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\(Q=3x^2\left(x+y+2\right)-y^2\left(x+y+2\right)+2\left(x+y+2\right)+2016\)
\(=2016\)
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x+y=-2
Q=-6x2+2y2+6x2-2y2+2x+2y+2020
Q=2(x+y)+2020 = 2.(-2)+2020=2020-4=2016
Tick mk nhé
\(B=4x^2+6x^2y^2+2y^4+20y^2\)
\(=4x^2+4x^2y^2+2x^2y^2+2y^4+20y^2\)
\(=4x^2.\left(x^2+y^2\right)+2y^2.\left(x^2+y^2\right)+20y^2\)
\(=\left(x^2+y^2\right).\left(4x^2+2y^2\right)+20y^2\)
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\(x^4+4x^3y+6x^2y^2+4xy^3+y^4-x-y-10\)
\(=\left(x^4+2x^3y+x^2y^2\right)+\left(2x^3y+4x^2y^2+2xy^3\right)+\left(x^2y^2+2xy^3+y^4\right)-\left(x+y\right)-10\)
\(=x^2\left(x^2+2xy+y^2\right)+2xy\left(x^2+2xy+y^2\right)+y^2\left(x^2+2xy+y^2\right)-\left(x+y\right)-10\)
\(=\left(x^2+2xy+y^2\right)\left(x^2+2xy+y^2\right)-\left(x+y\right)-10\)
\(=\left(x+y\right)^2\left(x+y\right)^2-\left(x+y\right)-10\)
\(=\left(x+y\right)^4-\left(x+y\right)-10\)
\(=2^4-2-10\) \(=4\)
a) M + (5x2 - 2xy) = 6x2 + 9xy - y2
=> M = (6x2 + 9xy - y2) - (5x2 - 2xy)
=> M = 6x2 + 9xy - y2 - 5x2 + 2xy = (6x2 - 5x2) + (9xy + 2xy) - y2 = x2 + 11xy - y2
b) Sửa đề lại đi nhé
c) (25x2y - 13x2y + y3) - M = 11x2y - 2y2
=> M = (25x2y - 13x2y + y3) - (11x2y - 2y2)
=> M = 25x2y - 13x2y + y3 - 11x2y + 2y2
=> M = x2y + y3 + 2y2
d) M = 0 - (12x4 - 15x2y + 2xy2 + 7) = -12x4 + 15x2y - 2xy2 - 7
a) Ta có : M = 6x2 + 9xy - y2 - (5x2 - 2xy)
= 6x2 + 9xy - y2 - 5x2 + 2xy
= x2 + 11xy - y2
b) Ta có M = x2 - 7xy + 8y2 - (3xy - 24y2)
= x2 - 7xy + 8y2 - 3xy + 24y2
= x2 - 10xy + 32y2
c) Ta có M = 25x2.y- 13x2y + y3 - (11x2y - 2y2)
= 25x2.y- 13x2y + y3 - 11x2y + 2y2
= x2y + y3 + 2y2
d) Ta có M = -(12x4 - 15x2y + 2xy2 + 7)
= -12x4 + 15x2y - 2xy2 - 7
Lời giải:
Ta có:
\(x^4+4x^3y+6x^2y^2+4xy^3+y^4-x-y-10\)
\(=(x^4+x^3y)+3(x^3y+2x^2y^2+xy^3)+(xy^3+y^4)-(x+y)-10\)
\(=x^3(x+y)+3xy(x^2+2xy+y^2)+y^3(x+y)-(x+y)-10\)
\(=x^3(x+y)+3xy[x(x+y)+y(x+y)]+y^3(x+y)-(x+y)-10\)
\(=x^3(x+y)+3xy(x+y)^2+y^3(x+y)-(x+y)-10\)
\(=2x^3+6xy(x+y)+2y^3-2-10\)
\(=2[x^3+3xy(x+y)+y^3]-12\)
\(=2[x^2(x+y)+y^2(x+y)+2xy(x+y)]-12\)
\(=2(x+y)(x^2+y^2+2xy)-12=2(x+y)(x+y)^2-12\)
\(=2(x+y)^3-12=2.2^3-12=4\)
Nếu bạn đã biết hằng đẳng thức thì:
\(x^4+4x^3y+6x^2y^2+4xy^3+y^4-x-y-10\)
\(=(x+y)^4-(x+y)-10=2^4-2-10=4\)
Lời giải:
Ta có:
x4+4x3y+6x2y2+4xy3+y4−x−y−10
=(x4+x3y)+3(x3y+2x2y2+xy3)+(xy3+y4)−(x+y)−10
=x3(x+y)+3xy(x2+2xy+y2)+y3(x+y)−(x+y)−10
=x3(x+y)+3xy[x(x+y)+y(x+y)]+y3(x+y)−(x+y)−10
=x3(x+y)+3xy(x+y)2+y3(x+y)−(x+y)−10
=2x3+6xy(x+y)+2y3−2−10
=2[x3+3xy(x+y)+y3]−12
=2[x2(x+y)+y2(x+y)+2xy(x+y)]−12
=2(x+y)(x2+y2+2xy)−12=2(x+y)(x+y)2−12
=2(x+y)3−12=2.23−12=4
Nếu bạn đã biết hằng đẳng thức thì:
x4+4x3y+6x2y2+4xy3+y4−x−y−10
=(x+y)4−(x+y)−10=24−2−10=4
Câu 1 :
\(3\left(x-3\right)\left(x+7\right)+\left(1-4\right)\left(x+4\right)+18\)
\(=3\left(x^2+4x-21\right)-3\left(x+4\right)\)
\(=3x^2+12x-63-3x-12=3x^2+9x-75\)
Thay x = 1/2 vào ta được
\(\dfrac{3.1}{4}+\dfrac{9}{2}-75=-\dfrac{279}{4}\)
Câu 2 :
\(5x^2+5xy+5x=5x\left(x+y+1\right)\)
Thay x = 60 ; y = 50 ta được
\(300\left(60+50+1\right)=33300\)
Câu 3 :
\(4x^2y^2+2xy^2+6x^2y=2xy\left(2xy+y+3x\right)\)
Thay x = 10 ; y = 1/2 ta được
\(\dfrac{2.10.1}{2}\left(\dfrac{2.10.1}{2}+\dfrac{1}{2}+30\right)=405\)
1: \(=3\left(x^2+4x-21\right)+x^2-16+18\)
\(=3x^2+12x-63+x^2+2\)
\(=4x^2+12x-61\)
\(=4\cdot\dfrac{1}{4}+12\cdot\dfrac{1}{2}-61=1-61+6=-54\)
2: \(=5\cdot60^2+5\cdot60\cdot50+5\cdot60=33300\)
3: \(=4\cdot10^2\cdot\dfrac{1}{4}+2\cdot10\cdot\dfrac{1}{4}+6\cdot100\cdot\dfrac{1}{2}=405\)
= (x -3)2 +( y+1)2 =0
x = 3
y = -1