Question

a, cho x+y=1.Tính x³+y³+3xy
b, cho x-y=1.Tính x³-y³-3xy

Answer 1

a) Vì x + y = 1 => ( x + y )³ = 1

=> x³ + 3x²y + 3xy² + y³ = 1

=> x³ + y³ + 3xy ( x + y ) = 1

=> x³ + y³ +3xy = 1 (do x+y=1)

b) x-y=1 => (x-y)³=1

=> x³ - 3x²y + 3xy² -y³ = 1

=> x³ -y³ - 3xy (x - y) = 1 

=> x³ - y³ -3xy =1 (do x-y=1) 

Answer 2

x + y = 1

=> (x + y)³ = 1

<=> x³ + y³ + 3x²y + 3xy² = 1

<=> x³ + y³ + 3xy (x+y) = 1

<=> x³ + y³ + 3xy = 1

Vậy ... = 1

 

x - y = 1

=> (x - y)³ = 1

<=> x³ - y³ - 3x²y + 3xy² = 1

<=> x³ - y³ - 3xy (x - y) = 1

<=> x³ - y³ - 3xy = 1

Vậy ... = 1

Answer 3

a) Ta có: x³+y³+3xy=(x+y)³-3xy(x+y)+3xy

=1³-3xy+3xy

=1

b) Ta có: x³-y³-3xy=(x-y)³+3xy(x-y)-3xy

=1³+3xy-3xy

=1