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NN
2 tháng 7 2019
a) a4 + a2 - 2
a4 + 2a2 - a2 - 2
a2.( a2 + 2 ) - ( a2 + 2 )
( a2 - 1 ).( a2 + 2 )
( a + 1 ).( a - 1 ).( a2 +2 )
b) x4 + 4x2 - 5
x4 + 5x2 - x2 - 5
x2.( x2 + 5 ) - ( x2 + 5 )
( x2 - 1 ).( x2 + 5 )
( x + 1 ).( x - 1 ).( x2 + 5 )
c) x3 - 19x - 30
x3 + 2x2 - 2x2 + 4x - 15x - 30
x2( x + 2 ) - 2x.( x + 2 ) - 15.( x + 2 )
( x + 2 ).( x2 - 2x - 15 )
d) x3 - 7x - 6
x3 - 3x2 + 3x2 - 9x + 2x - 6
x2.( x - 3 ) + 3x.( x - 3 ) + 2.( x - 3 )
( x - 3 ).( x2 + 3x +2 )
( x - 3 ).( x2 + 2x + x + 2 )
( x - 3 ).( x.( x + 2 ) + ( x + 2 )
( x + 1 ).( x + 2 ).( x - 3 )
e) x3 - 5x2 - 14x
x3 - 7x2 + 2x2 - 14x
x2.( x - 7 ) + 2x.( x - 7 )
( x - 7 ).( x2 + 2x )
x.( x + 2 ).( x - 7 )
8 tháng 10 2016
b) 3x4-3x3+9x3-9x2-24x2+24x-48x+48
=3x3(x-1)+9x2(x-1)-24x(x-1)-48(x-1)
=(x-1)(3x3+9x2-24x-48)
=3(x-1)(x3+3x2-8x-16)
5 tháng 1 2022
\(\dfrac{7x^2-14x+7}{3x^2-3x}=\dfrac{7\left(x^2-2x+1\right)}{3x\left(x-1\right)}=\dfrac{7\left(x-1\right)^2}{3x\left(x-1\right)}=\dfrac{7\left(x-1\right)}{3x}\)
29 tháng 11 2017
\(a,\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7x+7}{3x}\)
\(b,\dfrac{2a^2-2ab}{ac+ad-bc-bd}=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}=\dfrac{2a\left(a-b\right)}{\left(a-b\right)\left(c+d\right)}=\dfrac{2a}{c+d}\)
\(c,\dfrac{x^2-xy}{y^2-x^2}=\dfrac{x\left(x-y\right)}{-\left(x-y\right)\left(x+y\right)}=\dfrac{x}{-x-y}\)
29 tháng 11 2017
a) \(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
b) \(\dfrac{2a^2-2ab}{ac+ad-bc-bd}=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}\) ( có sửa đề )
\(=\dfrac{2a\left(a-b\right)}{\left(c+d\right)\left(a-b\right)}=\dfrac{2a}{c+d}\)
c) \(\dfrac{x^2-xy}{y^2-x^2}=\dfrac{-x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{-x}{\left(x+y\right)}\)
4 tháng 8 2020
Bài 4:
a) Ta có: \(a^4+a^2+1\)
\(=a^4+2a^2+1-a^2\)
\(=\left(a^2+1\right)^2-a^2\)
\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)
b) Ta có: \(a^4+a^2-2\)
\(=a^4+2a^2-a^2-2\)
\(=a^2\left(a^2+2\right)-\left(a^2+2\right)\)
\(=\left(a^2+2\right)\left(a^2-1\right)\)
\(=\left(a^2+2\right)\left(a-1\right)\left(a+1\right)\)
c) Ta có: \(x^4+4x^2-5\)
\(=x^4+5x^2-x^2-5\)
\(=x^2\left(x^2+5\right)-\left(x^2+5\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
d) Ta có: \(x^3-19x-30\)
\(=x^3-25x+6x-30\)
\(=x\left(x^2-25\right)+6\left(x-5\right)\)
\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
e) Ta có: \(x^3-7x-6\)
\(=x^3-4x-3x-6\)
\(=x\left(x^2-4\right)-3\left(x+2\right)\)
\(=x\left(x-2\right)\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x-3\right)\)
\(=\left(x+2\right)\left(x^2-3x+x-3\right)\)
\(=\left(x+2\right)\left[x\left(x-3\right)+\left(x-3\right)\right]\)
\(=\left(x+2\right)\left(x-3\right)\left(x+1\right)\)
f) Ta có: \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x^2-7x+2x-14\right)\)
\(=x\left[x\left(x-7\right)+2\left(x-7\right)\right]\)
\(=x\left(x-7\right)\left(x+2\right)\)
S
1
PA
9 tháng 9 2016
\(A=3x^2-14x^2+4x+3\)
Giả sử:
\(A=\left(3x+a\right)\left(x^2+bx+c\right)\)
\(=3x^3+3bx^2+3cx+ax^{2\:}+abx+ac\)
\(=3x^3+\left(3b+a\right)x^2+\left(3c+ab\right)x+ac\)
Ta có:
\(\begin{cases}3b+a=-14\\3c+ab=4\\ac=3\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=-5\\c=3\end{cases}\)
Vậy \(A=\left(3x+1\right)\left(x^2-5x+3\right)\)
NV
3
20 tháng 8 2019
x4+7x3+14x2+14x+4
=x4+7x3+4x2+10x2+14x+4
=(x4+4x2+4)+(7x3+14x)+10x2
=(x2+2)2+7x(x2+2)+10x2
=(x2+2)2+2x(x2+2)+5x(x2+2)+10x2
=(x2+2)(x2+2+2x)+5x(x2+2+2x)
=(x2+2+2x)(x2+2+5x)
\(\dfrac{7x^3+14x^2+7x}{14x^2+14x}=\dfrac{7x\left(x^2+2x+1\right)}{14x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{2\left(x+1\right)}=\dfrac{x+1}{2}\)
\(\dfrac{7x^3+14x^2+7x}{14x^2+14x}=\dfrac{7x\left(x^2+2x+1\right)}{14x\left(x+1\right)}=\dfrac{7x\left(x+1\right)^2}{14x\left(x+1\right)}=\dfrac{\left(x+1\right)}{2}\)