Tìm x:
(3x-1)(3x+1)+9(x-2)^2=0
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H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
c) \(x^2-9=2\cdot\left(x+3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left[x-3-2\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3-2x-6\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
d) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x^2-8x\right)+\left(3x-24\right)=0\)
\(\Leftrightarrow x\left(x-8\right)+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)
1 tháng 9 2023
a) \(x^2-9=2\left(x+3\right)^2\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)=2\left(x+3\right)^2\)
\(\Leftrightarrow2\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[2\left(x+3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[2x+6-x+3\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+9\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+9=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x-8\right)x+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
c) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
H
18 tháng 10 2023
\(\dfrac{1}{2}-3x+\left|x-1\right|=0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}-0\\ \Rightarrow3x+\left|x-1\right|=\dfrac{1}{2}\\ \Rightarrow\left|x-1\right|=\dfrac{1}{2}-3x\\ \Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}-3x\\x-1=-\dfrac{1}{2}+3x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+3x=\dfrac{1}{2}+1\\x-3x=-\dfrac{1}{2}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=\dfrac{3}{2}\\2x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
__
\(\dfrac{1}{2}\left|2x-1\right|+\left|2x-1\right|=x+1\\ \Rightarrow\left|2x-1\right|\cdot\left(\dfrac{1}{2}+1\right)=x+1\\ \Rightarrow\left|2x-1\right|\cdot\dfrac{3}{2}=x+1\\ \Rightarrow\left|2x-1\right|=x+1:\dfrac{3}{2}\\ \Rightarrow\left|2x-1\right|=x+\dfrac{2}{3}\\ \Rightarrow\left[{}\begin{matrix}2x-1=x+\dfrac{2}{3}\\2x-1=-x-\dfrac{2}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-x=\dfrac{2}{3}+1\\2x+x=-\dfrac{2}{3}+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\3x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{1}{9}\end{matrix}\right.\)
5 tháng 11 2023
2: \(3x\left(x-4\right)+2x-8=0\)
=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
3: 4x(x-3)+x2-9=0
=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(4x+x+3\right)=0\)
=>\(\left(x-3\right)\left(5x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4: \(x\left(x-1\right)-x^2+3x=0\)
=>\(x^2-x-x^2+3x=0\)
=>2x=0
=>x=0
5: \(x\left(2x-1\right)-2x^2+5x=16\)
=>\(2x^2-x-2x^2+5x=16\)
=>4x=16
=>x=4
19 tháng 12 2021
g: \(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4=0\)
\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+36=0\)
\(\Leftrightarrow\left(x+3\right)^2\left(x^2+6x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\sqrt{5}-3\\x=-\sqrt{5}-3\end{matrix}\right.\)
MA
1
7 tháng 11 2021
a) 15x2-3x=0
=>3x(5x-1)=0
=>2 TH
=>*3x=0 *5x-1=0
=>x=0 =>5x=1=>x=1/5
vậy x=0 hoặc x=1/5
b) (3x-2) (x+3)+ (x2-9)=0
=>(3x-2)(x+3)+(x-3)(x+3)=0
=>(x+3).(3x-2+x-3)=0
=>(x+3).(4x-5)=0
=> 2 TH
*x+3=0=>x=0-3=>x=-3
*4x-5=0=>4x=5=>x=5/4
vậy x=-3 hoặc x=5/4
c) (x-1)3- (x+1) (2-3x)=-3
\(\Rightarrow\left(x-1\right)^3-\left(x+1\right)\left(2-3x\right)+3=0\)
\(\Rightarrow\left(x^3-3x^2+3x-1\right)-\left(2x-3x^2+2-3x\right)+3=0\)
\(\Rightarrow x^3-3x^2+3x-1-2x+3x^2-2+3x+3=0\)
\(\Rightarrow x^3-3x^2+3x^2+3x-2x+3x-1-2+3=0\)
\(\Rightarrow x^3+4x=0\)
\(\Rightarrow x\left(x^2+4\right)=0\)
=> 2 TH
*x=0
*x^2+4=0
vì: x^2>0
do đó:x^2+4>0
=> x^2+4 ko có gt nào x t/m y/cầu đề bài
vậy x=0
4 tháng 5 2020
Làm 1 câu thuii nha mik nhát quá!! nhưng các bài còn lại tương tự nha!!
a. \(\left(x+1\right)\left(3-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\3-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
Vậy..
hok tốt!!
HA
4 tháng 5 2020
\(\left(x+1\right)\left(3-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
vậy x=-1 hoặc x=3
\(\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\2x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}}\)
vậy x=2 hoặc x=1/2
câu c tương tự
\(\left(x^2+1\right)\left(81-x^2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\81-x^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\x^2=81\end{cases}}}\Leftrightarrow\orbr{\begin{cases}x\in\varnothing\\x=\pm9\end{cases}}\)
vậy x=9 hoặc x=-9
3 tháng 10 2021
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
VH
10 tháng 4 2022
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ML
12 tháng 8 2015
2)a)3x(x-5)-(x-1)(2+3x)=30
<=>3x2-15x-3x2+x+2=30
<=>-14x+2=30
<=>-14x=30-2
<=>-14x=28
<=>x=-2
b)(x+2)(x+3)-(x-2)(x+5)=0
<=>x2+5x+6-x2-3x+10=0
<=>2x+16=0
<=>2x=-16
<=>x=-8
c)(3x+2)(2x+9)-(x+2)(6x+1)=9
<=>6x2+31x+18-6x2-13x-2=9
<=>18x+16=9
<=>18x=9-16
<=>18x=-7
<=>x=-7/18
(3x-1)(3x+1)+9(x-2)2 = 0
9x2 + 9(x2 - 4x +4)=0
9x2 + 9x2 - 36x + 36 = 0
18x2 -36x +36 = 0
x2 - 2x + 2 = 0
x2 - 2x +1 +1=0
(x-1)2 + 1 = 0
(x-1)2 > =0 nên (x-1)2 +1> 0 => x không có GTTM