrac{1}{8}\right)^2=\dfrac{33}{64}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{8}=\dfrac{\sqrt{33}}{8}\\x-\dfrac{1}{8}=-\dfrac{\sqrt{33}}{8}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{33}}{8}\\x=\dfrac{1-\sqrt{33}}{8}\end{matrix}\right.\)
b)
\(x^2-4x+3=0\\ \Leftrightarrow x^2-4x+4=-3+4=1\\ \Leftrightarrow\left(x-2\right)^2=1\\ \Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
23 tháng 10 2017
c)
\(3x^2-12x=0\\ \Leftrightarrow x^2-4x=0\\ \Leftrightarrow x^2-4x+4=4\\ \Leftrightarrow\left(x-2\right)^2=4\\ \Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
d) TH1:
\(x^2+3x+4=0\\ \Leftrightarrow x^2+2.1,5x+\left(1,5\right)^2=\left(1,5\right)^2-4=-\dfrac{7}{4}\\ \Leftrightarrow\left(x+1,5\right)^2=-\dfrac{7}{4}\left(vô\:lí\right)\)
do đó pt trên vô nghiệm
TH2:
\(x^2+3x-4=0\\ \Leftrightarrow x^2+2.\dfrac{3}{2}x+\dfrac{3}{2}=4+\dfrac{3}{2}=\dfrac{25}{4}\\ \Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=\dfrac{5}{2}\\x+\dfrac{3}{2}=-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{2}=1\\x=-\dfrac{8}{2}=-4\end{matrix}\right.\)
ND
0
ND
1
27 tháng 7 2017
x2 - 4 = 0
x2 = 4
\(\orbr{\begin{cases}x^2=2^2\\x^2=\left(-2\right)^2\end{cases}}\)
\(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
3x2 - 75 = 0
3x2 = 75
x2 = 25
\(\orbr{\begin{cases}x^2=5^2\\x^2=\left(-5\right)^2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
( x + 2 )2 = 25
\(\orbr{\begin{cases}\left(x+2\right)^2=5^2\\\left(x+2\right)^2=\left(-5\right)^2\end{cases}}\)
\(\orbr{\begin{cases}x+2=5\\x+2=-5\end{cases}}\)
\(\orbr{\begin{cases}x=3\\x=-7\end{cases}}\)
11 tháng 10 2020
a) \(x\left(x-2\right)-7x+14=0\)
\(\Leftrightarrow x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) \(x^2+12x-13=0\)
\(\Leftrightarrow\left(x^2-x\right)+\left(13x-13\right)=0\)
\(\Leftrightarrow x\left(x-1\right)+13\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-13\end{cases}}\)
d) \(4x^2-4x=8\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
e) \(x^2-6x=1\)
\(\Leftrightarrow\left(x-3\right)^2=10\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{10}\\x-3=-\sqrt{10}\end{cases}}\Rightarrow\orbr{\begin{cases}x=3+\sqrt{10}\\x=3-\sqrt{10}\end{cases}}\)
11 tháng 10 2020
a) x( x - 2 ) - 7x + 14 = 0
<=> x( x - 2 ) - 7( x - 2 ) = 0
<=> ( x - 2 )( x - 7 ) = 0
<=> \(\orbr{\begin{cases}x-2=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)
b) x2( x - 3 ) + 12 - 4x = 0
<=> x2( x - 3 ) - 4( x - 3 ) = 0
<=> ( x - 3 )( x2 - 4 ) = 0
<=> \(\orbr{\begin{cases}x-3=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) x2 + 12x - 13 = 0
<=> x2 - x + 13x - 13 = 0
<=> x( x - 1 ) + 13( x - 1 ) = 0
<=> ( x - 1 )( x + 13 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\x+13=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-13\end{cases}}\)
d) 4x2 - 4x = 8
<=> 4( x2 - x ) = 8
<=> x2 - x = 2
<=> x2 - x - 2 = 0
<=> x2 + x - 2x - 2 = 0
<=> x( x + 1 ) - 2( x + 1 ) = 0
<=> ( x + 1 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
e) x2 - 6x = 1
<=> x2 - 6x + 9 = 1 + 9
<=> ( x - 3 )2 = 10
<=> ( x - 3 )2 = ( ±√10 )2
<=> \(\orbr{\begin{cases}x-3=\sqrt{10}\\x-3=-\sqrt{10}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3+\sqrt{10}\\x=3-\sqrt{10}\end{cases}}\)
4x2-12x=0
=>4x(x-3)=0
=>x=0 hoặc x-3=0
=>x=0 hoặc x=3