Tìm X
3× ( X + 5 ) - 9 = 27
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22 tháng 8 2021
\(x^3+27=-x^2+9\Leftrightarrow x^3+x^2+18=0\Leftrightarrow\left(x+3\right)\left(x^2-2x+6\right)=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)(do \(x^2-2x+6=\left(x-1\right)^2+5\ge5>0\))
22 tháng 8 2021
Ta có: \(x^3+27=-x^2+9\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-3\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
3 tháng 9 2023
a: 49x^2-25=0
=>(7x-5)(7x+5)=0
=>7x-5=0 hoặc 7x+5=0
=>x=5/7 hoặc x=-5/7
b: Đề thiếu vế phải rồi bạn
c: (3x-2)^2-9(x+4)(x-4)=2
=>9x^2-12x+4-9(x^2-16)=2
=>9x^2-12x+4-9x^2+144=2
=>-12x+148=2
=>-12x=-146
=>x=146/12=73/6
d: x^3-6x^2+12x-8=0
=>(x-2)^3=0
=>x-2=0
=>x=2
e: x^3-9x^2+27x-27=0
=>(x-3)^3=0
=>x-3=0
=>x=3
H9
HT.Phong (9A5)
CTVHS
3 tháng 9 2023
a) \(-25+49x^2=0\)
\(\Leftrightarrow49x^2-25=0\)
\(\Leftrightarrow\left(7x\right)^2-5^2=0\)
\(\Leftrightarrow\left(7x-5\right)\left(7x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-5=0\\7x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}7x=5\\7x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{7}\\x=-\dfrac{5}{7}\end{matrix}\right.\)
b) \(16x^2-25\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left[5\left(x-2\right)\right]^2=0\)
\(\Leftrightarrow\left(4x-5x+10\right)\left(4x+5x-10\right)=0\)
\(\Leftrightarrow\left(10-x\right)\left(9x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}10-x=0\\9x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{10}{9}\end{matrix}\right.\)
c) \(\left(3x-2\right)^2-9\left(x+4\right)\left(x+4\right)=2\)
\(\Leftrightarrow9x^2-12x+4-9\left(x^2+8x+16\right)=2\)
\(\Leftrightarrow9x^2-12x+4-9x^2-72x-144=2\)
\(\Leftrightarrow-84x-140=2\)
\(\Leftrightarrow-84x=142\)
\(\Leftrightarrow x=-\dfrac{142}{84}\)
\(\Leftrightarrow x=-\dfrac{71}{42}\)
d) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-2^3=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
e) \(-27+27x-9x^2+x^3=0\)
\(\Leftrightarrow x^3-9x^2+27x-27=0\)
\(\Leftrightarrow\left(x-3\right)^3=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
NH
2
LP
13 tháng 8 2021
a)\(3x\left(x-1\right)+2x^2\left(x-1\right)=0\\ \Leftrightarrow x\left(x-1\right)\left(3+2x\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\3+2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=\dfrac{-3}{2}\end{matrix}\right.\)
13 tháng 8 2021
a: Ta có: \(3x^2-3x+2x^3-2x^2=0\)
\(\Leftrightarrow2x^3+x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+x-3\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b: Ta có: \(x^3+27=-x^2+9\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-3\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
CM
11 tháng 9 2019
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
27 tháng 10 2021
a: x=3
b: \(2x-1=2\)
hay \(x=\dfrac{3}{2}\)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
LM
1
29 tháng 10 2021
g) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Rightarrow-2\left(2x-5\right)=0\Rightarrow x=\dfrac{5}{2}\)
i) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-2x\right)=0\Rightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
CM
7 tháng 5 2018
a) x = 4
b) x = 3
c) x = 2
d) x = 1
e) x = 3
f) x = 2
g) x = 4
h) x = 3
19 tháng 12 2021
a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
3×(x+5)-9=27
3×(x+5)=27+9
3×(x+5)=36
(x+5)=36:3
(x+5)=12
x=12-5
x=7
vậy x=7
Giải:
3 x ( x + 5 ) - 9 = 27
3 x ( x + 5 ) = 27 + 9
3 x ( x + 5 ) = 36
x + 5 = 36 : 3
x + 5 = 12
x = 12 - 5
x = 7
Vậy x = 7
học tốt!!!