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AL
4
23 tháng 10 2021
\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
RH
31 tháng 8 2021
a) (x - 3)2 - 5.(x - 2) + 5 = 0.
<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0
<=> x^2 - 11x + 24 = 0
<=> (x-3)(x-8)=0
<=> x = 3 hoặc x = 8
26 tháng 10 2021
a: \(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
c: \(\Leftrightarrow\left(x-1\right)\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
26 tháng 10 2021
a) \(x^2-6x=0\\ \Leftrightarrow x\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c) \(9x^2\left(x-1\right)=x-1\\ \Leftrightarrow\left(9x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(3x+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
d) \(x^2-4=\left(x-2\right)^2\\ \Leftrightarrow\left(x-2\right)\left(x+2-x+2\right)=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x=2\)
e) \(\Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
f) \(x^3-0,36=0\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
g) \(\Leftrightarrow\left(5x-1\right)\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2018\end{matrix}\right.\)
h) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
13 tháng 12 2023
a: \(3\left(x-3\right)-6x=0\)
=>\(3x-9-6x=0\)
=>-3x-9=0
=>3x+9=0
=>3x=-9
=>\(x=-\dfrac{9}{3}=-3\)
b: Đề thiếu vế phải rồi bạn
c: \(2\left(x-3\right)+3x=9\)
=>2x-6+3x=9
=>5x-6=9
=>5x=6+9=15
=>x=15/5=3
d: \(x\left(x-11\right)+2\left(x-11\right)=0\)
=>\(\left(x-11\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)
e: \(x\left(x+2\right)+8=x^2\)
=>\(x^2+2x+8=x^2\)
=>2x+8=0
=>2x=-8
=>x=-8/2=-4
f: \(8\left(x+1\right)+2x=-2\)
=>\(8x+8+2x=-2\)
=>10x=-2-8=-10
=>\(x=-\dfrac{10}{10}=-1\)
g: 12-3(x+2)=0
=>3(x+2)=12
=>x+2=12/3=4
=>x=4-2=2
E
15 tháng 9 2021
a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0
TH2 3x-4=0 <=>x=4/3
KL:.....
b) (x+3)(x−1)+2x(x+3)=0.
<=> (x+3)(x-1+2x)=0
TH1: x+3=0 <=> x=-3
TH2 x-1=0 <=> x=1
KL:.....
c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 \)
KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)
KL:....
15 tháng 9 2021
a) \(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(9x^2+6x+1=0\)
\(\Leftrightarrow\left(3x+1\right)^2=0\)
\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
d) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
1 tháng 10 2021
a) \(\left(5-x\right)\left(x-1\right)=-2x\left(x-1\right)\)
\(\Rightarrow\left(5-x\right)\left(x-1\right)+2x\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(5-x+2x\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
b) \(\left(x+3\right)^2-\left(x-13\right)\left(x+13\right)=0\)
\(\Rightarrow x^2+6x+9-x^2+169=0\)
\(\Rightarrow6x=-178\Rightarrow x=-\dfrac{89}{3}\)
6 tháng 11 2021
\(a,\Leftrightarrow\left(x+3\right)\left(x+3-x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow x=0\left(x^2+4>0\right)\)
NH
6 tháng 11 2021
\(a,x^2+2.x.3+3^2-\left(x^2-3^2\right)=0\)
\(x^2+6x+9-x^2+9=0\)
\(6x+18=0\)
\(6x=-18\)
\(x=-3\)
Vậy x=-3
\(b,5x^3+20x=0\)
\(5x\left(x^2+4\right)=0\)
\(Th1:5x=0=>x=0\)
\(Th2:x^2+4=0\)
\(x^2=-4\)(vô lý)
Vậy x=0
(2-x)^3+(2+x)^3-12x(x+1)=0
=>\(8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x\left(x+1\right)=0\)
=>\(12x^2+16-12x^2-12x=0\)
=>16-12x=0
=>4-3x=0
=>x=4/3