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b. \(N=x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2012\)\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(3x^2+6xy+3y^2\right)+\left(3x+3y\right)+2012\)
\(=\left(x+y\right)^3-3\left(x^2+2xy+y^2\right)+3\left(x+y\right)+2012\)
\(=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2012\) (*)
Thay x + y =101 vào biểu thức (*) ta được:
\(N=101^3-3.101^2+3.101+2012\)
= 1002013
Câu a ko hỉu đề!
Câu b:
Ta có: N = \(x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2012\)
= \(\left(x^3+3x^2y+3xy^2+y^3\right)-3\left(x^2+2xy+y^2\right)+3\left(x+y\right)+2012\)
= \(\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2012\)
= \(\left(x+y-1\right)^3+2013\)
Thay x + y = 101 vào N ta được:
N = 1003 + 2013 = 1002013
P = 3x2 - 2x + 3y2 - 2y + 6xy +2018
P = 3(x2 + y2 + 2xy) - 2(x + y) + 2018
P = 3[(x + y)2 - 2xy + 2xy] -2.5 + 2018
P = 3[ 52 +0] - 10 + 2018
P = 3.25 + 2008
P = 75 + 2008
P = 2083
P = 3x2 - 2x + 3y2 - 2y + 6xy - 100
= (3x2 + 6xy + 3y2) - (2x + 2y) - 100
= 3(x2 + 2xy + y2) - 2(x + y) - 100
= 3(x + y)2 - 2.5 - 100
= 3. 52 -10 - 100
= 75 - 10 - 100 = -35
Q = x3 + y3 - 2x2 - 2y2 + 3xy(x + y) - 4xy + 3(x+y) +10
= x3 + y3 - 2x2 - 2y2 + 3x2y + 3xy2 - 4xy + 3.5 + 10
= (x3 + 3x2y + 3xy2 + y3) - (2x2 + 4xy + 2y2) + 15 + 10
= (x + y)3 - 2(x2 + 2xy + y2) + 25
= 53 - 2(x + y)2 +25
= 125 - 2. 52 + 25
= 125 - 50 + 25 = 100
) \(\dfrac{x^3+8y^3}{2y+x}\)
\(=\dfrac{x^3+\left(2y\right)^3}{x+2y}\)
\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)
\(=x^2+2xy+4y^2\)
b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)
\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)
\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)
\(=\dfrac{3a-1}{2\left(a-4\right)}\)
c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)
\(=\dfrac{\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2}\)
d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)
\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)
\(=x^2-10x+25+7x+14-x^2-2x\)
\(=39-5x\)
e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)
\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)
\(=\dfrac{3x+2x+1}{x-2}\)
\(=\dfrac{5x+1}{x-2}\)
h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)
\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
\(B=x^3+3x^2+3x^2y+3xy^2+y^3+3y^2+6xy+3x+3y+2019\)
\(=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2019\)
\(=\left[\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+1\right]+2018\)
\(=\left(x+y-1\right)^3+2018\)
Mà \(x+y=101\)
\(B=\left(101-1\right)^3+2018=1002018\)
Đang 3x2+3y2 sao lại ra -3(x+y)2 ?? Phải là +3(x2+y2) chứ :v Không nhớ hằng đẳng thức 1 và 3 à :v với cả 6xy đâu?
a, Ta có : \(\frac{3y}{4}=\frac{3y}{4}.1=\frac{3y}{4}.\frac{2x}{2x}=\frac{6xy}{8x}\) ( đpcm )
b, Ta có : \(6x^2y=6x^2y\)
=> \(3x^2.2y=\left(-3x^2\right).\left(-2y\right)\)
=> \(\frac{-3x^2}{2y}=\frac{3x^2}{-2y}\) ( đpcm )
c, Ta có : \(6x-6y=6x-6y\)
=> \(6x-6y=-6y+6x\)
=> \(6\left(x-y\right)=-6\left(y-x\right)\)
=> \(2\left(x-y\right).3=-2\left(y-x\right).3\)
=> \(\frac{2\left(x-y\right)}{3\left(y-x\right)}=\frac{-2}{3}\) ( đpcm )
b: \(=3\left[\left(x+y\right)^2-2xy\right]-2\left[\left(x-y\right)^3+3xy\left(x-y\right)\right]\)
\(=3\left(1-2xy\right)-2\left(1+3xy\right)\)
\(=3-6xy-2-6xy=-12xy+1\)
c: \(=\left(x+y\right)^3-3\left(x^2+y^2+2xy\right)+3\left(x+y\right)+2012\)
\(=101^2-3\cdot101^2+3\cdot101+2012\)
=1002013