a) \(A+B=x^2-3xy-y^2+1+\left(2x^2+y^2-7xy-5\right)\)
\(=x^2-3xy-y^2+1+2x^2+y^2-7xy-5\)
\(=\left(x^2+2x^2\right)+\left(-3xy-7xy\right)+\left(-y^2+y^2\right)+\left(1-5\right)\)
\(=3x^2-10xy-4\)
b) \(C=B-A\)
\(=2x^2+y^2-7xy-5-\left(x^2-3xy-y^2+1\right)\)
\(=2x^2 +y^2-7xy-5-x^2+3xy+y^2-1\)
\(=\left(2x^2-x^2\right)+\left(y^2+y^2\right)+\left(-7xy+3xy\right)+\left(-5-1\right)\)
\(=x^2+2y^2-4xy-6\)
c) Thay \(x=2;y=\dfrac{-1}{2}\) vào \(C\), ta được:
\(C=2^2+2\cdot\left(\dfrac{-1}{2}\right)^2-4\cdot2\cdot\left(\dfrac{-1}{2}\right)-6\)
\(=4+\dfrac{1}{2}+4-6\)
\(=\dfrac{5}{2}\)
Vậy: ...
#\(Toru\)
a) \(A+B=\left(x^2-3xy-y^2+1\right)+\left(2x^2+y^2-7xy-5\right)\)
\(=3x^2-10xy-4\)
b) \(B-A=\left(2x^2+y^2-7xy-5\right)-\left(x^2-3xy-y^2+1\right)\)
\(=x^2-4xy+2y^2-6\)
c) Tại x=2; \(y=-\dfrac{1}{2}\)
C\(=\left(2^2\right)-4.2.\dfrac{-1}{2}+2\left(\dfrac{-1}{2}\right)^2-6\)
\(=4+4+\dfrac{1}{2}-6=2+\dfrac{1}{2}=\dfrac{5}{2}\)
