№1 (3х+4)(3х-4)= (4х+5)(4х+5)= (2х-3)(2х-3)= (2х+1)(4х²-2х+1)= - вопрос №901002613434454523232142122 от urannoss127 16.08.2021 09:12

Question

Алгебра: №1 (3х+4)(3х-4)= (4х+5)(4х+5)= (2х-3)(2х-3)= (2х+1)(4х²-2х+1)= (х-2)(х²+2х+4)= №2 применить формулу (а±в)²=а²±2ав+в² для нахождения значения степени 51²,39²,103²,78²

urannoss127 · Answer

(3х + 4)(3х - 4) = (3х)² - 4² = 9х² - 16
(4х + 5)(4х + 5) = (4х + 5)² = (4х)² + 2*4х*5 + 5² = 16х² + 40х + 25
(2х - 3)(2х - 3) = (2х - 3)² = (2х)² - 2*2х * 3 + 3² = 4х² - 12х + 9
(2х + 1)(4х² - 2х + 1) = (2х + 1)( (2х)² - 2*х*1 + 1²) = (2х)³ + 1³ = 8х³ + 1
(х - 2)(х² + 2х + 4) = (х - 2)(х² + х*2 + 2²) = х³ - 2³ = х³ - 8

51² = (50 + 1)² = 50² + 2*50*1 + 1² = 2500 + 100 + 1= 2601
39² = (40 - 1)² = 40² - 2*40*1 + 1² = 1600 - 80 + 1 = 1521
103² = (100 + 3)² = 100² + 2*100*3 + 3² = 10000 + 600 + 9 = 10609
78² = ( 80 - 2)² = 80² - 2*80*2 + 2² = 6400 - 320 + 4 = 6084