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3 tháng 7 2018
1) \(x^2+6x+8\)
\(=x^2+2x+4x+8\)
\(=x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+4\right)\left(x+2\right)\)
2) \(x^2-5x-14\)
\(=x^2-7x+2x-14\)
\(=x\left(x-7\right)+2\left(x-7\right)\)
\(=\left(x-7\right)\left(x+2\right)\)
3 tháng 7 2018
3) \(2x^2+5x+3\)
\(=2x^2+2x+3x+3\)
\(=2x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+3\right)\)
4) \(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
TT
11 tháng 10 2020
a/ Sai đề à??
\(\left(2x^3-3\right)^2-\left(4x^2-9\right)=0\)
\(\Leftrightarrow4x^6-12x^3+9-4x^2+9=0\)
\(\Leftrightarrow4x^6-13x^2-4x^2+18=0\)
b/ \(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x^2+3+2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\) (do \(x^2+3+2x>0\forall x\))
d/ \(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
14 tháng 12 2018
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
10 tháng 10 2017
******************************************************
a) \(x^3-5x^2+8x-4=x^3-x^2-4x^2+4x+4x-4\)
\(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
b) \(x^3-3x+2=x^3+2x^2-2x^2-4x+x+2\)
\(=x^2\left(x+2\right)-2x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x+1\right)=\left(x+2\right)\left(x-1\right)^2\)
c) \(x^3-5x^2+3x+9=x^3+x^2-6x^2-6x+9x+9\)
\(=x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+9\right)=\left(x+1\right)\left(x-3\right)^2\)
d) \(x^3+8x^2+17x+10=x^3+2x^2+6x^2+12x+5x+10\)
\(=x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+6x+5\right)=\left(x+2\right)\left(x+5\right)\left(x+1\right)\)
e) \(x^3+3x^2+6x+4=x^3+x^2+2x^2+2x+4x+4\)
\(=x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+2x+4\right)\)
f) \(x^3+3x^2+3x+2=x^3+2x^2+x^2+2x+x+2\)
\(=x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+x+1\right)\)
10 tháng 7 2018
a) 5x - 5y + ax - ay = 5(x - y) + a(x - y)
= (5 + a)(x - y)
b) x3 - x + 3x2y + 3xy2 + y3 - y
= (x3 + 3x2y + 3xy2 + y3) - (x + y)
= (x + y)3 - (x + y)
= (x + y)[(x + y)2 - 1]
= (x + y)(x + y + 1)(x + y - 1)
c) x2 - 2x - 3 = x2 + x - 3x -3
= x(x + 1) - 3(x + 1)
= (x - 3)(x + 1)
e) 6x - 9 - x2 = 3x - 9 + 3x - x2
= 3(x - 3) + x(3 - x)
= 3(x - 3) - x(x - 3)
= (3 - x)(x - 3)
26 tháng 10 2022
b: \(=x^4+x^2+36-2x^3+12x^2-12x+x^2-6x+9\)
\(=x^4-2x^3+14x^2-18x+45\)
\(=x^4+9x^2-2x^3-18x+5x^2+45\)
\(=\left(x^2+9\right)\left(x^2-2x+5\right)\)
d: \(=2x^4+2x^3+6x^2-x^3-x^2-3x+x^2+x+3\)
\(=\left(x^2+x+3\right)\left(2x^2-x+1\right)\)
e: \(=3x^4-3x^3-3x^2-2x^3+2x^2+2x+2x^2-2x-2\)
\(=\left(x^2-x-1\right)\left(3x^2-2x+1\right)\)
A
17 tháng 7 2019
ý bạn là như thế này đúng không ạ:
a/ \(x^2-6x+5=0\)
\(x^2-5x-x+5=0\)
\(x\left(x-5\right)-\left(x-5\right)=0\)
\(\left(x-5\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}x-5=0\rightarrow x=5\\x-1=0\rightarrow x=1\end{cases}}\)
b/\(2x^2+7x+9=0\)
?!
c/ \(4x^2-7x+3=0\)
\(4x^2-4x-3x+3=0\)
\(4x\left(x-1\right)-3\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x-3\right)=0\)
\(\orbr{\begin{cases}x-1=0\Rightarrow x=1\\4x-3=0\Rightarrow x=\frac{3}{4}\end{cases}}\)
d/ \(2\left(x+5\right)=2x+10\)
-,- mik ko rõ đề ạ, sai thì ibox ạ.Cảm ơn
T
17 tháng 7 2019
\(x^2-6x+5=0\)
<=> \(x^2-x-5x+5=0\)
<=> \(x\left(x-1\right)-5\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x-5\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Vậy phương trình có nghiệm là x=1 và x=5
\(2x^2+7x-9=0\) ( nếu là 9 thì ko ra kq đc nên mình đổi thành -9 nha )
<=> \(2x^2-2x+9x-9=0\)
<=> \(2x\left(x-1\right)+9\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(2x+9\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\2x+9=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=\frac{-9}{2}\end{matrix}\right.\)
\(4x^2-7x+3=0\)
<=> \(4x^2-4x-3x+3=0\)
<=>\(4x\left(x-1\right)-3\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(4x-3\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
\(2\left(x+5\right)=x^2+5x\)
<=> \(2\left(x+5\right)-x^2-5x=0\)
<=>\(2\left(x+5\right)-x\left(x+5\right)=0\)
<=>\(\left(x+5\right)\left(2-x\right)=0\)
<=>\(\left\{{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
\(5x\left(x-3\right)-x^2+6x-9\\ =5x\left(x-3\right)-\left(x^2-6x+9\right)\\ =5x\left(x-3\right)-\left(x-3\right)^2\\ =\left(x-3\right)\left(5x-x+3\right)\\ =\left(x-3\right)\left(4x+3\right)\)