:根号x+根号(x+7)+2*根号(x^2+7x)=35-2x用换元法做

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:根号x+根号(x+7)+2*根号(x^2+7x)=35-2x
用换元法做

√x + √(x+7) + 2√(x^2+7x) = 35-2x
√x + √(x+7) + 2√(x(x+7)) = 35-2x
2x + 2√(x(x+7)) + (√x+√(x+7)) = 35
x + 2√x√(x+7) + (x+7) + (√x+√(x+7)) = 35+7
(√x)^2 + 2√x√(x+7) + (√(x+7))^2 + (√x+√(x+7)) = 42
(√x+√(x+7))^2 + (√x+√(x+7)) - 42 = 0
解得
√x+√(x+7) = -7或√x+√(x+7)=6
因为√x+√(x+7) > 0故√x+√(x+7)=6
√(x+7)=6-√x
两边同时平方得:
x+7 = 36-12√x + x
化简得12√x = 29
x=841/144
这样可以的哦,这题还算简单不需要用换元法的~
希望对你有所帮助

√x + √(x+7) + 2√(x^2+7x) = 35-2x
√x + √(x+7) + 2√(x(x+7)) = 35-2x
2x + 2√(x(x+7)) + (√x+√(x+7)) = 35
x + 2√x√(x+7) + (x+7) + (√x+√(x+7)) = 35+7
(√x)^2 + 2√x√(x+7) + (√(x+7))^2 + (√x+√(...

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√x + √(x+7) + 2√(x^2+7x) = 35-2x
√x + √(x+7) + 2√(x(x+7)) = 35-2x
2x + 2√(x(x+7)) + (√x+√(x+7)) = 35
x + 2√x√(x+7) + (x+7) + (√x+√(x+7)) = 35+7
(√x)^2 + 2√x√(x+7) + (√(x+7))^2 + (√x+√(x+7)) = 42
(√x+√(x+7))^2 + (√x+√(x+7)) - 42 = 0
解得
√x+√(x+7) = -7或√x+√(x+7)=6
因为√x+√(x+7) > 0故√x+√(x+7)=6
√(x+7)=6-√x
两边同时平方得:
x+7 = 36-12√x + x
化简得12√x = 29
x=841/144
求采纳,O(∩_∩)O谢谢

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