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a, \(25x^2+5xy+\frac{1}{4}y^2=\left(5x\right)^2+2.5x.\frac{1}{2}y+\left(\frac{1}{2}y\right)^2\)
\(=\left(5x+\frac{1}{2}y\right)^2\)
b, \(9x^2+12x+4=\left(3x\right)^2+2.3x.2+2^2=\left(3x+2\right)^2\)
c, \(x^2-6x+5-y^2-4y=\left(x^2-6x+9\right)-\left(y^2+4y+4\right)\)
\(=\left(x-3\right)^2-\left(y+2\right)^2=\left(x-y-5\right)\left(x+y-1\right)\)
d, \(\left(2x-y\right)^2+4\left(x+y\right)^2-4\left(2x-y\right)\left(x+y\right)\)
\(=\left(2x-y\right)^2-2\left(2x-y\right)\left(2x+2y\right)+\left(2x+2y\right)^2\)
\(=\left(2x-y+2x+2y\right)^2=\left(4x+y\right)^2\)
\(a.=\left(2x\right)^2-2.2x.2y+\left(2y\right)^2=\left(2x-2y\right)^2\)
\(b.=\left(3x\right)^2-2.3x.2+2^2=\left(3x-2\right)^2\)
a. 4x2+4y2-8xy=(2x)2+(2y)2-8xy
=(2x-2y)2
b.9x2-12x+4=(3x)2-12x+22
=(3x-2)2
c.xy2+1/4x2y4+1=xy2+(1/2xy2)2+1
=(1/2xy2+2)2
Bài 1:
a) \(x^2-10x+26+y^2+2y\)
\(=x^2-2.x.5+25+y^2+2y+1\)
\(=\left(x-5\right)^2+\left(y+1\right)^2\)
b) Sửa đề \(z^2-6z+5-t^2-4t\)
\(=z^2-2.z.3+9-4-t^2-4t\)
\(=\left(z-3\right)^2-\left(t^2+4t+4\right)\)
\(=\left(z-3\right)^2-\left(t+2\right)^2\)
c) \(\left(x+y-4\right)\left(x+y+4\right)\)
\(=\left(x+y\right)^2-4^2\)
d) \(a^2-b^2+c^2-2ac-d^2+2bd\)
\(=\left(a^2-2ac+c^2\right)-\left(b^2-2bd+d^2\right)\)
\(=\left(a-c\right)^2-\left(b-d\right)^2\)
e) \(\left(a-b-c\right)\left(a+b-c\right)\)
\(=\left(a-c-b\right)\left(a-c+b\right)\)
\(=\left(a-c\right)^2-b^2\)
f) \(4a^2+2b^2-4ab-2b+1\)
\(=\left(2a\right)^2-2.2a.b+b^2+b^2-2b+1\)
\(=\left(2a-b\right)^2+\left(b-1\right)^2\)
Bài 2:
a) Sửa đề \(4x^2-4xy+y^2\)
\(=\left(2x\right)^2-2.2x.y+y^2\)
\(=\left(2x-y\right)^2\)
b) \(y^2-6y+9\)
\(=y^2-2.y.3+3^2\)
\(=\left(y-3\right)^2\)
c) \(a^2+a+\dfrac{1}{4}\)
\(=a^2+2a.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\)
\(=\left(a+\dfrac{1}{2}\right)^2\)
d) \(a^2-12a+36\)
\(=a^2-2.a.6+6^2\)
\(=\left(a-6\right)^2\)
i) \(x^2-xy+\dfrac{1}{4}y^2\)
\(=x^2-2.x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\)
\(=\left(x-\dfrac{1}{2}y\right)^2\)
e) \(9x^2-24x+16\)
\(=\left(3x\right)^2-2.3x.4+4^2\)
\(=\left(3x-4\right)^2\)
f) \(x^2-3x+\dfrac{9}{4}\)
\(=x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\)
\(=\left(x-\dfrac{3}{2}\right)^2\)
g) \(1-2xy^2+x^2y^4\)
\(=1-2xy^2+\left(xy^2\right)^2\)
\(=\left(1-xy^2\right)^2\)
h) \(\left(2a-b\right)^2+2\left(2a-b\right)+1\)
\(=\left(2a-b+1\right)^2\)
Bài 3:
a) \(A=\dfrac{1}{4}x^2-xy+y^2\)
\(A=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.y+y^2\)
\(A=\left(\dfrac{1}{2}x-y\right)^2\)
Thay x = 2012 và y = 1004 vào A ta được
\(A=\left(\dfrac{1}{2}.2012-1004\right)^2\)
\(A=\left(1006-1004\right)^2\)
\(A=2^2=4\)
b) \(B=9x^2-3xy+\dfrac{1}{4}y^2\)
\(B=\left(3x\right)^2-2.3x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\)
\(B=\left(3x-\dfrac{1}{2}y\right)^2\)
Thay x = 231 và y = 1384 vào B ta được
\(B=\left(3.231-\dfrac{1}{2}.1384\right)^2\)
\(B=\left(693-692\right)^2\)
\(B=1^2=1\)
a, Đề sai bạn ơi phải là cộng 16 chứ không phải cộng 4
b,B= (x-2y+1)^2
a/ 9x2-12xy+4y2 = (3x - 2y)2
b/ 25x2-10x+1 = (5x - 1)2
c/ 9x2-12x+4 = (3x - 2)2
d/ 4x2+20x+25 = (2x + 5)2
e/ x4-4x2+4 = (x2 - 2)2
a)x^2.16-4xy+4y^2
<=>16.x^2-2x2y+(2y)^2
<=>16(x-2y)^2
b)x^5-x^4+x^3-x^2
<=>(x^5-x^4)+(x^3-x^2)
<=>x^4(x-1)+x^2(x-1)
<=>(x-1)(x^4+x^2)
c)x^5+x^3-x^2-1
<=>(x^5+x^3)-(x^2+1)
<=>x^3(x^2+1)-(x^2+1)
<=>(x^2+1)(x^3-1)
d)x^4-3x^3-x+3
<=>(x^4-3x^3)-(x-3)
<=>x^3(x-3)-(x_3)
<=>(x-3)(x^3-1)
\(a,x^2.16-4xy+4y^2\)
\(=16.x^2-4xy+4y^2\)
\(=16.\left[x^2-4xy+\left(2y\right)^2\right]\)
\(=16.\left(x-2y\right)^2\)
\(b,x^5-x^4+x^3-x^2\)
\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^4+x^2\right)\)
\(=x^2\left(x-1\right)\left(x^2+1\right)\)
\(c,x^5+x^3-x^2-1\)
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^3-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
\(d,x^4-3x^3-x+3\)
\(=x^3\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3-1\right)\)
\(=\left(x-3\right)\left(x-1\right)\left(x^2+x+1\right)\)
a) ( 2x + 1 )2 + 10( 2x + 1 ) + 25
= ( 2x + 1 )2 + 2.( 2x + 1 ).5 + 52
= [ ( 2x + 1 ) + 5 ]2
= ( 2x + 1 + 5 )2
= ( 2x + 6 )2
b) x2 + 2x( y - 2 ) + y2 - 4y + 4
= x2 + 2x( y - 2 ) + ( y2 - 4y + 4 )
= x2 + 2x( y - 2 ) + ( y - 2 )2
= [ x + ( y - 2 ) ]2
= ( x + y - 2 )2
c) x2 + 12x + 40 + y2 + 4y
= ( x2 + 12x + 36 ) + ( y2 + 4y + 4 )
= ( x + 6 )2 + ( y + 2 )2 ( cấy ni không viết được ;-; )
d) x2 - 8x - 20 - y2 - 12y
= ( x2 - 8x + 16 ) - ( y2 + 12y + 36 )
= ( x - 4 )2 - ( y + 6 )2
= [ ( x - 4 ) - ( y + 6 ) ][ ( x - 4 ) + ( y + 6 ) ]
= ( x - 4 - y - 6 )( x - 4 + y + 6 )
= ( x - y - 10 )( x + y + 2 )
e) x2 + y2 + 4x + 4y + 2( x + 2 )( y + 2 ) + 8
= ( x2 + 4x + 4 ) + 2( x + 2 )( y + 2 ) + ( y2 + 4y + 4 )
= ( x + 2 )2 + 2( x + 2 )( y + 2 ) + ( y + 2 )2
= [ ( x + 2 ) + ( y + 2 ) ]2
= ( x + 2 + y + 2 )2
= ( x + y + 4 )2
a, \(y^2+4y+4=y^2+2.y.2+2^2=\left(y+2\right)^2\)
\(x^2+8x+16=x^2+2.x.4+4^2=\left(x+4\right)^2\)
\(4x^2+4xy^2+y^4=\left(2x\right)^2+2.2x.y^2+\left(y^2\right)^2=\left(2x+y^2\right)^2\)
b, \(1024-64b^2+b^4=32^2-2.32.b^2+\left(b^2\right)^2=\left(32-b^2\right)^2\)
\(9x^2+24x+16=\left(3x\right)^2+2.3x.4+4^2=\left(3x+4\right)^2\)
\(3660,25x^2-121xy^2+y^4=\left(60,5x\right)^2-2.60,5x.y^2+\left(y^2\right)^2=\left(60,5x-y^2\right)^2\)