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聯立方程(數學)
22. (5^x)+2y=125 (25^2x+y)=1 24. 2^3x=4^y+1 5^x-2=5^2-y 25. (3^2x-1)-3^5-y =0 (9^x)-3(81^y)=0
最佳解答:
22. 5^(x + 2y) = 125 ... (1) 25^(2x + y) = 1... (2) (1) => 5^(x + 2y) = 5^3 x + 2y = 3 x = 3 - 2y ... (3) (2) => 25^(2x + y) = 25^0 2x + y = 0 Use (3), 2(3 - 2y) + y = 0 6 - 4y + y = 0 3y = 6 y = 2 Sub into (3), x = -1 24. 2^3x=4^(y + 1) ... (1) 5^(x - 2)=5^(2 - y) ... (2) (1) => 2^3x = (2^2)^(y + 1) 2^3x = 2^(2y + 2) 3x = 2y + 2 ... (3) (2) => x - 2 = 2 - y x = 4 - y ... (4) Sub into (3), 3(4 - y) = 2y + 2 12 - 3y = 2y + 2 10 = 5y y = 2 Sub into (4), x = 2 25. 3^(2x - 1) - 3^(5 - y) = 0... (1) (9^x) - 3(81^y) = 0 ... (2) (1) => 3^(2x - 1) = 3^(5 - y) 2x - 1 = 5 - y y = 6 - 2x ... (3) (2) => 9^x = 3(81^y) 3^2x = (3)(3^4y) 3^2x = 3^(4y + 1) 2x = 4y + 1 Use (3) 2x = 4(6 - 2x) + 1 2x = 24 - 8x + 1 10x = 25 x = 2.5 Sub into (3), y = 1
其他解答:
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聯立方程(數學)
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發問:22. (5^x)+2y=125 (25^2x+y)=1 24. 2^3x=4^y+1 5^x-2=5^2-y 25. (3^2x-1)-3^5-y =0 (9^x)-3(81^y)=0
最佳解答:
22. 5^(x + 2y) = 125 ... (1) 25^(2x + y) = 1... (2) (1) => 5^(x + 2y) = 5^3 x + 2y = 3 x = 3 - 2y ... (3) (2) => 25^(2x + y) = 25^0 2x + y = 0 Use (3), 2(3 - 2y) + y = 0 6 - 4y + y = 0 3y = 6 y = 2 Sub into (3), x = -1 24. 2^3x=4^(y + 1) ... (1) 5^(x - 2)=5^(2 - y) ... (2) (1) => 2^3x = (2^2)^(y + 1) 2^3x = 2^(2y + 2) 3x = 2y + 2 ... (3) (2) => x - 2 = 2 - y x = 4 - y ... (4) Sub into (3), 3(4 - y) = 2y + 2 12 - 3y = 2y + 2 10 = 5y y = 2 Sub into (4), x = 2 25. 3^(2x - 1) - 3^(5 - y) = 0... (1) (9^x) - 3(81^y) = 0 ... (2) (1) => 3^(2x - 1) = 3^(5 - y) 2x - 1 = 5 - y y = 6 - 2x ... (3) (2) => 9^x = 3(81^y) 3^2x = (3)(3^4y) 3^2x = 3^(4y + 1) 2x = 4y + 1 Use (3) 2x = 4(6 - 2x) + 1 2x = 24 - 8x + 1 10x = 25 x = 2.5 Sub into (3), y = 1
其他解答:
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