Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a ) \(x^2-x+1\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
a) x^4 - x^3 - x + 1
= x^3 ( x - 1 ) - ( x- 1 )
= ( x^3 - 1 )(x - 1)
= ( x- 1 )^2 (x^2 + x + 1 )
a)x4-x3-x+1
=x3(x-1)-(x-1)
=(x-1)(x3-1)
=(x-1)(x-1)(x2+x+1)
=(x-1)2(x2+x+1)
b)5x2-4x+20xy-8y
(sai đề)
\(C=2\left(x-\frac{5}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\Rightarrow C_{min}=\frac{7}{8}\)
\(D=\left(x^2+4xy+4y^2\right)+\left(y^2+y+\frac{1}{4}\right)+\frac{8083}{4}\)
\(D=\left(x+2y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{8083}{4}\ge\frac{8083}{4}\)
\(E=\frac{1}{2}\left(4x^2+y^2+\frac{9}{4}-4xy-6x+3y\right)+\frac{1}{2}\left(y^2+y+\frac{1}{4}\right)+\frac{15}{4}\)
\(E=\frac{1}{2}\left(2x-y-\frac{3}{2}\right)^2+\frac{1}{2}\left(y+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\)
\(A=-\left(x-2\right)^2+11\le11\)
\(B=-\left(x+\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)
\(C=-\left(x-3y\right)^2-\left(y-2\right)^2+11\le11\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
a) \(5ax-15ay+20a\)
\(=5a\left(x-3y+4\right)\)
b) \(6xy-12x-8y\)
\(=6\left(xy-2x-3y\right)\)
c) \(3ab\left(x-y\right)+3a\left(y-x\right)\)
\(=3a\left(x-y\right)\left(b-1\right)\)
d) \(x^2-xy+2x-2y\)
\(=\left(x+2\right)\left(x-y\right)\)
e) \(ax^2-5x^2-ax+5x+a-5\)
\(=\left(a-5\right)\left(x^2-x+1\right)\)
a, \(5ax-15ay+20a=5a\left(x-5y+4\right)\)
b, sai
c, \(3ab\left(x+y\right)+3a\left(y-x\right)=3ab\left(x+y\right)-3a\left(x+y\right)=\left(3ab-3a\right)\left(x+y\right)\)
d, \(x^2-xy+2x-2y=x\left(x+2\right)-y\left(x+2\right)=\left(x-y\right)\left(x+2\right)\)
Tượng tự ...
a) 5ax - 15ay + 20a = 5a( x - 3y + 4 )
b) 6xy - 12x - 8y = 2( xy - 6x - 4y )
c) 3ab( x - y ) + 3a( y - x ) = 3ab( x - y ) - 3a( x - y ) = ( x - y )( 3ab - 3a ) = 3a( x - y )( b - 1 )
d) x2 - xy + 2x - 2y = x( x - y ) + 2( x - y ) = ( x - y )( x + 2 )
e) ax2 - 5x2 - ax + 5x + a - 5 = x2( a - 5 ) - x( a - 5 ) + ( a - 5 ) = ( a - 5 )( x2 - x + 1 )
g) x2y - 4xy2 + 4y3 - 36yz2 = y( x2 - 4xy + 4y2 - 36z2 ) = y[ ( x2 - 4xy + 4y2 ) - 36z2 ] = y[ ( x - 2y )2 - ( 6z )2 ] = y( x - 2y - 6z )( x - 2y + 6z )
h) 4xy - x2 - 4y2 + m2 - 6m + 9
= ( m2 - 6x + 9 ) - ( x2 - 4xy + 4y2 )
= ( m - 3 )2 - ( x - 2y )2
= ( m - 3 - x + 2y )( m - 3 + x - 2y )
i) x2 + x - 12 = x3 - 3x + 4x - 12 = x( x - 3 ) + 4( x - 3 ) = ( x - 3 )( x + 4 )
k) 5x2 + 14x - 3 = 5x2 - x + 15x - 3 = x( 5x - 1 ) + 3( 5x - 1 ) = ( 5x - 1 )( x + 3 )
m) x2 - 5xy + 4y2 = x2 - xy - 4xy + 4y2 = x( x - y ) - 4y( x - y ) = ( x - y )( x - 4y ) < đã sửa đề >
n) 3x2 - 5xy + 2y2 + 4x - 4y = ( 3x2 - 5xy + 2y2 ) + ( 4x - 4y ) = ( 3x2 - 3xy - 2xy + 2y2 ) + 4( x - y ) = [ 3x( x - y ) - 2y( x - y ) ] + 4( x - y ) = ( x - y )( 3x - 2y ) + 4( x - y ) = ( x - y )( 3x - 2y + 4 )
f) 2x3 + 4x2y + 2xy2 = 2x( x2 + 2xy + y2 ) = 2x( x + y )2