Bài 1: Tìm x
a) 3(x-1)^2.3x(x-5)=0
b) (x+3)^2-5x-15=0
c) 2x^5-4x^3+2x=0
K
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LT
1
Những câu hỏi liên quan
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=\(\dfrac{19}{24}\)
30 tháng 7 2021
a) \(\left(2x-1\right)^2-25=0\)
⇔ \(\left(2x-1\right)^2-5^2=0\)
⇔ \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)
⇒ \(2x-1-5=0\) hoặc \(2x-1+5=0\)
⇔ \(x=3\) hoặc \(x=-2\)
TC
30 tháng 7 2021
Bài 1: Tìm x
a) (2x-1) ² - 25 = 0
<=> (2x-1)2 = 25
<=> 2x-1 = 5 hay 2x-1 =-5
<=> 2x= 6 hay 2x=-4
<=> x=3 hay x= -2
Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0
<=> (x-1)(3x+1)=0
<=> x-1=0 hay 3x+1=0
<=> x=1 hay 3x=-1
<=> x=1 hay x=\(\dfrac{-1}{3}\)
Vậy S={1;\(\dfrac{-1}{3}\)}
c) 2(x+3) - x ² - 3x = 0
<=> 2(x+3)- x(x+3)=0
<=> (x+3)(2-x)=0
<=> x+3=0 hay 2-x=0
<=> x=-3 hay x=2
Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0
<=> x(x-2)+3(x-2)=0
<=> (x-2)(x+3)=0
<=> x-2=0 hay x+3=0
<=> x=2 hay x=-3
Vậy S={2;-3}
e) 4x ² - 4x +1 = 0
<=> (2x-1)2=0
<=> 2x-1=0
<=> 2x=1
<=> x=\(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
f) x +5x2 = 0
<=> x(1+5x)=0
<=>x=0 hay 1+5x=0
<=> x=0 hay 5x=-1
<=> x=0 hay x= \(\dfrac{-1}{5}\)
Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0
<=> x2-x+3x-3=0
<=> x(x-1)+3(x-1)=0
<=> (x-1)(x+3)=0
<=> x-1=0 hay x+3=0
<=> x=1 hay x=-3
Vậy S={1;-3}
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 \(x=0;x=2\)
26 tháng 9 2021
e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
f: Ta có: \(x^3-6x^2+12x-19=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)
\(\Leftrightarrow\left(x-2\right)^3=11\)
hay \(x=\sqrt[3]{11}+2\)
8 tháng 4 2021
a) Ta có: (5x-1)(x-3)<0
nên 5x-1 và x-3 trái dấu
Trường hợp 1:
\(\left\{{}\begin{matrix}5x-1>0\\x-3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x< 3\end{matrix}\right.\Leftrightarrow\dfrac{1}{5}< x< 3\)
Trường hợp 2:
\(\left\{{}\begin{matrix}5x-1< 0\\x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x>3\end{matrix}\right.\Leftrightarrow loại\)
Vậy: S={x|\(\dfrac{1}{5}< x< 3\)}
T
22 tháng 10 2023
\(a,(x-2)^2-25=0\\\Leftrightarrow (x-2)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
\(---\)
\(b,4x(x-2)+x-2=0\\\Leftrightarrow4x(x-2)+(x-2)=0\\\Leftrightarrow(x-2)(4x+1)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(---\)
\(c,4x(x-2)-x(3+4x)(?)\)
\(d,(2x-5)^2-3x(5-2x)=0\\\Leftrightarrow(2x-5)^2+3x(2x-5)=0\\\Leftrightarrow(2x-5)(2x-5+3x)=0\\\Leftrightarrow(2x-5)(5x-5)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\5x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=1\end{matrix}\right.\)
\(---\)
\(e,x^2-25-(x+5)=0(sửa.đề)\\\Leftrightarrow(x^2-5^2)-(x+5)=0\\\Leftrightarrow (x-5)(x+5)-(x+5)=0\\\Leftrightarrow(x+5)(x-5-1)=0\\\Leftrightarrow(x+5)(x-6)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
\(---\)
\(f,5x(x-3)-x+3=0\\\Leftrightarrow5x(x-3)-(x-3)=0\\\Leftrightarrow(x-3)(5x-1)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
\(Toru\)
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}
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
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) \(2x^5-4x^3+2x=0\)
\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)
\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)
\(\Rightarrow2x\left(x^2-1\right)^2=0\)
\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
\(\text{#}Toru\)