b) 9x\(^3\)-9x\(^2\)y-4x+4y
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a) \(x-xy+y-y^2=x\left(1-y\right)+y\left(1-y\right)=\left(x+y\right)\left(1-y\right)\)
b) \(x^2-2x-y^2+1=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
c) \(4x^2-4xy+y^2=\left(2x\right)^2-2.2x.y+y^2=\left(2x-y\right)^2\)
d) \(9x^3-9x^2y-4x+4y=9x^2\left(x-y\right)-4\left(x-y\right)=\left(9x^2-4\right)\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\)
e) \(x^3+2+3\left(x^3-2\right)=x^3+2+3x^3-6=4x^3-4=4\left(x^3-1\right)=4\left(x-1\right)\left(x^2+x+1\right)\)
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
Bài 2:
\(A=x^2+4y^2-2x+10-4xy-4y\)
\(=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
Thay x + 2y = 5 vào biểu thức A ta được: \(A=5^2-2.5+10=25\)
\(B=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)\left(y-1\right)+y^2-2y+1\)
\(=x^2+4xy+4y^2-2xy+2x-4y^2+4y+y^2-2y+1\)
\(=x^2+2xy+y^2+2x+2y+1\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1\)
Thay x + y = 5 vào biểu thức B ta được: \(B=5^2+2.5+1=25+10+1=36\)
\(C=x^2-y^2-4x=\left(x^2-4x+4\right)-y^2-4\)
\(=\left(x-2\right)^2-y^2-4\) \(=\left(x-y-2\right)\left(x-2+y\right)-4\)
Thay x + y = 2 vào C ta được: \(C=\left(x-2-y\right)\left(2-2\right)-4=0-4=-4\)
\(D=x^2+y^2+2xy-4x-4y-3\)
\(=\left(x+y\right)^2-4\left(x+y\right)-3\) Thay x + y = 4 vào D ta được:
\(D=4^2-4.4-3=16-16-3=-3\)
Bài 3:
a) \(N=-9x^2+12x-5=-\left(9x^2-12x+4\right)-1\)
\(=-\left(3x-2\right)^2-1\)
Do \(\left(3x-2\right)^2\ge0\) nên \(-\left(3x-2\right)^2-1< 0\)
Vậy N < 0
b) ghi đề cẩn thận lại đi, mk k hiểu
a) $x^3-3x^2y+4x-12y$
$=(x^3-3x^2y)+(4x-12y)$
$=x^2(x-3y)+4(x-3y)$
$=(x-3y)(x^2+4)$
b) $4x^2-y^2+4y-4$
$=4x^2-(y^2-4y+4)$
$=(2x)^2-(y^2-2\cdot y\cdot2+2^2)$
$=(2x)^2-(y-2)^2$
$=[2x-(y-2)][2x+(y-2)]$
$=(2x-y+2)(2x+y-2)$
c) $9x^2-6x-y^2+2y$
$=(9x^2-y^2)-(6x-2y)$
$=[(3x)^2-y^2]-2(3x-y)$
$=(3x-y)(3x+y)-2(3x-y)$
$=(3x-y)(3x+y-2)$
$\text{#}Toru$
1: \(\left(x-2\right)^2-4x+8\)
\(=\left(x-2\right)\left(x-2-4\right)\)
\(=\left(x-2\right)\left(x-6\right)\)
3: \(a^3+6a^2+9a-ab^2\)
\(=a\left(a^2+6a+9-b^2\right)\)
\(=a\left(a+3-b\right)\left(a+3+b\right)\)
a: \(A=4\cdot\dfrac{5}{2}\sqrt{x}-\dfrac{8}{3}\cdot\dfrac{3}{2}\sqrt{x}-\dfrac{4}{3x}\cdot\dfrac{3x}{8}\cdot\sqrt{x}\)
\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)
\(=\dfrac{11}{2}\sqrt{x}\)
b: \(B=\dfrac{y}{2}+\dfrac{3}{4}\cdot\left|2y-1\right|-\dfrac{3}{2}\)
\(=\dfrac{y}{2}+\dfrac{3}{4}\left(1-2y\right)-\dfrac{3}{2}\)
=1/2y+3/4-3/2y-3/2
=-y-3/4
9x3 - 9x2y - 4x + 4y
= ( 9x3 - 9x2y ) - ( 4x - 4y )
= 9x2 ( x - y ) - 4 ( x - y )
= ( 9x2 - 4 ) ( x - y )
= [ ( 3x )2 - 22 ] ( x - y )
= ( 3x - 2 ) ( 3x + 2 ) ( x - y )
9x3 - 9x2y - 4x + 4y
= 9x2( x - y ) - 4( x - y )
= ( x - y )( 9x2 - 4 )
= ( x - y )( 3x - 2 )( 3x + 2 )