Giải các phương trình sau:

a) \(\frac{{x - 1}}{{59}} + \frac{{x - 2}}{{58}} + \frac{{x - 3}}{{57}} = \frac{{x - 4}}{{56}} + \frac{{x - 5}}{{55}} + \frac{{x - 6}}{{54}}\);

b) (3x – 2)² – (x + 3)² = 0;

c) x² – 9 = 2x + 6;

d) x³ + 9x² + 27x + 27 = 0.

Lời giải

a) \(\frac{{x - 1}}{{59}} + \frac{{x - 2}}{{58}} + \frac{{x - 3}}{{57}} = \frac{{x - 4}}{{56}} + \frac{{x - 5}}{{55}} + \frac{{x - 6}}{{54}}\)

\( \Leftrightarrow \frac{{x - 1}}{{59}} - 1 + \frac{{x - 2}}{{58}} - 1 + \frac{{x - 3}}{{57}} - 1 = \frac{{x - 4}}{{56}} - 1 + \frac{{x - 5}}{{55}} - 1 + \frac{{x - 6}}{{54}} - 1\)

\( \Leftrightarrow \frac{{x - 1 - 59}}{{59}} + \frac{{x - 2 - 58}}{{58}} + \frac{{x - 3 - 57}}{{57}} = \frac{{x - 4 - 56}}{{56}} + \frac{{x - 5 - 55}}{{55}} + \frac{{x - 6 - 54}}{{54}}\)

\( \Leftrightarrow \frac{{x - 60}}{{59}} + \frac{{x - 60}}{{58}} + \frac{{x - 60}}{{57}} - \frac{{x - 60}}{{56}} - \frac{{x - 60}}{{55}} - \frac{{x - 60}}{{54}} = 0\)

\( \Leftrightarrow \left( {x - 60} \right)\left( {\frac{1}{{59}} + \frac{1}{{58}} + \frac{1}{{57}} - \frac{1}{{56}} - \frac{1}{{55}} - \frac{1}{{54}}} \right) = 0\)    (1)

Ta có \(\frac{1}{{59}} < \frac{1}{{56}}\). Suy ra \(\frac{1}{{59}} - \frac{1}{{56}} < 0\).

Chứng minh tương tự \(\frac{1}{{58}} - \frac{1}{{55}} < 0\) và \(\frac{1}{{57}} - \frac{1}{{54}} < 0\).

Vì vậy \(\frac{1}{{59}} + \frac{1}{{58}} + \frac{1}{{57}} - \frac{1}{{56}} - \frac{1}{{55}} - \frac{1}{{54}} < 0\).

Khi đó phương trình (1) tương đương với: x – 60 = 0

⇔ x = 60.

Vậy x ∈ {60}.

b) (3x – 2)² – (x + 3)² = 0

⇔ (3x – 2 – x – 3)(3x – 2 + x + 3) = 0

⇔ (2x – 5)(4x + 1) = 0

⇔ 2x – 5 = 0 hoặc 4x + 1 = 0

⇔ \(x = \frac{5}{2}\) hoặc \(x = - \frac{1}{4}\).

Vậy \(x \in \left\{ {\frac{5}{2}; - \frac{1}{4}} \right\}\).

c) x² – 9 = 2x + 6

⇔ x² – 2x – 15 = 0

⇔ x² + 3x – 5x – 15 = 0

⇔ x(x + 3) – 5(x + 3) = 0

⇔ (x – 5)(x + 3) = 0

⇔ x – 5 = 0 hoặc x + 3 = 0

⇔ x = 5 hoặc x = –3

Vậy x ∈ {5; –3}.

d) x³ + 9x² + 27x + 27 = 0

⇔ (x + 3)³ = 0

⇔ x + 3 = 0

⇔ x = –3.