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a)x3 + 3x2 + 3x
=x3 + 3x2 + 3x+1-1
=(x+1)3-1.Với x=99
=>A=(99+1)3-1=1003-1
=1 000 000 -1 = 999 999
\(P=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2017\)
\(=\left(x+y-1\right)^3+2018\)
\(=100^3+2018\)
\(P=x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2015\)
\(\Leftrightarrow P=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(3x^2+6xy+3y^2\right)+\left(3x+3y\right)+2015\)
\(\Leftrightarrow P=\left(x+y\right)^3-3\left(x^2+2xy+y^2\right)+3\left(x+y\right)+2015\)
\(\Leftrightarrow P=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2015\)
\(\Leftrightarrow P=101^3-3.101^2+3.101+2015\)
\(P=x^3-3x^2+3x^2y+3xy^2+y^3-3y^2-6xy+3x+3y+2015\)
\(\Leftrightarrow P=x^3+3x^2y+3xy^2+y^3-3x^2-6xy-3y^2+3x+3y+2015\)
\(\Leftrightarrow P=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(3x^2+6xy+3y^2\right)+\left(3x+3y\right)+2015\)
\(\Leftrightarrow P=\left(x^3+3x^2y+3xy^2+y^3\right)-3\left(x^2+2xy+y^2\right)+3\left(x+y\right)+2015\)
\(\Leftrightarrow P=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2015\)
\(\Leftrightarrow P=101^3-3.101^2+3.101+2015\)
\(\Leftrightarrow P=1030301-30603+303+2015\)
\(\Leftrightarrow P=999698+303+2015\)
\(\Leftrightarrow P=1000001+2015\)
\(\Leftrightarrow P=1002016\)
a: \(A=x^3+3x^2+3x+1-1\)
\(=\left(x+1\right)^3-1\)
\(=100^3-1=999999\)
b: \(B=3\left[\left(x+y\right)^2-2xy\right]-2\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]\)
\(=3\left(1-2xy\right)-2\left(1-3xy\right)\)
\(=3-6xy-2+6xy=1\)
c: \(C=\left(x^3+3x^2y+3xy^2+y^3\right)-3\left(x^2+2xy+y^2\right)+3\left(x+y\right)+2017\)
\(=101^3-3\cdot101^2+3\cdot101+2017\)
\(=101^3-3\cdot101^2+3\cdot101-1+2018\)
\(=100^3+2018=1002018\)
\(B=x^3-3x^2+3xy^2+3x^2y+y^3-3y^2-6xy+3x+3y+2012\\ =\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2012\\ =\left[\left(x+y\right)^3-3\left(x+y\right)^3+3\left(x+y\right)-1\right]+2013\\ =\left(x+y-1\right)^3+2013\)thay x+y=101 vào ta có
\(B=\left(101-1\right)^3+2013=1002013\)
A=3(x2+2xy+y2)-2(x+y)-100=3(x+y)2-2.5-100=3.52-110=-35
B=x3+3x2y+3xy2+y3-2(x2+2xy+y2)+3(x+y)+10=(x+y)3-2(x+y)2+3.5+10=53-2.52+25=100
trả lời:
A=3(x2+2xy+y2)-2(x+y)-100
=3(x+y)2-2.5-100
=3.52-110
=-35
B=x3+3x2y+3xy2+y3-2(x2+2xy+y2)+3(x+y)+10
=(x+y)3-2(x+y)2+3.5+10
=53-2.52+25
=100
học tốt
P = 3x2 - 2x + 3y2 - 2y + 6xy - 100
= 3( x2 + 2xy + y2 ) - 2( x + y ) - 100
= 3( x + y )2 - 2( x + y ) - 100
Với x + y = 5
=> P = 3.52 - 2.5 - 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( x + y ) + 10
= ( x3 + 3x2y + 3xy2 + y3 ) - ( 2x2 + 4xy + 2y2 ) + 3( x + y )
= ( x + y )3 - 2( x2 + 2xy + y2 ) + 3( x + y ) + 10
= ( x + y )3 - 2( x + y )2 + 3( x + y ) + 10
Với x + y = 5
=> Q = 53 - 2.52 + 3.5 + 10 = 100
a. \(P=3x^2-2x+3y^2-2y+6xy-100\)
\(\Leftrightarrow P=\left(3x^2+6xy+3y^2\right)-\left(2x+2y\right)-100\)
\(\Leftrightarrow P=3\left(x+y\right)^2-2\left(x+y\right)-100\)
\(\Leftrightarrow P=3.5^2-2.5-100\)
\(\Leftrightarrow P=-35\)
b. \(Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(2x^2+4xy+2y^2\right)+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=5^3-2.5^2+3.5+10\)
\(\Leftrightarrow Q=100\)
Ta có: \(3x^2-4xy+y^2=3x-3y\)
\(\Leftrightarrow2x^2-2xy+\left(x^2-2xy+y^2\right)=3\left(x-y\right)\)
\(\Leftrightarrow2x\left(x-y\right)+\left(x-y\right)^2-3\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(2x+x-y-3\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(3x-y-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=y\\3x-y=3\end{cases}}\)
Vì x và y là 2 số thực phân biệt nên TH x=y không xảy ra\(\Rightarrow3x-y=3\)
Lại có: \(9x^2-6xy+y^2+y-3x+4=\left(3x-y\right)^2+y-3x+4\)
\(=\left(3x-y\right)^2-\left(3x-y\right)+4\)
Ta thay \(3x-y=3\)vào biểu thức trên:
\(\Rightarrow\left(3x-y\right)^2-\left(3x-y\right)+4=3^2-3+4=9+1=10\)
Vậy giá trị cần tìm của biểu thức đó là 10.
1; \(x^2\) + 3\(x^2\) + 3\(x\) = 4\(x^2\) + 3\(x\) (1)
Thay \(x=99\) vào (1) ta có:
4.992 + 3.99 = 4.9801 + 297 = 39204 + 297 = 39501