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a)A=3(x2+2xy+y2)-2(x+y)-100=3(x+y)2-2.5-100=3.52-110=-35
b)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
(l ike nha)
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. \(Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(2x^2+4xy+2y^2\right)+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=5^3-2.5^2+3.5+10\)
\(\Leftrightarrow Q=100\)
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
B=....
<=>B=x^3+y^3+3xy(x+y)-2(x^2+y^2+2xy)+3(x+y)+10
<=>B=(x+y)^3-2(x+y)^2+3(x+y)+10
tại x+y=5 thay vao B ta đc:
B=5^3-2.5^2+3.5+10
B=100
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
\(M=x^3+y^3-2\left(x^2+y^2\right)+3xy\left(x+y\right)-4xy+3x+10+3y\)
\(=x^3+y^3-2x^2-2y^2+3x^2y+3xy^2-4xy+3x+10+3y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-2\left(x^2+2xy+y^2\right)+3\left(x+y\right)+10\)
\(=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)
Ta có: x + y = 5
\(\Rightarrow\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10=5^3-2.5^2+3.5+10=125-50+15+10=100\)
Vậy M = 100.