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nguyenbahiep1

[TEX]\sqrt{3x+4} = u \\ dk : u \geq 0 \\ u^2 -4 = 3x \\ 4x^2 - 5x - 4 +\sqrt{3x+4} =0 \Leftrightarrow 4.\frac{(u^2-4)^2}{9}- \frac{5.(u^2-4)}{3} - 4 + u = 0 \\ 4(u^4-2u^2+1) - 15(u^2-4) - 36 + 9u \\ 4u^4-23u^2 +9u + 28 = 0 \\ (u+1)(4u^3-4u^2-19u+28) = 0 [/TEX]
 
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