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1) Ta có: \(x^2-4xy+4y^2\)
\(=x^2-2.x.2y+\left(2y\right)^2\)
\(=\left(x-2y\right)^2\)
Phép tính trở thành: \(\left(x-2y\right)^2:\left(x-2y\right)=x-2y\)
2) Ta có: \(25x^2+2xy+\dfrac{1}{25}y^2\)
\(=\left(5x\right)^2+2.5x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\)
\(=\left(5x+\dfrac{1}{5}y\right)^2\)
Phép tính trở thành: \(\left(5x+\dfrac{1}{5}y\right)^2:\left(5x+\dfrac{1}{5}y\right)=5x+\dfrac{1}{5}y\)
1) (x² - 4xy + 4y²) : (x - 2y)
= (x - 2y)² : (x - 2y)
= x - 2y
2) (25x² + 2xy + 1/25 y²) : (5x + 1/5 y)
= 5x + 1/5 y)² : (5x + 1/5 y)
= 5x + 1/5 y
a: \(\dfrac{1}{2}x^2\cdot2x^3-4x^2+3=x^5-4x^2+3\)
b: \(2y\left(xy-1\right)\left(xy+1\right)=2y\left(x^2y^2-1\right)=2x^2y^3-2y\)
h: \(=\left(x+3\right)\cdot\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
\(\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\)
\(\Leftrightarrow x^3+8y^3=0\)
\(\Leftrightarrow x^3=-8y^3\)
\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=16\)
\(\Leftrightarrow x^3-8y^3=16\)
\(\Leftrightarrow-8y^3-8y^3=16\)
\(\Leftrightarrow y^3=-1\Rightarrow y=-1\Rightarrow x=2\)
`-1/3x^5y^2:(-2xy)-(x^2+2x+1):(x+1)`
`=-1/3:(-2).(x^5:x).(y^2:y)-(x+1)^2:(x+1)`
`=-1/6x^4y-(x+1)`
`=-1/6x^4y-x-1`
\(\dfrac{-1}{3}x^5y^2:\left(-2xy\right)-\left(x^2+2x+1\right):\left(x+1\right)\)
\(=\dfrac{1}{6}x^4y-x-1\)
a: \(=\left(3-x\right)\left(x+1\right)\)
b: \(=3x\left(x-y\right)-5\left(x-y\right)\)
=(x-y)(3x-5)
c: \(=x\left(x-y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(x-10\right)\)
a) \(=x\left(3-x\right)+\left(3-x\right)=\left(3-x\right)\left(x+3\right)\)
b) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(=x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(x-10\right)\)
d) \(=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)
e) \(=\left(x-y\right)\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(x-y-4\right)\)
f) \(=9-\left(4x^2-4xy+y^2\right)=9-\left(2x-y\right)^2=\left(3-2x+y\right)\left(3+2x-y\right)\)
g) \(=y\left(y^2-2xy+x^2-y\right)\)
h) \(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) \(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(2x+y\right)\)
trở thành một hằng đảng thức. Giá trị trong ô vuông là:
[(x2-2xy+2xy2).(x+2y)-(x2+4y2).(x-y)]2xy
=( x3 + 2x2y-2x2y-4xy2+2x2y2+4xy3-x3+x2y-4xy2+4y3 )2xy
=2xy(2x2y2-8xy2+4xy3+x2y+4y3)
= 4x3y3-16x2y3+8x2y4+2x3y2+8xy4
Trả lời:
[ ( x2 - 2xy + 2xy2 ) ( x + 2y ) - ( x2 + 4y2 ) ( x - y ) ] 2xy
= [ ( x3 + 2x2y - 2x2y - 4xy2 + 2x2y2 + 4xy3 ) - ( x3 - x2y + 4xy2 - 4y3 ) ] 2xy
= ( x3 + 2x2y - 2x2y - 4xy2 + 2x2y2 + 4xy3 - x3 + x2y - 4xy2 + 4y3 ) 2xy
= ( x2y - 8xy2 + 2x2y2 + 4xy3 + 4y3 ) 2xy
= 2x3y2 - 16x2y3 + 4x3y3 + 8x2y4 + 8xy4