Bài 1 trang 23 Tài liệu dạy – học Toán 9 tập 2

Đề bài

Dùng phương pháp thế để giải các hệ phương trình sau:

a) (left{ begin{array}{l}4x – y = 2x + 3y = 7end{array} right.)

b) (left{ begin{array}{l}4x – y = 82x – y = 10end{array} right.)

c) (left{ begin{array}{l}3x – 2y = 16x – 2y = 4end{array} right.)

d) (left{ begin{array}{l}2x + 5y = 16x – 15y = 4end{array} right.)

e) (left{ begin{array}{l}dfrac{x}{2} – dfrac{y}{3} = 13x – 2y = 6end{array} right.)

f) (left{ begin{array}{l}dfrac{{2x}}{3} – dfrac{{5y}}{3} = 14x – 10y = 6end{array} right.)

h) (left{ begin{array}{l}x + ysqrt 3 = 0xsqrt 3 + 2y = 2end{array} right.)

Lời giải chi tiết

(begin{array}{l}a),,left{ begin{array}{l}4x – y = 2x + 3y = 7end{array} right. Leftrightarrow left{ begin{array}{l}y = 4x – 2x + 3left( {4x – 2} right) = 7end{array} right. Leftrightarrow left{ begin{array}{l}y = 4x – 2x + 12x – 6 = 7end{array} right. Leftrightarrow left{ begin{array}{l}y = 4x – 213x = 13end{array} right. Leftrightarrow left{ begin{array}{l}x = 1y = 2end{array} right.end{array})

Vậy (left( {x;y} right) = left( {1;2} right)) là nghiệm của hệ phương trình.

(begin{array}{l}b),,left{ begin{array}{l}4x – y = 82x – y = 10end{array} right. Leftrightarrow left{ begin{array}{l}y = 4x – 82x – left( {4x – 8} right) = 10end{array} right. Leftrightarrow left{ begin{array}{l}y = 4x – 82x – 4x + 8 = 10end{array} right. Leftrightarrow left{ begin{array}{l}y = 4x – 8 – 2x = 2end{array} right. Leftrightarrow left{ begin{array}{l}x = – 1y = – 12end{array} right.end{array})

Vậy (left( {x;y} right) = left( { – 1; – 12} right)) là nghiệm của hệ phương trình.

(begin{array}{l}c),,left{ begin{array}{l}3x – 2y = 16x – 2y = 4end{array} right. Leftrightarrow left{ begin{array}{l}3x – 2y = 13x – y = 2end{array} right. Leftrightarrow left{ begin{array}{l}y = 3x – 23x – 2left( {3x – 2} right) = 1end{array} right. Leftrightarrow left{ begin{array}{l}y = 3x – 23x – 6x + 4 = 1end{array} right. Leftrightarrow left{ begin{array}{l}y = 3x – 2 – 3x = – 3end{array} right. Leftrightarrow left{ begin{array}{l}x = 1y = 1end{array} right.end{array})

Vậy (left( {x;y} right) = left( {1;1} right)) là nghiệm của hệ phương trình.

(begin{array}{l}d),,left{ begin{array}{l}2x + 5y = 16x – 15y = 4end{array} right. Leftrightarrow left{ begin{array}{l}x = dfrac{{1 – 5y}}{2}6.dfrac{{1 – 5y}}{2} – 15y = 4end{array} right. Leftrightarrow left{ begin{array}{l}x = dfrac{{1 – 5y}}{2}3left( {1 – 5y} right) – 15y = 4end{array} right. Leftrightarrow left{ begin{array}{l}x = dfrac{{1 – 5y}}{2}3 – 15y – 15y = 4end{array} right. Leftrightarrow left{ begin{array}{l}x = dfrac{{1 – 5y}}{2} – 30y = 1end{array} right. Leftrightarrow left{ begin{array}{l}y = dfrac{{ – 1}}{{30}}x = dfrac{{1 – 5.dfrac{{ – 1}}{{30}}}}{2} = dfrac{{1 + dfrac{1}{6}}}{2} = dfrac{7}{{12}}end{array} right.end{array})

Vậy (left( {x;y} right) = left( {dfrac{7}{{12}};dfrac{{ – 1}}{{30}}} right)) là nghiệm của hệ phương trình.

(e),,left{ begin{array}{l}dfrac{x}{2} – dfrac{y}{3} = 13x – 2y = 6end{array} right. Leftrightarrow left{ begin{array}{l}3x – 2y = 63x – 2y = 6end{array} right. )

Hệ phương trình vô số nghiệm.

(f),,left{ begin{array}{l}dfrac{{2x}}{3} – dfrac{{5y}}{3} = 14x – 10y = 6end{array} right. Rightarrow left{ begin{array}{l}2x – 5y = 32x – 5y = 3end{array} right. ) Hệ phương trình vô số nghiệm.

(begin{array}{l}g),,,left{ begin{array}{l}x + ysqrt 3 = 0xsqrt 3 + 2y = 2end{array} right. Leftrightarrow left{ begin{array}{l}x = – ysqrt 3 – ysqrt 3 .sqrt 3 + 2y = 2end{array} right. Leftrightarrow left{ begin{array}{l}x = – ysqrt 3 – 3y + 2y = 2end{array} right. Leftrightarrow left{ begin{array}{l}x = – ysqrt 3 y = – 2end{array} right. Leftrightarrow left{ begin{array}{l}x = 2sqrt 3 y = – 2end{array} right.end{array})

Vậy (left( {x;y} right) = left( {2sqrt 3 ; – 2} right)) là nghiệm của hệ phương trình.