Question

Calcule o valor numérico de (x + y)² e o de x² + y², em cada caso.
a) x = 3 e y = 4
b) x =-7 e y = 7
d) x =0 e y = 9
e) x = y = 1
c) x = 6 e y = -5
f) x = 1,1 e y = 0,4​

Answer 1

a) (3+4)^2=7^2=49\ e\ 3^2=9(3+4)

2

=7

2

=49 e 3

2

=9

b) (-7+7)^2=0^2=0\ e\ (-7)^2=49(−7+7)

2

=0

2

=0 e (−7)

2

=49

c) (6-5)^2=1^2=1\ e\ 6^2=36(6−5)

2

=1

2

=1 e 6

2

=36

d) (0+9)^2=9^2=81\ e\ 0^2=0(0+9)

2

=9

2

=81 e 0

2

=0

e) (1+1)^2=2^2=4\ e\ 1^2=1(1+1) ko

2

=2

2

=4 e 1

2

=1

f)

\begin{gathered}(1,1+0,4)^2=(1,5)^2=(\frac{15}{10})^2=\frac{225}{100}=\frac{45}{20}=\frac{9}{4}\ e\ \\(1,1)^2=(\frac{11}{10})^2=\frac{121}{100}\end{gathered}