F) 9x2 -1
K
Khách
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NG
a,x3+3x2+3x+1
b,x2+6x+9
c,-x3+9x2-27x+27
d,x2+4x+4
k,10x-25-x2
f,(x+y)2-9x2
g,8x3+42x2y+16xy2+6xy+y3
2
H9
HT.Phong (9A5)
CTVHS
3 tháng 7 2023
a) \(x^3+3x^2+3x+1=x^2+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x-1\right)^3\)
b) \(x^2+6x+9=x^2+2\cdot3\cdot x+3^2=\left(x+3\right)^2\)
c) \(-x^3+9x^2-27x+27\)
\(=-\left(x^3-9x^2+27x-27\right)\)
\(=-\left(x^3-3\cdot3\cdot x^2+3\cdot3^2\cdot x-3^3\right)=-\left(x-3\right)^3\)
d) \(x^2+4x+4=x^2+2\cdot2\cdot x+2^2=\left(x+2\right)^2\)
k) \(10x-25-x^2=-x^2+10x-25=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2\cdot5\cdot x+5^2\right)=-\left(x-5\right)^2\)
f) \(\left(x+y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left[\left(x-y\right)-3x\right]\left[\left(x-y\right)+3x\right]\)
\(=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
C
1
AH
Akai Haruma
Giáo viên
9 tháng 9 2021
Lời giải:
Khi $m=-3$ thì $f(x)=5x^3-9x^2+2x-3$
$f(x)=5x^3-9x^2+2x-3=5x^2(x-1)-4x(x-1)-2(x-1)-5$
$=(x-1)(5x^2-4x-2)-5$
Như vậy, với mọi số tự nhiên $x\neq 1$, để $f(x)\vdots x-1$ thì $5\vdots x-1$ hay $x-1$ là ước của $5$
$\Rightarrow x-1\in\left\{\pm 1;\pm 5\right\}$
$\Leftrightarrow x\in\left\{2;0;-4;6\right\}$
Mà $x$ tự nhiên nên $x\in\left\{0;2;6\right\}$
6 tháng 7 2021
a) Ta có: \(x^3+x^2+4\)
\(=x^3+2x^2-x^2+4\)
\(=x^2\left(x+2\right)-\left(x+2\right)\left(x-2\right)\)
\(=\left(x+2\right)\left(x^2-x+2\right)\)
b) Ta có: \(9x^2+12x-5\)
\(=9x^2+15x-3x-5\)
\(=3x\left(3x+5\right)-\left(3x+5\right)\)
\(=\left(3x+5\right)\left(3x-1\right)\)
c) Ta có: \(x^4+1997x^2+1996x+1997\)
\(=x^4+x^2+1+1996x^2+1996x+1996\)
\(=\left(x^4+2x^2+1-x^2\right)+1996\left(x^2+x+1\right)\)
\(=\left[\left(x^2+1\right)^2-x^2\right]+1996\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+1996\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)
d) Ta có: \(x^2-x-2001\cdot2002\)
\(=x^2-2002x+2001x-2001\cdot2002\)
\(=x\left(x-2002\right)+2001\left(x-2002\right)\)
\(=\left(x-2002\right)\left(x+2001\right)\)
CM
29 tháng 12 2018
Đáp án D
f ' x = x 2 x 2 − 3 x x 2 − 9 x 2 + 4 x + 3 khi đó f ' x đổi dấu khi đi qua điểm x = 0 , x = − 1
Suy ra f(x) có 2 điểm cực trị
NT
1
26 tháng 10 2021
a: \(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
c: \(\Leftrightarrow\left(x-1\right)\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
26 tháng 10 2021
a) \(x^2-6x=0\\ \Leftrightarrow x\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c) \(9x^2\left(x-1\right)=x-1\\ \Leftrightarrow\left(9x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(3x+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
d) \(x^2-4=\left(x-2\right)^2\\ \Leftrightarrow\left(x-2\right)\left(x+2-x+2\right)=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x=2\)
e) \(\Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
f) \(x^3-0,36=0\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
g) \(\Leftrightarrow\left(5x-1\right)\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2018\end{matrix}\right.\)
h) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
9 tháng 12 2021
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
9\(x^2\) - 1
= (3\(x\) - 1)(3\(x\) + 1)