(2x-7)^2-11(2x-7)+30=0. Created by nagira77. algebra-ru

(2x-7)^2-11(2x-7)+30=0...

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(2x-7)²-11(2x-7)+30=0,
4х²-28х+49-22х+77+30=0,
4х²-50х+156=0,
2(2х²-25х+78)=0,
2(2х²-12х-13х+78)=0,
2(2х(х-6)-13(х-6))=0,
2(2х-13)(х-6)=0,
(2х-13)(х-6)=0,
2х-13=0, х-6=0,
2х=13, х=6.
х=13:2,
х=6,5.
Ответ: 6; 6,5.

1 votes Thanks 4

A1dar Наиболее рациональное решение
(2x - 7)² - 11(2x - 7) + 30 = 0
Замена t = 2x - 7; t ≥ 0
t² - 11t + 30 = 0
D = 11² - 4 · 30 = 1
t₁ = (11 + 1) / 2 = 6
t₂ = (11 - 1) / 2 = 5
Возвращаемся к замене
2x - 7 = 6 ⇔ 2x = 13 ⇔ x = 6,5
2x - 7 = 5 ⇔ 2x = 12 ⇔ x = 6
Ответ: 6; 6,5

0 votes Thanks 1

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