Giải các bất phương trình sau:

a) \[8x + 2 < 7x - 1\].

b) \(3x - 8 > 4x - 12\).

c) \(3\left( {x - 2} \right) - 5 \ge 3\left( {2x - 1} \right)\).

d) \(5x - 7\left( {2x - 5} \right) < 2\left( {x - 1} \right)\).

e) \[{\left( {x - 1} \right)^2} < x\left( {x + 3} \right)\].

f) \[\left( {x + 3} \right)\left( {x - 1} \right) < {\left( {x + 1} \right)^2} - 4\].

g) \(\left( {x + 2} \right)\left( {x + 4} \right) > \left( {x - 2} \right)\left( {x + 8} \right) + 26\).

h) \[{\left( {x - 4} \right)^2} - \left( {x + 5} \right)\left( {x - 5} \right)\;\; \ge \;\; - 8x + 41\].

i) \(\frac{{x + 1}}{3} + \frac{x}{2} \ge 4\).

j) \[\frac{{2x - 1}}{3} - \frac{{x + 2}}{2} \ge \frac{{5x + 4}}{6}\].

Xét \(\Delta ABC\) vuông tại \(A,\) ta có:

⦁ \(B{C^2} = A{B^2} + A{C^2}\) (định lí Pythagore);

⦁ \[\widehat {A\,} + \widehat {B\,} + \widehat {C\,} = 180^\circ \] (tổng ba góc của một tam giác).

a) Ta có \(B{C^2} = A{B^2} + A{C^2} = {3^2} + {4^2} = 25,\) suy ra \(BC = 5\).

\(\sin B = \frac{{AC}}{{BC}} = \frac{4}{5},\) suy ra \(\widehat {B\,} \approx 53^\circ 8'.\)

Suy ra \[\widehat {C\,} = 180^\circ - \widehat {A\,} - \widehat {B\,} \approx 180^\circ - 90^\circ - 53^\circ 8' = 36^\circ 52'.\]

Vậy \(AB = 3,\,\,AC = 4,\,\,BC = 5;\,\,\widehat {A\,\,} = 90^\circ ,\,\,\widehat {B\,} \approx 53^\circ 8',\,\,\widehat {C\,} \approx 36^\circ 52'.\)

b) Ta có \(A{C^2} = B{C^2} - A{B^2} = {11^2} - {6^2} = 85,\) suy ra \(AC = \sqrt {85} \approx 9,22.\)

\(\cos B = \frac{{AB}}{{BC}} = \frac{6}{{11}},\) suy ra \(\widehat {C\,} \approx 56^\circ 57'.\)

Suy ra \[\widehat {B\,} = 180^\circ - \widehat {A\,} - \widehat {C\,} \approx 180^\circ - 90^\circ - 56^\circ 57' = 33^\circ 3'.\]

Vậy \[AB = 6,\,\,AC \approx 9,22,\,\,BC = 11;\,\,\widehat {A\,\,} = 90^\circ ,\,\,\widehat {B\,} \approx 56^\circ 57',\,\,\widehat {C\,} \approx 33^\circ 3'.\]

c) Ta có \(\sin C = \frac{{AB}}{{BC}},\) suy ra \(BC = \frac{{AB}}{{\sin C}} = \frac{9}{{\sin 32^\circ }} \approx 16,98.\)

\(AC = AB \cdot \cot C = 9 \cdot \cot 32^\circ \approx 14,40.\)

\[\widehat {B\,} = 180^\circ - \widehat {A\,} - \widehat {C\,} = 180^\circ - 90^\circ - 32^\circ = 58^\circ .\]

Vậy \[AB = 9,\,\,AC \approx 14,40,\,\,BC \approx 16,98;\,\,\widehat {A\,\,} = 90^\circ ,\,\,\widehat {B\,} = 58^\circ ,\,\,\widehat {C\,} = 32^\circ .\]

d) Ta có \(AB = BC \cdot \sin C = 12 \cdot \sin 37^\circ \approx 7,22.\)

\(AC = BC \cdot \cos C = 12 \cdot \cos 37^\circ \approx 9,58.\)

\[\widehat {B\,} = 180^\circ - \widehat {A\,} - \widehat {C\,} = 180^\circ - 90^\circ - 37^\circ = 53^\circ .\]

Vậy \[AB \approx 7,22,\,\,AC \approx 9,58,\,\,BC = 12;\,\,\widehat {A\,\,} = 90^\circ ,\,\,\widehat {B\,} = 53^\circ ,\,\,\widehat {C\,} = 37^\circ .\]

e) Ta có: \(AB = AC \cdot \tan C = 4 \cdot \tan 40^\circ \approx 3,36.\)

\(AC = BC \cdot \cos C,\) suy ra \(BC = \frac{{AC}}{{\cos C}} = \frac{4}{{\cos 40^\circ }} \approx 5,22\).

\[\widehat {B\,} = 180^\circ - \widehat {A\,} - \widehat {C\,} = 180^\circ - 90^\circ - 40^\circ = 50^\circ .\]

Vậy \[AB \approx 3,36,\,\,AC = 4,\,\,BC \approx 5,22;\,\,\widehat {A\,\,} = 90^\circ ,\,\,\widehat {B\,} = 50^\circ ,\,\,\widehat {C\,} = 40^\circ .\]