Question

d) [(-3).(-3)²-(-3)³-(-3)*(-3)+275]-(-1)
6. Calculaţi A = ab² + a²b² - ab, ştiind că:
a) a=-2, b=3;
+2020

Answer 1

Răspuns:

Explicație pas cu pas:

D) [(-3).(-3)²·(-3)³·(-3)·(-3)+27⁵]·(-1)⁽¹⁺²⁺³⁺   ⁺²⁰²⁰⁾=[(-3)⁽¹⁺²⁺³⁺⁴⁺⁵⁾+(3³)⁵]·(-1)⁽¹⁺²⁺³⁺   ⁺²⁰²⁰⁾=[(-3)¹⁵+3¹⁵]·(-1)⁽¹⁺²⁺³⁺   ⁺²⁰²⁰⁾=[-3¹⁵+3¹⁵]·(-1)⁽¹⁺²⁺³⁺   ⁺²⁰²⁰⁾=

0·(-1)⁽¹⁺²⁺³⁺   ⁺²⁰²⁰⁾=0

6. Calculaţi A = ab² + a²b² - ab, ştiind că:

A = ab² + a²b² - ab=ab(b+ab-1)

a) a=-2, b=3; A =ab(b+ab-1)=-2·3·(3-2·3-1)=-6·(3-7)=-6·(-4)=24

b) a=3, b=-1; A =ab(b+ab-1)=3·(-1)·[-1+3·(-1)-1]=3·(-1-3-1)=3·(-5)=-15    

c) a=-4, b=-3; A =ab(b+ab-1)=(-4)·(-3)·[-3+(-4)·(-3)-1]=12·(-3+12-1)=12·8=96

d) a=5, b=-4; A =ab(b+ab-1)=5·(-4)·[-4+5·(-4)-1]=-20·(-4-20-1)=-20·(-25)=500

e) a=-1, b=0; A =ab(b+ab-1)=(-1)·0·[0+(-1)·0-1]=0    

f) a=b=-2; A =ab(b+ab-1)=(-2)·(-2)·[-2+(-2)·(-2)-1]=4·(-2+4-1)=4·1=4  

g) a=0, b=-10; A =ab(b+ab-1)=0·(-10)·[-10+0·(-10)-1]=0  

h) a=3, b=5; A =ab(b+ab-1)=3·5·(5+3·5-1)=15·19=285