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DT
2
3 tháng 7 2019
\(a,4x^2-12x+y=9\left(x-2\right)\)
y ở đâu ở đây ???
\(b,x^3+3x^2+3x+1=0\)
\(\Rightarrow\left(x+1\right)^3=0\Rightarrow x=-1\)
3 tháng 7 2019
1) b) x^3 + 3x^2 + 3x + 1 = 0
<=> (x + 3)^3 = 0
<=> x = -1
=> x = -1
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
27 tháng 9 2020
a) \(3x^3-12x=0\)
=> \(3x\left(x^2-4\right)=0\)
=> \(\orbr{\begin{cases}3x=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
=> \(x^2\left(x-3\right)+\left(-4x+12\right)=0\)
=> \(x^2\left(x-3\right)-4x+12=0\)
=> \(x^2\left(x-3\right)-4\left(x-3\right)=0\)
=> \(\left(x-3\right)\left(x^2-4\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}\)
c) \(\left(3x-1\right)^2-\left(2x-3\right)^2=0\)
=> \(\left[3x-1-\left(2x-3\right)\right]\left(3x-1+2x-3\right)=0\)
=> \(\left(3x-1-2x+3\right)\left(3x-1+2x-3\right)=0\)
=> \(\left(x+2\right)\left(5x-4\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{4}{5}\end{cases}}\)
d) \(x^2-4x-21=0\)
=> \(x^2+3x-7x-21=0\)
=> \(x\left(x+3\right)-7\left(x+3\right)=0\)
=> (x + 3)(x - 7) = 0 => x = -3 hoặc x = 7
e) 3x2 - 7x - 10 = 0
=> 3x2 + 3x - 10x - 10 = 0
=> 3x(x + 1) - 10(x + 1) = 0
=> (x + 1)(3x - 10) = 0
=> x = -1 hoặc x = 10/3
F
27 tháng 9 2020
a) \(3x^3-12x=0\)
\(\Leftrightarrow3x\left(x^2-4\right)=0\)
\(\Leftrightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{-2;0;2\right\}\)
b) \(x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\in\left\{-2;2;3\right\}\)
c) \(\left(3x-1\right)^2-\left(2x-3\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5x-4\right)=0\)
\(\Leftrightarrow x\in\left\{-2;\frac{4}{5}\right\}\)
XO
27 tháng 9 2020
Ta có : 3x3 - 12x = 0
=> 3x(x2 - 4) = 0
=> x(x - 2)(x + 2) = 0
=> \(x\in\left\{0;2;-2\right\}\)
b) x2(x - 3) + 12 - 4x = 0
=> x2(x - 3) - 4(x - 3) = 0
=> (x2 - 4)(x - 3) = 0
=> \(\orbr{\begin{cases}x^2-4=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=4\\x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm2\\x=3\end{cases}}\)
Vậy \(x\in\left\{-2;2;3\right\}\)
c) (3x - 1)2 - (2x - 3)2 = 0
=> (3x - 1 - 2x + 3)(3x - 1 + 2x - 3) = 0
=> (x + 2)(5x - 4) = 0
=> \(\orbr{\begin{cases}x+2=0\\5x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=0,8\end{cases}}\)
Vậy \(x\in\left\{-2;0,8\right\}\)
d) x2 - 4x - 21 = 0
=> x2 - 7x + 3x - 21 = 0
=> x(x - 7) + 3(x - 7) = 0
=> (x + 3)(x - 7) = 0
=> \(\orbr{\begin{cases}x+3=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=7\end{cases}}\)
Vậy \(x\in\left\{-3;7\right\}\)
e) 3x2 - 7x - 10 = 0
=> 3x2 + 3x - 10x - 10 = 0
=> 3x(x + 1) - 10(x + 1) = 0
=> (3x - 10)(x + 1) = 0
=> \(\orbr{\begin{cases}3x-10=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=-1\end{cases}}\)
Vậy \(x\in\left\{\frac{10}{3};-1\right\}\)
PV
1
15 tháng 10 2017
Bài 4 :
\(\left(5x-20\right)+\left(3x^2-12x\right)=0\)
\(\Leftrightarrow5\left(x-4\right)+3x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(5+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5+3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(x=4\) hoặc \(x=-\dfrac{5}{3}\)
Bài 5 :
\(\left(1-x\right)-3x^2+3x=0\)
\(\Leftrightarrow\left(1-x\right)-\left(3x^2-3x\right)=0\)
\(\Leftrightarrow\left(1-x\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow\left(1-x\right)+3x\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\1+3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=-\dfrac{1}{3}\)
Bài 6 :
\(\left(4x+20\right)-\left(x+5\right)^2=0\)
\(\Leftrightarrow4\left(x+5\right)-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(4-x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\-x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)
Vậy \(x=-5\) hoặc \(x=-1\)
22 tháng 2 2020
\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
\(TH1:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
\(TH2:x+6=0\Leftrightarrow x=-6\)
\(TH3:x^2+5=0\Leftrightarrow x^2=5\Leftrightarrow x=\sqrt{5}\)( ns vô nghiệm cx ko sai nha )
\(\left(2x+5\right)^2=\left(3x-1\right)^2\)
\(2x+5=3x-1\)
\(2x-3x=-1-5\)
\(-1x=-6\)
\(x=6\)
15 tháng 2 2017
a) Gần giống cho nó giống luôn.
cần thêm (-x^3+2x^2-x) là giống
\(\left(x-1\right)^4+x^3-2x^2+x=\left(x-1\right)^4+x\left(x^2-2x+1\right)=\left(x-1\right)^4+x\left(x-1\right)^2\)
\(\left(x-1\right)^2\left[\left(x-1\right)^2+x\right]\)
\(\left[\begin{matrix}x-1=0\Rightarrow x=0\\\left(x-1\right)^2+x=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\end{matrix}\right.\)
Nghiệm duy nhất: x=1
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)
\(x^2-4x+3=0\\ \Leftrightarrow x^2-4x+4=-3+4=1\\ \Leftrightarrow\left(x-2\right)^2=1\\ \Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
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:
\(x^2+3x-4=0\\ \Leftrightarrow x^2+2.\dfrac{3}{2}x+\dfrac{3}{2}=4+\dfrac{3}{2}=\dfrac{25}{4}\\ \Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=\dfrac{5}{2}\\x+\dfrac{3}{2}=-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{2}=1\\x=-\dfrac{8}{2}=-4\end{matrix}\right.\)