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14 tháng 9 2021
a) \(2x\left(x-3\right)+3\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(x\left(3x-1\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)+5\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
15 tháng 10 2021
a:Ta có: \(16-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
ZH
1
17 tháng 11 2021
\(a,\Leftrightarrow\left(x^2-8x+16\right)-10=0\\ \Leftrightarrow\left(x-4\right)^2-10=0\\ \Leftrightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\\ b,\Leftrightarrow10\left(2x-1\right)+6x=9x\\ \Leftrightarrow20x-10-3x=0\\ \Leftrightarrow17x=10\Leftrightarrow x=\dfrac{10}{17}\)
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) \(2x^5-4x^3+2x=0\)
\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)
\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)
\(\Rightarrow2x\left(x^2-1\right)^2=0\)
\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
\(\text{#}Toru\)
ZH
1
5 tháng 10 2021
\(a,\Rightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(2x-1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow\left(x^2-8x+16\right)-10=0\\ \Rightarrow\left(x-4\right)^2-10=0\\ \Rightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\)
K
30 tháng 6 2021
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
TK
1
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 \(x=0;x=2\)
KV
1 tháng 1
a) x² - 4 = 0
x² = 4
x = 2 hoặc x = -2
b) 2x(x + 5) - 3(5 + x) = 0
(x + 5)(2x - 3) = 0
X + 5 = 0 hoặc 2x - 3 = 0
*) x + 5 = 0
x = -5
*) 2x - 3 = 0
2x = 3
x = 3/2
c) x³ - 6x² + 11x - 6 = 0
x³ - x² - 5x² + 5x + 6x - 6 = 0
(x³ - x²) - (5x² - 5x) + (6x - 6) = 0
x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0
(x - 1)(x² - 5x + 6) = 0
(x - 1)(x² - 2x - 3x + 6) = 0
(x - 1)[(x² - 2x) - (3x - 6)] = 0
(x - 1)[x(x - 2) - 3(x - 2)] = 0
(x - 1)(x - 2)(x - 3) = 0
x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0
*) x - 1 = 0
x = 1
*) x - 2 = 0
x = 2
*) x - 3 = 0
x = 3
Vậy x = 1; x = 2; x = 3
NT
3
27 tháng 6 2021
\(a,=3x-9-4x+12=-x+3=0\)
\(\Leftrightarrow x=3\)
Vậy ..
\(b,=\left(x+2\right)\left(x+2-x+2\right)=4\left(x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy ..
\(c,=x^3-3x^2+3x-1=\left(x-1\right)^3=0\)
\(\Leftrightarrow x=1\)
Vậy ..
\(d,\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy ..
\(e,=\left(2x-3-5\right)\left(2x-3+5\right)=\left(2x-8\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)
Vậy ...
27 tháng 6 2021
a) Ta có: 3(x-3)-4x+12=0
\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
hay x=3
Vậy: S={3}
b) Ta có: \(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow x^2+4x+4-x^2+4=0\)
\(\Leftrightarrow4x=-8\)
hay x=-2
Vậy: S={-2}
c) Ta có: \(x^3+3x=3x^2+1\)
\(\Leftrightarrow x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
Vậy: S={1}
d) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: S={0;2;-2}
\(a,\Leftrightarrow\left(x-5\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)