Rozwiąż równanie:
(√3y+2)²-(√3y+2)(√3y-2)=0
tylko proszę w miarę czytelnie ;)
3y + 4√3y + 4 - (3y - 4) = 0
3y + 4√3y + 4 - 3y + 4 = 0
4√3y + 8 = 0
4(√3y + 2) = 0
4 ≠ 0 więc √3y + 2 = 0
√3y + 2 = 0
√3y = -2
(√3y)² = -2²
3y = 4
y = 4/3 = 1⅓
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(√3y+2)²-(√3y+2)(√3y-2)=0
3y + 4√3y + 4 - (3y - 4) = 0
3y + 4√3y + 4 - 3y + 4 = 0
4√3y + 8 = 0
4(√3y + 2) = 0
4 ≠ 0 więc √3y + 2 = 0
√3y + 2 = 0
√3y = -2
(√3y)² = -2²
3y = 4
y = 4/3 = 1⅓