{toán 9} giải phương trình vô tỉ

N

nguyenbahiep1

1.[TEX]\sqrt{3x+1}[/TEX] = -4x^2+13x-5

[laTEX]dk: -4x^2+13x-5 \geq 0 \\ \\ (3x+1)= (-4x^2+13x-5)^2 \\ \\ (4x^2-15x+8)(4x^2-11x+3) = 0 [/laTEX]
 
N

nguyenbahiep1

2.[TEX]x+\sqrt{4-x^2}[/TEX] = [TEX]2+3x\sqrt{4-x^2}[/TEX]

[laTEX]TXD: -2 \leq x \leq 2 \\ \\ x+\sqrt{4-x^2} = u \Leftrightarrow u^2 = 4 +2x\sqrt{4-x^2} \\ \\ \frac{u^2-4}{2}= x\sqrt{4-x^2}[/laTEX]
 
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