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Bài làm :
Ta có :
\(Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\):
\(Q=x^3+y^3-2x^2-2y^2+3x^2y+3xy^2-4xy+3\left(x+y\right)+10\)
\(Q=\left(x^3+y^3+3x^2y+3xy^2\right)-\left(2x^2+2y^2+4xy\right)+3\left(x+y\right)+10\)
\(Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)
Thay x+y=5 vào biểu thức trên ; ta được :
\(Q=5^3-2.5^2+3.5+10=125-50+15+10=100\)
Vậy Q=100
\(Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=x^3+y^3-2x^2-2y^2+3x^2y+3xy^2-4xy+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(2x^2+4xy+2y^2\right)+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)
Thay x + y = 5 vào pt ta được :
\(Q=5^3-2.5^2+3.5+10=125-50+15+10=100\)
Vậy Q = 100 <=> x + y = 5
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) \(A=x^2+2xy+y^2-4x-4y+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
b) \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2.7+37=100\)
c) \(C=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10=25\)
a) \(A=x^2+2xy+y^2-4x-4v+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
Câu 2:
\(B=x^2+2x+y^2-2x-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2\cdot7+37=49+37+14=100\)
Câu 3:
\(C=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2\cdot5+10=25\)
\(a,P=3x^2-2x+3y^2-2y+6xy-100\)
\(P=3\left(x^2+y^2\right)-\left[2\left(x+y\right)\right]+6xy-100\)
\(P=3\left(x^2+y^2+2xy-2xy\right)-2.5+6xy-100\)
\(P=3\left(x+y\right)^2-6xy-10+6xy-100\)
\(P=3.25-10-100\)
\(P=-35\)
\(b,Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(Q=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x^2+y^2+2xy-2xy\right)+3xy.5-4xy+3.5+10\)\(Q=5.\left(x^2+y^2+2xy-3xy\right)-2\left(x+y\right)^2+4xy+15xy-4xy+25\)
\(Q=5.5-15xy-2.25+15xy+25\)
\(Q=25-50+25=0\)
a) P= 3x2 -2x + 3y2-2y + 6xy -100
= (3x2+ 3y2 + 6xy) - 2(x+y) -100
=3(x2 + y2 +2xy) - 2(x+y) -100
=3(x+y)2 - 2(x+y) -100
=3 . 52 -2 .5 -100
=35
b) Q=x3 + y3 -2x2 -2y2 + 3xy (x+y) -4xy + 3(x+y) + 10
=(x3 +y3) + 3xy (x+y) + 3(x+y) -4xy -2x2 -2y2 + 10
=(x+y) (x2 -xy +y2 ) + 3xy (x+y) + 3 (x+y) - 2 (2xy + x2 +y2 ) + 10
=(x+y) (x2 -xy +y2 + 3xy ) + 3(x+y) -2 (2xy + x2 + y2 ) + 10
=(x+y) (x2 +2xy +y2 ) + 3(x+y) - 2(x+y)2 + 10
= (x+y)3 + 3(x+y) - 2 (x+y)2 + 10
=53 + 3.5 -2. 52+ 10
=100
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. \(Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(2x^2+4xy+2y^2\right)+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10\)
\(\Leftrightarrow Q=5^3-2.5^2+3.5+10\)
\(\Leftrightarrow Q=100\)
\(A=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(A=\left(x^3+y^3\right)-2\left(x^2+y^2\right)+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(A=\left(x+y\right)^3-3xy\left(x+y\right)-2\left(\left(x+y\right)^2-2xy\right)+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(A=\left(x+y\right)^3-3xy\left(x+y\right)-2\left(x+y\right)^2+4xy+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)
\(A=\left(5\right)^3-3xy\left(5\right)-2\left(5\right)^2+4xy+3xy\left(5\right)-4xy+3\left(5\right)+10\)
\(A=125-15xy-50+4xy+15xy-4xy+15+10\)
\(A=100\)