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a) \(=x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)
b) \(=x^2-\left(3y\right)^2=\left(x-3y\right)\left(x+3y\right)\)
c) \(=\left(2x-1\right)^2-\left(2y\right)^2=\left(2x-1-2y\right)\left(2x-1+2y\right)\)
d) \(=x^2-10xy+\left(5y\right)^2=\left(x-5y\right)^2\)
e) \(=\left(3x\right)^2-6x+1=\left(3x-1\right)^2\)
f) \(=\left(5x\right)^2+20x+4=\left(5x+2\right)^2\)
a: 2x^2y-50xy=2xy(x-25)
b: 5x^2-10x=5x(x-2)
c: 5x^3-5x=5x(x^2-1)=5x(x-1)(x+1)
d: \(x^2-xy+x=x\left(x-y+1\right)\)
e: x(x-y)-2(y-x)
=x(x-y)+2(x-y)
=(x-y)(x+2)
f: 4x^2-4xy-8y^2
=4(x^2-xy-2y^2)
=4(x^2-2xy+xy-2y^2)
=4[x(x-2y)+y(x-2y)]
=4(x-2y)(x+y)
f1: x^2ỹ-y^2+y
=(x-y)(x+y)+(x+y)
=(x+y)(x-y+1)
a)\(6x-9-x^2\)
\(=-\left(x^2+6x+9\right)\)
\(=-\left(x+3\right)^2\)
b)\(x^2+4y^2+4xy\)
\(=\left(x+2y\right)^2\)
c)\(x^2+8x+16\)
\(=\left(x+4\right)^2\)
d)\(9x^2-12xy+4y^2\)
\(=\left(3x-2y\right)^2\)
e)\(-25x^2y^2+10xy-1\)
\(=-\left(25x^2y^2-10xy+1\right)\)
\(=-\left(5xy-1\right)^2\)
f)\(4x^2-4x+1\)
\(=\left(2x-1\right)^2\)
j)\(x^2+6x+9\)
\(=\left(x+3\right)^2\)
h)\(9x^2-6x+1\)
\(=\left(3x-1\right)^2\)
#H
a, 6x - 9 - x2 = - x2 + 6x - 9 = - (x2 - 6x + 9) = - (x - 3)2
b, x2 + 4y2 + 4xy = x2 + 2. x . 2y + (2y)2 = (x + 2y)2
c, x2 + 8x + 16 = x2 + 2 . x . 4 + 42 = (x + 4)2
d, 9x2 - 12xy + 4y2 = (3x)2 - 2 . 3x . 2y + (2y)2 = (3x - 2y)2
e, - 25x2y2 + 10xy - 1 = - (25x2y2 - 10xy + 1) = - [(5xy)2 - 2 . 5xy + 1] = - (5xy - 1)2
f, 4x2 - 4x + 1 = (2x)2 - 2 . 2x + 1 = (2x - 1)2
j, x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
h, 9x2 - 6x + 1 = (3x)2 - 2 . 3x + 1 = (3x - 1)2
Bài 1:
a: \(\left(\dfrac{1}{3}x+2\right)\left(3x-6\right)\)
\(=x^2-3x+6x-12\)
\(=x^2+3x-12\)
b: \(\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)
c: \(\left(-2xy+3\right)\left(xy+1\right)\)
\(=-2x^2y^2-2xy+3xy+3\)
\(=-2x^2y^2+xy+3\)
d: \(x\left(xy-1\right)\left(xy+1\right)\)
\(=x\left(x^2y^2-1\right)\)
\(=x^3y^2-x\)
Bài 2:
a: Ta có: \(M=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(=27x^3+8\)
\(=27\cdot\dfrac{1}{27}+8=9\)
b: Ta có: \(N=\left(5x-2y\right)\left(25x^2+10xy+4y^2\right)\)
\(=125x^3-8y^3\)
\(=125\cdot\dfrac{1}{125}-8\cdot\dfrac{1}{8}\)
=0
a) 5xy ( x - y ) - 2x + 2y
= 5xy ( x - y ) - 2 ( x - y )
= ( x - y ) ( 5xy - 2 )
b) 6x-2y-x(y-3x)
= 2 ( y - 3x ) - x ( y - 3x )
= ( y - 3x ( ( 2 - x )
c) x2 + 4x - xy-4y
= x ( x + 4 ) - y ( x + 4 )
( x + 4 ) ( x - y )
d) 3xy + 2z - 6y - xz
= ( 3xy - 6y ) + ( 2z - xz )
= 3y ( x - 2 ) + z ( x - 2 )
= ( x - 2 ) ( 3y + z )
a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)
b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)
c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)
d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)
11)
a,4-9x^2=0
(2-3x)(2+3x)=0
2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3
b,x^2 +x+1/4=0
(x+1/2)^2 =0
x+1/2=0
x=-1/2
c,2x(x-3)+(x-3)=0
(x-3)(2x+1)=0
x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2
d,3x(x-4)-x+4=0
3x(x-4)-(x-4)=0
(x-4)(3x-1)=0
x-4=0=>x=4 hoặc 3x-1=0=>x=1/3
e,x^3-1/9x=0
x(x^2-1/9)=0
x(x+1/3)(x-1/3)=0
x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3
f,(3x-y)^2-(x-y)^2 =0
(3x-y-x+y)(3x-y+x-y)=0
2x(4x-2y)=0
4x(2x-y)=0
x=0hoặc 2x-y=0=>x=y/2
a) \(4x^2-12x+9\)
\(=\left(2x\right)^2-2.2x.3+3^2\)
\(=\left(2x-3\right)^2\)
b) \(4x^2+4x+1\)
\(=\left(2x\right)^2+2.2x.1+1^2\)
\(=\left(2x+1\right)^2\)
c) \(1+12x+36x^2\)
\(=1^2+2.1.6x+\left(6x\right)^2\)
\(=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2\)
\(=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2\)
\(=\left(3x-4y\right)^2\)
e) \(\frac{x^2}{4}+2xy+4y^2\)
\(=\left(\frac{x}{2}\right)^2+2.\frac{x}{2}.2y+\left(2y\right)^2\)
\(=\left(\frac{x}{2}+2y\right)^2\)
f) \(-x^2+10x-25\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2.5x+5^2\right)\)
\(=-\left(x-5\right)^2\)
g) \(-16a^4b^6-24a^5b^5-9a^6b^4\)
\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)
\(=-a^4b^4\left[\left(4b\right)^2+2.4b.3a+\left(3a\right)^2\right]\)
\(=-a^4b^4\left(4b+3a\right)^2\)
h) \(25x^2-20xy+4y^2\)
\(=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
i) \(25x^4-10x^2y+y^2\)
\(=\left(5x^2\right)^2-2.5x^2y+y^2\)
\(=\left(5x^2-y\right)^2\)
a) \(-y^2+\dfrac{1}{9}\)
= \(-\left(y^2-\dfrac{1}{9}\right)\)
= \(-\left(y-\dfrac{1}{3}\right)\left(y+\dfrac{1}{3}\right)\)
b) \(4\left(x-3\right)^2-9\left(x+1\right)^2\)
= \(\left(2x-3\right)^2-\left(3x+3\right)^2\)
= \(\left(2x-3+3x+3\right)\left(2x-3-3x-3\right)\)
= \(5x\left(-x-6\right)\)
c) \(25x^2-20xy+4y^2\)
= \(\left(5x-2y\right)^2\)
d) \(-9x^2+12xy-4y^2\)
= \(-\left(9x^2-12xy+4y^2\right)\)
= \(-\left(3x-2y\right)^2\)
e) \(25x^2-\dfrac{1}{8}x^2y^2\)
= \(\left(5x-\dfrac{\sqrt{2}}{4}xy\right)\left(5x+\dfrac{\sqrt{2}}{4}xy\right)\)
f) \(9x^2+6x+1\)
= \(\left(3x+1\right)^2\)
okey! Vì you t sẽ chăm thêm 1 lần nữa!!!^^
\(a.-y^2+\dfrac{1}{9}=\left(\dfrac{1}{3}\right)^2-y^2=\left(\dfrac{1}{3}-y\right)\left(\dfrac{1}{3}+y\right)\)
\(b,4\left(x-3\right)^2-9\left(x+1\right)^2=\left[2\left(x-3\right)\right]^2-\left[3\left(x+1\right)\right]^2=\left(2x-6\right)^2-\left(3x+3\right)^2=\left(2x-6-3x-3\right)\left(2x-6+3x+3\right)=\left(-x-9\right)\left(5x-3\right)\)\(c,25x^2-20xy+4y^2=\left(5x-2y\right)^2\)
\(d,-9x^2+12xy-4y^2=-\left(3x-2y\right)^2\)
\(e,25x^2-\dfrac{1}{8}x^2y^2=\left(5x-\dfrac{\sqrt{2}}{4}xy\right)\left(5x+\dfrac{\sqrt{2}}{4}xy\right)\)\(f,9x^2+6x+1=\left(3x+1\right)^2\)