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1 Điền vào chỗ chấm để có hằng đẳng thức thích hợp :
a) x2 - 4y2 = ........(x-2y)(x+2y)......................
b) 1/42 + 2xy + 4y2 =.............(1/4+2y)2..........................
c) 64x3 - 1 =..................(4x-1)(16x2+4x+1)................
d) 25 + 10y + y2 =........(5+y)2....................
a) \(2x^2y^2-\frac{4}{3}x^2y+2xy\)
\(=xy\left(2xy-\frac{4}{3}x+2\right)\)
b) 2xy2.(x + 5y) - 4xy(5y + x)
= (5y + x)(2xy2 - 4xy)
= 2xy(5y + x)(y - 2)
c) 25 - 4x2 - y2 + 4xy
= 25 - (4x2 - 4xy + y2)
= 52 - (2x + y)2
= (5 - 2x - y)(5 + 2x + y)
d) x2 + 4x - 2xy - 4y +y2
= (x2 - 2xy + y2) + (4x - 4y)
= (x - y)2 + 4(x - y)
= (x - y)(x - y + 4)
e) 12y3 - 3x2y + 12xy - 12y
= 3y(4y2 - x2 + 4x - 4)
= 3y[4y2 - (x - 2)2]
= 3y(2y - x + 2)(2y + x - 2)
f) 64x4 + y4
= (8x2)2 + 16x2y2 + y4 - 16x2y2
= (8x2 + y2)2 - (4xy)2
= (8x2 + y2 - 4xy)(8x2 + y2 + 4xy)
a) \(2x^2y^2-\frac{4}{3}x^2y+2xy\)
b) \(2xy^2\left(x+5y\right)-4xy\left(5y+x\right)\)
\(=\left(x+5y\right)\left(2xy^2-4xy\right)\)
\(=2\left(x+5y\right)\left(xy^2-2xy\right)\)
c) \(25-4x^2-y^2+4xy\)
\(=25-\left(4x^2+y^2-4xy\right)\)
\(=5^2-\left[\left(2x\right)^2-2.2x.y+y^2\right]\)
\(=5^2-\left(2x-y\right)^2\)
\(=\left(5-2x+y\right)\left(5+2x-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y\right)+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(12y^3-3x^2y+12xy-12y\)
f) \(64x^4+y^4\)
\(=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2+4xy\right)\left(8x^2+y^2-4xy\right)\)
c)\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)
d)\(\dfrac{a^2}{4}-2a+4=\left(\dfrac{a}{2}-2\right)^2\)
e) \(4y^2-9x^2=\left(2y-3x\right)\left(2y+3x\right)\)
f)\(9y^2-\dfrac{1}{4}=\left(3y-\dfrac{1}{2}\right)\left(3y+\dfrac{1}{2}\right)\)
g)\(8x^3+8a^3=\left(2x+2a\right)\left(4x^2-4xa+4a^2\right)\)
B1:
\(=x^2+2x-5x-10+3\left(x^2-2^2\right)-\left(9x^2-2.3x.\frac{1}{2}+\frac{1}{4}\right)+5x^2\)
\(=-10-12-\frac{1}{4}=-22\frac{1}{4}\)
Bài 1.
( x - 5 )( x + 2 ) + 3( x - 2 )( x + 2 ) - ( 3x - 1/2 )2 + 5x2
= x2 - 3x - 10 + 3( x2 - 4 ) - ( 9x2 - 3x + 1/4 ) + 5x2
= 6x2 -- 3x - 10 + 3x2 - 12 - 9x2 + 3x - 1/4
= -89/4 không phụ thuộc vào biến
=> đpcm
Bài 2 < mình viết luôn nhé >
a) ( x + 2y2 )2 = x2 + 4xy2 + 4y4
b) ( a - 5/2b )2 = a2 - 5ab + 25/4b2
c) ( m + 1/2 )2 = m2 + m + 1/4
d) x2 - 16y4 = ( x + 4y2 )( x - 4y2 )
e) 25a2 - 1/4b2 = ( 5a + 1/2b )( 5a - 1/2b )
a, bằng cách tìm nhân tử chung
1,\(x^2-3x\)
=x.(\(\left(x-3\right)\)
2,\(15x^2-6x\)
=3x.(5x-2)
3,\(4x\left(x-y\right)\)\(+2y\left(x-y\right)\)
=(x-y).(4x+2y)
=2(x-y).(x+y)
=2(\(x^2-y^2\left(\right)\)
b, dùng hằng đẳng thức
1,\(64x^2-25y^2\)
=\(\left(8x\right)^2-\left(5y\right)^2\)
=(8x-5y)(8x+5y)
2,\(9x^2-30x-25\)
=\(\left(3x-5\right)^2\)
3,
\(\dfrac{1}{4}x^2+2x+4\)
=\(\left(\dfrac{1}{2}x+2\right)^2\)
4,\(25a^2-2a+\dfrac{1}{25}\)
=(\(\left(5a-\dfrac{1}{5}\right)^2\)
a) \(M=10x^2+6y+4y^2+4xy+2\)
\(=\left(10x^2+4xy+\dfrac{2}{5}y^2\right)+\left(\dfrac{18}{5}y^2+6y+\dfrac{5}{2}\right)-\dfrac{1}{2}\)
\(=10\left(x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)+\dfrac{18}{5}\left(y^2+\dfrac{5}{3}y+\dfrac{25}{36}\right)-\dfrac{1}{2}\)
\(=10\left(x+\dfrac{1}{5}y\right)^2+\dfrac{18}{5}\left(y+\dfrac{5}{6}\right)^2-\dfrac{1}{2}\ge-\dfrac{1}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{5}y=0\\y+\dfrac{5}{6}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{5}{6}\end{matrix}\right.\)
b) \(H=-x^2+2xy-4y^2+2x+10y-8\)
\(=-x^2+2x\left(y+1\right)-\left(y^2+2y+1\right)-\left(3y^2-12y+7\right)\)
\(=-x^2+2x\left(y+1\right)-\left(y+1\right)^2-3\left(y^2-4y+4\right)+5\)
\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+5\le5\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
c) \(K=2x^2+2xy-2x+2xy+y^2\)
bn xem lại cái đề nhé, sao lại có 2 lần 2xy
b)(y-2)^3=y^3-8+12y-6y^2
c)8x^3+y^3=(2x+y)(4x^2+y^2-4xy)
2)
=(xy+2/3)^2
1 ) ...
2 ) \(\dfrac{x^2}{4}+2xy+4y^2=\left(\dfrac{x}{2}\right)^2+2.\dfrac{x}{2}.2y+\left(2y\right)^2=\left(\dfrac{x}{2}+2y\right)^2\)
3 ) \(64x^3-1=\left(4x\right)^3-1=\left(4x-1\right)\left[\left(4x\right)^2+4x+1\right]=\left(4x-1\right)\left(16x^2+4x+1\right)\)
4 ) \(25-10y+y^2=5^2-2.5y+y^2=\left(5-y\right)^2\)
1 ) \(x^2-4xy^2\)
\(=x^2-2x.2y^2+\left(2y^2\right)^2-\left(2y^2\right)^2\)
\(=\left(x-2y^2\right)^2-\left(2y^2\right)^2\)
\(=\left(x-2y^2-2y^2\right)\left(x-2y^2+2y^2\right)\)
\(=\left(x-4y^2\right)x\)