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a. \(2a^2+5ab-3b^2-7b-2\)
\(=\left(2a^2+6ab+2a\right)-\left(ab+3b^2+b\right)-\left(2a+6b+2\right)\)
\(=2a\left(a+3b+1\right)-b\left(a+3b+1\right)-2\left(a+3b+1\right)\)
\(=\left(2a-b-2\right)\left(a+3b+1\right)\)
b. \(2x^2-7xy+x+3y^2-3y\)
\(=\left(2x^2-xy\right)-\left(6xy-3y^2\right)+\left(x-3y\right)\)
\(=x\left(2x-y\right)-3y\left(2x-y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y\right)+\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x-y+1\right)\)
c. \(6x^2-xy-2y^2+3x-2y\)
\(=\left(6x^2+3xy\right)-\left(4xy-2y^2\right)+\left(3x-2y\right)\)
\(=3x\left(2x+y\right)-2y\left(2x+y\right)+\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(2x+y\right)+\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(2x+y+1\right)\)
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)
Ta có:
a) 6x2y - 3y2 - 2x2 + y = (6x2y - 2x2) - (3y2 - y) = 2x2(3y - 1) - y(3y - 1) = (2x2 - y)(3y - 1)
b) 2x2 + x - 4xy - 2y + 2x + 1 = (x2 + x) - (4xy + 2y) + (x2 + 2x + 1) = x(x + 1) - 2y(2x + 1) + (x + 1)2
= (x + x + 1)(x + 1) - 2y(2x + 1) = (2x + 1)(x + 1) - 2y(2x + 1) = (2x + 1)(x + 1 - 2y)
c) 16x2y - 4xy2 - 4x3 + x2y = 4xy(4x - y) - x2(4x - y) = (4xy - x2)(4x - y)
d) 4x2 - 20x + 25 - 36y2 = (2x - 5)2 - (6y)2 = (2x - 5 - 6y)(2x - 5 + 6y)
e) x2 - 4y2 + 6x - 4y + 8 = (x2 + 6x + 9) - (4y2 + 4y + 1) = (x + 3)2 - (2y + 1)2 = (x + 3 - 2y - 1)(x + 3 + 2y + 1) = (x + 2 - 2y)(x + 4 + 2y)
g) Ta có : x10 + x5 + 1
= (x10 - x) + (x5 - x2) + (x2 + x + 1)
= x(x9 - 1) + x2(x3 - 1) + (x2 + x + 1)
= x(x3 - 1)(x6 + x3 + 1) + x2(x3 - 1) + (x2 + x + 1)
= (x7 + x4 + x)(x - 1)(x2 + x + 1) + x2(x - 1)(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x8 - x7 + x 5 - x4 + x2 - x + x4 + x3 + x2 + 1)
= (x2 + x + 1)(x8 - x7 + x5 + x3 - x + 1)
h) TT trên (dài dòng ktl)
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 \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
\(a)\) \(x^2-2x-4y^2-4y\)
\(=\)\(\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\)\(\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\)\(\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\)\(\left(x-2y-2\right)\left(x+2y\right)\)
\(=\)\(2\left(x-y\right)\left(x+2y\right)\)
Chúc bạn học tốt ~
a) Ta có x2 - 2x - 4y2 - 4y
= x2 - 2x + 1 - 4y2 - 4y - 1
= (x - 1)2 - (4y2 + 4y + 1)
= (x - 1)2 - (2y + 1)2
= (x - 1 - 2y - 1)(x - 1 + 2y + 1)
= (x - 2y - 1)(x + 2y)
Bài làm
a) 4x - 8y
<=> 4( x - 2y )
b) 12x( x - 2y ) - 8y( x - 2y )
<=> ( 12x - 8y )( x - 2y )
<=> 4( 3x - 2y )( x - 2y )
c) 2x + 2y - x2 - xy
= 2( x + y ) - x( x + y )
= ( x + y )( 2 - x )
d) x2 - 4y2
<=> ( x - 2y )( x + 2y )
e) x3 + x2y - 4x - 4y
<=> x2( x + y ) - 4( x + y )
<=> ( x - 2 )( x + 2 )( x + y )
g) 3x2 - 6xy + 3y2 - 12x3
<=>3( x2 - 3xy + y2 - 4x3 )
# Học tốt #
a)4(x-2y)
b)(x-2y)(12x-8y)
=4(x-2y)(3x-2y)
c)2(x+y)-x(x+y)
=(2-x)(x+y)
d)(x-2y)(x+2y)
e)x2(x+y)-4(x+y)
=(x+y)(x2-4)
=(x+y)(x-2)(x+2)
g)3(x2-2xy+y2-4z3)
=3[(x-y)2-4z3]
????????????phải là 4z2chứ nhỉ.....
a) \(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
b) \(5x^2-5xy-3x+3y\)
\(=5x\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-3\right)\)
c) \(x^2-2x-4y^2+1\)
\(=\left(x-1\right)^2-4y^2\)
\(=\left(x-2y-1\right)\left(x+2y-1\right)\)
c) 2x^3y - 2xy^3 - 4xy^2 - 2xy
= 2xy ( x^2 - y^2 - 2y - 1 )
= 2xy ( x^2 - ( y^2 + 2y + 1 )
= 2xy ( x^2 - ( y + 1 )^2 )
= 2x ( x - y - 1 )( x + y + 1 )
sai bạn ơi !
đáp án là
= 2xy (x + y + 1) (x - y + 1)
that pun cho ban Nguyen Dieu Thao :((
a) \(3x^2-3y^2+2x+2y\)
\(=3\left(x^2-y^2\right)+2\left(x+y\right)\)
\(=3\left(x+y\right)\left(x-y\right)+2\left(x+y\right)\)
\(=\left(x+y\right)\left(3x-3y+2\right)\)
d) \(4x^2-4x-3\)
\(=\left(2x\right)^2-2.2x+1-4\)
\(=\left(2x-1\right)^2-2^2\)
\(=\left(2x-3\right)\left(2x+1\right)\)