A) 5(x+3)-2x(3+x)=0
b)4x(x-2004)-x+2004
c)(x+1)2=x+1
tìm x nhé
K
Khách
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7 tháng 11 2021
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
7 tháng 11 2021
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
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=\(\dfrac{19}{24}\)
LH
1
20 tháng 10 2021
a: \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b: \(\left(x+1\right)^2-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
SL
2
LC
24 tháng 9 2016
Tìm x
a) 5(x+3)-2x(3+x)=0
\(\Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=\frac{5}{2}\end{array}\right.\)
b) 4x(x-2004)-x+2004
\(\Leftrightarrow4x\left(x-2004\right)-\left(x-2004\right)=0\)
\(\Leftrightarrow\left(x-2007\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2007\\x=\frac{1}{4}\end{array}\right.\)
c) (x+1)2=x+1
\(\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)
24 tháng 9 2016
c) \(\left(x+1\right)^2=x+1\)
\(\Rightarrow\left(x+1\right)^2-\left(x+1\right)=0\)
\(\Rightarrow x+1.\left(x+1-1\right)=0\)
\(\Rightarrow\left(x+1\right).x=0\)
\(\Rightarrow x+1=0\) hoặc \(x=0\)
+) \(x+1=0\Rightarrow x=-1\)
Vậy x = 0 hoặc x = -1
19 tháng 10 2021
\(a,x\left(x+9\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Rightarrow x\left(x^2+4x+4\right)=0\\ \Rightarrow x\left(x+2\right)^2=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ c,\Rightarrow\left(x-5-4\right)\left(x-5+4\right)=0\\ \Rightarrow\left(x-9\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\\ d,\Rightarrow3\left(x+2\right)-x\left(x+2\right)=0\\ \Rightarrow\left(x+2\right)\left(3-x\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ e,\Rightarrow x^3+6x^2+12x+8-x^3-6x^2=4\\ \Rightarrow12x=-4\Rightarrow x=-\dfrac{1}{3}\\ g,\Rightarrow\left(x+2\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
CM
8
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) \(2x^5-4x^3+2x=0\)
\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)
\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)
\(\Rightarrow2x\left(x^2-1\right)^2=0\)
\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
\(\text{#}Toru\)
10 tháng 11 2021
\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
a/ \(5\left(x+3\right)-2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=\frac{5}{2}\end{array}\right.\)
b/ \(4x\left(x-2004\right)-x+2004=0\)
\(\Leftrightarrow4x\left(x-2004\right)-\left(x-2004\right)=0\)
\(\Leftrightarrow\left(x-2007\right)\left(4x-1\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2007\\x=\frac{1}{4}\end{array}\right.\)
c/ \(\left(x+1\right)^2=x+1\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)
A) 5(x+3)-2x(3+x)=0
=> 5(x+3)-2x(x+3)=0
=> (5-2x)(x+3)=0
\(\Rightarrow\left[\begin{array}{nghiempt}5-2x=0\\x+3=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)