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a ) có \(x^2+y^2+4x-2xy+4y+2019=\left(x-y\right)^2+4\left(x-y\right)+2019=49+28+2019=2096\)
b) \(x^3-3xy\left(x-y\right)-y^3-x^2+2xy-y^2=\left(x-y\right)^3-\left(x-y\right)^2=343-49=294\)
c)\(x^2\left(x+1\right)-y^2\left(y-1\right)+xy-3xy\left(x-y+1\right)=x^3-y^3+x^2+y^2+xy-3x^2y+3xy^2-3xy=\left(x-y\right)^3+\left(x-y\right)^2=343+49=392\)
a. Có \(x+y=2\Rightarrow x^2+2xy+y^2=4\Rightarrow x^2+y^2=4-2.\left(-3\right)=10\)
\(x^4+y^4=\left(x^2\right)^2+\left(y^2\right)^2=\left(x^2+y^2\right)^2-2x^2y^2\)
\(=10^2-2.\left(-3\right)^2=82\)
b. Ta có \(x+y=1\Rightarrow x^2+y^2=1-2xy\)
\(x^3+y^3+3xy=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)
\(=1.\left(1-2xy-xy\right)+3xy=1\)
Các câu còn lại tương tự
c) \(C=\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left[\left(a+b\right)^2-ab\right]=3\left(9^2-ab\right)\)
\(\left(a+b\right)^2=81\Leftrightarrow a^2+2ab+b^2=81\Leftrightarrow a^2+b^2=81-2ab\)
\(\left(a-b\right)^2=9\Leftrightarrow a^2+b^2=9+2ab\)
=> \(81-2ab=9+2ab\Rightarrow4ab=72\Leftrightarrow ab=18\)
\(\Leftrightarrow C=3\left(81-18\right)=189\)
\(D=\left(x^2+2xy+y^2\right)-4\left(x+y+1\right)\)
\(D=\left(x+y\right)^2-4.4=3^2-16=9-16=-7\)
\(A=x^2+2xy+y^2-4x-4y+1=\left(x+y\right)^2-4\left(x+y\right)+1=3^2-12+1=-2\)
\(B=x^2-2xy+y^2-5x+5y+6=\left(x-y\right)^2-5\left(x-y\right)+6=7^2-5.7+6=20\)
a)Ta có
A=\(x^2+2xy+y^2-4x-4y+1\)
=>A=\(\left(x+y\right)^2-4\left(x+y\right)+1\)
Mà x+y=3 nên
A=\(3^2-4\cdot3+1\)
A=-2
b)Ta có:
B=\(x^2-2xy+y^2-5x+5y+6\)
B=\(\left(x-y\right)^2-5\left(x-y\right)+6\)
Mà x-y=7 nên
B=\(7^2-5\cdot7+6\)
B=20
Bài 1.
A = x2 + 2xy + y2 = ( x + y )2 = ( -1 )2 = 1
B = x2 + y2 = ( x2 + 2xy + y2 ) - 2xy = ( x + y )2 - 2xy = (-1)2 - 2.(-12) = 1 + 24 = 25
C = x3 + 3xy( x + y ) + y3 = ( x3 + y3 ) + 3xy( x + y ) = ( x + y )( x2 - xy + y2 ) + 3xy( x + y )
= -1( 25 + 12 ) + 3.(-12).(-1)
= -37 + 36
= -1
D = x3 + y3 = ( x3 + 3x2y + 3xy2 + y3 ) - 3x2y - 3xy2 = ( x + y )3 - 3xy( x + y ) = (-1)3 - 3.(-12).(-1) = -1 - 36 = -37
Bài 2.
M = 3( x2 + y2 ) - 2( x3 + y3 )
= 3( x2 + y2 ) - 2( x + y )( x2 - xy + y2 )
= 3( x2 + y2 ) - 2( x2 - xy + y2 )
= 3x2 + 3y2 - 2x2 + 2xy - 2y2
= x2 + 2xy + y2
= ( x + y )2 = 12 = 1
Có (a+b+c)2 = 3(ab+bc+ac)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=3ab+3bc+3ac\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac-3ab-3bc-3ac\)\(=0\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ac\)\(=0\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2\)\(=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Rightarrow a=b=c\)