Tìm x:
3x^3-27x=0
K
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15 tháng 7 2021
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
TA
16 tháng 10 2019
1)3.x^2 - 75 = 0
3.x^2 - 3.25 = 0
3.(x^2-25)=0
x^2-5^2=0
(x-5)(x+5)=0
=> x-5=0 hoặc x+5=0
=> x=5 hoặc x=-5
KN
16 tháng 10 2019
1) \(3x^2-75=0\)
\(\Leftrightarrow3\left(x^2-25\right)=0\)
\(\Leftrightarrow x^2-25=0\)
\(\Leftrightarrow x^2=25\)
\(\Leftrightarrow x=\pm\sqrt{25}=\pm5\)
2) \(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
3) \(x^3+3x^2+3x=0\)
\(\Leftrightarrow x^3+3x^2+3x+1=1\)
\(\Leftrightarrow\left(x+1\right)^3=1^3\)
\(\Leftrightarrow x+1=1\Leftrightarrow x=0\)
24 tháng 8 2020
a, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)
\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)
b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P
Tự làm nốt nhé
NV
1
2 tháng 12 2016
a, 3x^2-27x=0
3x(x-9)=0
3x=0=>x=0
x-9=0=>x=9
b,2/3x(x^2-4)=0
2/3x=0=>x=0
x^2-4=0=>x=2
NV
2
PA
2 tháng 12 2016
\(3x^2-27x=0\)
\(3x\left(x-9\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x-9=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)
\(\frac{2}{3}x\left(x^2-4\right)=0\)
\(\frac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x-2=0\\x+2=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=2\\x=-2\end{array}\right.\)
HN
2 tháng 12 2016
a)\(3x^2-27x=0\)
\(3x\left(x-9\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x=0\\x-9=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)
b) \(\frac{2}{3}x\left(x^2-4\right)=0\)
\(\frac{2}{3}x\left(x+2\right)\left(x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\frac{2}{3}x=0\\x+2=0\\x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\\x=2\end{array}\right.\)
F
1
VQ
30 tháng 11 2017
a)x^2-5=0
x^2=5
x=2.236
b) 3x^3-27x=0
=)x=3
C)5x(x-1)-x+1=0
=)x=1
D)2(x+5)-x^2-5×=0
=)x=2
3x3 - 27 x = 0
=> 3x ( x2 - 9 ) = 0
\(=>\orbr{\begin{cases}3x=0\\x^2-9=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x^2=9\end{cases}}\)
=> x = + 3
hoặc x = 0
\(3x^2-27x=0\)
\(\Leftrightarrow3x\left(x+3\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)