Viết dưới dạng tích:
a) x2y2 - a4b6
b) 4x2y4 - ( 3xy2 - 1 )2
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a) \(8y^3+1\)
\(=\left(2y\right)^3+1^3\)
\(=\left(2y+1\right)\left(4y^2-2y+1\right)\)
b) \(y^3-8\)
\(=y^3-2^3\)
\(=\left(y-2\right)\left(y^2+2y+4\right)\)
`8y^3 + 1 = (2y+1)(4y^2 - 2y + 1)`
`y^3 -8 =(y-2)(y^2+2y+4)`
a: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
b: \(x^3-\dfrac{1}{8}=\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
c: \(8x^3+y^3=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
a) \(\left(x+3\right)\cdot\left(x^2-3x+9\right)\)
b) \(\left(x-\dfrac{1}{2}\right)\cdot\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
c) \(\left(2x+y\right)\cdot\left(4x^2-2xy+y^2\right)\)
a) \(=\left(x-2\right)^2\)
b) \(=\left(3x-2\right)^2\)
c) \(=\left(x-3y\right)^2\)
d) \(=\left(\dfrac{x}{2}+1\right)^2\)
e) \(=\left(x-4\right)^2\)
f) \(=\left(\dfrac{1}{2}xy^2+1\right)^2\)
g) \(=\left(x-1\right)\left(x+1\right)\)
h) \(=\left(5x-4\right)\left(5x+4\right)\)
Ta có P + Q=x2 y + xy2 - 5x2 y2 + x3 + 3xy2 - x2 y + x2 y2
= -4x2 y2 + x3 + 4xy2
Chọn B
a) 5.(-2).(-1)2 + 2.(-2).(-1) – 3.(-2).(-1)2
= 5.(-2).1 + 4 – 3.(-2).1
= -10 + 4 + 6
= 0
b) x2y2 + x4y4 + x6y6 tại x = 1 và y = -1
= 12(-1)2 + 14(-1)4 + 16(-1)6
= 1.1 + 1.1 + 1.1
= 1+1+1
= 3
a. (x + y)2 = x2 + 2xy + y2
b. (x - 2y)2 = x2 - 4xy - 4x2
c. (xy2 + 1)(xy2 - 1) = x2y4 - 1
d. (x + y)2(x - y)2 = (x2 + 2xy + y2)(x2 - 2xy + y2) = x4 - (2xy + y2)2 = x4 - (4x2y2 + y4) = x4 - 4x2y2 - y4
Chucs hocj toots
Câu 2:
a: \(x^2-4x+4=\left(x-2\right)^2\)
b: \(x^2+10x+25=\left(x+5\right)^2\)
d: \(9\left(x+1\right)^2-6\left(x+1\right)+1=\left(3x+2\right)^2\)
e: \(\left(x-2y\right)^2-8\left(x-2xy\right)+16x^2=\left(x-2y+4x\right)^2=\left(5x-2y\right)^2\)
a: \(\left(x+y+z\right)^2-\left(y+z\right)^2\)
\(=\left(x+y+z-y-z\right)\left(x+y+z+y+z\right)\)
\(=x\left(x+2y+3z\right)\)
b: \(\left(x+3\right)^2+4\left(x+3\right)+4\)
\(=\left(x+3+2\right)^2\)
\(=\left(x+5\right)\left(x+5\right)\)
c: \(25+10\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(x+1+5\right)^2\)
\(=\left(x+6\right)\left(x+6\right)\)
Ta có: P = x2y + xy2 – 5x2y2 + x3 và Q = 3xy2 – x2y + x2y2
⇒ P + Q = (x2y + xy2 – 5x2y2 + x3) + (3xy2 – x2y + x2y2)
= x2y + xy2 – 5x2y2 + x3 + 3xy2 – x2y + x2y2
= x3 +(– 5x2y2 + x2y2)+ (x2y – x2y) + (xy2+ 3xy2)
= x3 – 4x2y2 + 0 + 4xy2
= x3 – 4x2y2 + 4xy2
a) \(x^2y^2-a^4b^6=\left(xy\right)^2-\left(a^2b^3\right)^2=\left(xy-a^2b^3\right).\left(xy+a^2b^3\right)\)
b) \(4x^2y^4-\left(3xy^2-1\right)^2=\left(2xy^2\right)^2-\left(3xy^2-1\right)^2\)= \(\left(2xy^2-3xy^2+1\right).\left(2xy^2+3xy^2-1\right)\)
= \(\left(-xy^2+1\right).\left(5xy^2-1\right)\)