Phương trình......

P

phucvo29

[TEX]7 + 2\sqrt{x} - x = (2 + \sqrt{x})\sqrt{7-x}[/TEX]
[TEX]\Leftrightarrow (7-x) - \sqrt{x}\sqrt{7-x} + 2\sqrt{x} - 2\sqrt{7-x} = 0[/TEX]
[TEX]\Leftrightarrow (\sqrt{7-x}-\sqrt{x})(\sqrt{7-x} - 2) =0[/TEX]
[TEX]\Leftrightarrow \left[\begin{\sqrt{7-x} - \sqrt{x}=0}\\{\sqrt{7-x} - 2 = 0}[/TEX]
[TEX]\Leftrightarrow \left[\begin{7-x +x -2\sqrt{7x -x^2}=0}\\{x = 3}[/TEX]
[TEX]\Leftrightarrow \left[\begin{ 2\sqrt{7x -x^2}=7}\\{x = 3}[/TEX]
[TEX]\Leftrightarrow \left[\begin{x^2-7x+12,25=0}\\{x = 3}[/TEX]
[TEX]\Leftrightarrow \left[\begin{x=3,5}\\{x = 3}[/TEX]
 
L

lucifer_bg93

[TEX]7 + 2\sqrt{x} - x = (2 + \sqrt{x})\sqrt{7-x}[/TEX]
[TEX]\Leftrightarrow (7-x) - \sqrt{x}\sqrt{7-x} + 2\sqrt{x} - 2\sqrt{7-x} = 0[/TEX]
[TEX]\Leftrightarrow (\sqrt{7-x}-\sqrt{x})(\sqrt{7-x} - 2) =0[/TEX]
[TEX]\Leftrightarrow \left[\begin{\sqrt{7-x} - \sqrt{x}=0}\\{\sqrt{7-x} - 2 = 0}[/TEX]
[TEX]\Leftrightarrow \left[\begin{7-x +x -2\sqrt{7x -x^2}=0}\\{x = 3}[/TEX]
[TEX]\Leftrightarrow \left[\begin{ 2\sqrt{7x -x^2}=7}\\{x = 3}[/TEX]
[TEX]\Leftrightarrow \left[\begin{x^2-7x+12,25=0}\\{x = 3}[/TEX]
[TEX]\Leftrightarrow \left[\begin{x=3,5}\\{x = 3}[/TEX]

chỗ này bạn nhầm ah?
[tex]{\sqrt{7-x} - \sqrt{x}=0}[/TEX]
[tex]\Leftrightarrow \left[\begin{x\geq 0}\\{7-x =x} [/tex]
 
T

tieutumienque

hắn làm phức tạp thêm xo zui đó........keke..........................................................................................................................................
 
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