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23 tháng 10 2018
1. \(f\left(x\right)=x^3-7x-6\)
\(=x^3+x^2-x^2-x-6x-6\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+6\right)\)
\(=\left(x+1\right)\left(x^2+2x-3x-6\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x-3\right)\)
2/ \(f\left(x\right)=x^3+4x^2-7x-10\)
\(=x^3+5x^2-x^2-5x-2x-10\)
\(=x^2\left(x+5\right)-x\left(x+5\right)-2\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-x-2\right)\)
\(=\left(x+5\right)\left[\left(x^2-2x+x-2\right)\right]\)
\(=\left(x+5\right)\left[x\left(x-2\right)+\left(x-2\right)\right]\)
\(=\left(x+5\right)\left(x+1\right)\left(x-2\right)\)
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
S
23 tháng 12 2016
a, \(x^4+6x^3+7x^2-6x+1\)
\(=x^4-2x^2+1+6x^3+9x^2+6x\)
\(=\left(x^2-1\right)^2+6x\left(x^2-1\right)+9x^2\)
\(=\left(x^2-1+3x\right)^2\)
b, \(x^4-7x^3+14x^2-7x+1\)
\(=x^4+2x^2+1+7x^3+12x^2-7x\)
\(=\left(x^2+1\right)^2-7x\left(x^2+1\right)+12^2\)
\(=\left(x^2-1+3x\right)^2\)
c, \(12x^2-11x-36\)
\(=12x^2-27x+16x-36\)
\(=3x\left(4x-9\right)+4\left(4x-9\right)\)
\(=\left(4x-9\right)\left(3x+4\right)\)
AH
Akai Haruma
Giáo viên
10 tháng 10 2018
1)
\(15x^3+29x^2-8x-12=(15x^3+30x^2)-(x^2+2x)-(6x+12)\)
\(=15x^2(x+2)-x(x+2)-6(x+2)\)
\(=(x+2)(15x^2-x-6)=(x+2)(15x^2-10x+9x-6)\)
\(=(x+2)[5x(3x-2)+3(3x-2)]\)
\(=(x+2)(3x-2)(5x+3)\)
AH
Akai Haruma
Giáo viên
10 tháng 10 2018
2)
\(x^3+4x^2-29x+24=(x^3-x^2)+(5x^2-5x)-(24x-24)\)
\(=x^2(x-1)+5x(x-1)-24(x-1)\)
\(=(x-1)(x^2+5x-24)\)
\(=(x-1)(x^2-3x+8x-24)\)
\(=(x-1)[x(x-3)+8(x-3)]=(x-1)(x-3)(x+8)\)
NN
0
31 tháng 7 2022
a: \(A=x^2+3x+\dfrac{9}{4}+y^2-6y+9+1993.75\)
\(=\left(x+\dfrac{3}{2}\right)^2+\left(y-3\right)^2+1993.75>=1993.75\)
Dấu '=' xảy ra khi x=-3/2 và y=3
b: \(=3\left(x^2+\dfrac{7}{3}x+3\right)\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{59}{36}\right)\)
\(=3\left(x+\dfrac{7}{6}\right)^2+\dfrac{59}{12}>=\dfrac{59}{12}\)
Dấu '=' xảy ra khi x=-7/6
c: \(=4\left(x^2-\dfrac{15}{4}x+5\right)\)
\(=4\left(x^2-2\cdot x\cdot\dfrac{15}{8}+\dfrac{225}{64}+\dfrac{95}{64}\right)\)
\(=4\left(x-\dfrac{15}{8}\right)^2+\dfrac{95}{16}>=\dfrac{95}{16}\)
Dấu '=' xảy ra khi x=15/8
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 )
<=> \(2x^2-2x+9x-9=0\)
<=> \(2x\left(x-1\right)+9\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(2x+9\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\2x+9=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=\frac{-9}{2}\end{matrix}\right.\)
\(4x^2-7x+3=0\)
<=> \(4x^2-4x-3x+3=0\)
<=>\(4x\left(x-1\right)-3\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(4x-3\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
\(2\left(x+5\right)=x^2+5x\)
<=> \(2\left(x+5\right)-x^2-5x=0\)
<=>\(2\left(x+5\right)-x\left(x+5\right)=0\)
<=>\(\left(x+5\right)\left(2-x\right)=0\)
<=>\(\left\{{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
AN
6 tháng 10 2018
a)\(x^2-13x+36=x^2-4x-9x+36=x\left(x-4\right)-9\left(x-4\right)=\left(x-9\right)\left(x-4\right)\)
b)\(x^2+3x-18=x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x+6\right)\left(x-3\right)\)
c)\(x^2-5x-24=x^2+3x-8x-24=x\left(x+3\right)-8\left(x+3\right)=\left(x-8\right)\left(x+3\right)\)
d)\(3x^2-16x+5=3x^2-x-15x+5=x\left(3x-1\right)-5\left(3x-1\right)=\left(x-5\right)\left(3x-1\right)\)
e)\(8x^2+30x+7=8x^2+28x+2x+7=4x\left(2x+7\right)+\left(2x+7\right)=\left(4x+1\right)\left(2x+7\right)\)
g)\(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(2x-1\right)\left(x-3\right)\)
h)\(6x^2-7x+3=6x^2-9x-2x+3=3x\left(2x-3\right)-\left(2x-3\right)=\left(3x-1\right)\left(2x-3\right)\)
i)\(3x^2-14x+11=3x^2-3x-11x+11=3x\left(x-1\right)-11\left(x-1\right)=\left(3x-11\right)\left(x-1\right)\)
k)\(5x^2+8x-13=5x^2-5x+13x-13=5x\left(x-1\right)+13\left(x-1\right)=\left(5x+13\right)\left(x-1\right)\)
KB
6 tháng 10 2018
a ) \(x^2-13x+36=x^2-4x-9x+36=x\left(x-4\right)-9\left(x-4\right)=\left(x-9\right)\left(x-4\right)\)
b ) \(x^2+3x-18=x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x+6\right)\left(x-3\right)\)
c ) \(x^2-5x-24=x^2-3x+8x-24=x\left(x-3\right)+8\left(x-3\right)=\left(x+8\right)\left(x-3\right)\)
d ) \(3x^2-16x+5=3x^2-15x-x+5=3x\left(x-5\right)-\left(x-5\right)=\left(3x-1\right)\left(x-5\right)\)
e ) \(8x^2+30x+7=8x^2+2x+28x+7=2x\left(4x+1\right)+7\left(4x+1\right)=\left(2x+7\right)\left(4x+1\right)\)
g ) \(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(2x-1\right)\left(x-3\right)\)
h ) \(6x^2-7x-20=6x^2-15x+8x-20=3x\left(2x-5\right)+4\left(2x-5\right)=\left(3x+4\right)\left(2x-5\right)\)
i ) \(3x^2-14x+11=3x^2-3x-11x+11=3x\left(x-1\right)-11\left(x-1\right)=\left(3x-11\right)\left(x-1\right)\)
k ) \(5x^2+8x-13=5x^2-5x+13x-13=5x\left(x-1\right)+13\left(x-1\right)=\left(5x+13\right)\left(x-1\right)\)
\(6x^2-7x-20\)
\(=6x^2-15x+8x-20\)
\(=3x\left(2x-5\right)+4\left(2x-5\right)\)
\(=\left(2x-5\right)\left(3x+4\right)\)
6x2-7x-20
=6x2-15x+8x-20
=6x2+8x-15x-20
=3x(2x-5)-4(2x-5)
=(2x-5)(3x-4)