Giải phương trình

N

nguyenbahiep1

$\sqrt{4x^2+5x+1} - 2.\sqrt{x^2-x+1}=9x-3$

Giải

$TXD: x \geq \frac{-1}{4} , x \leq -1 \\ \frac{4x^2+5x+1-4(x^2-x+1)}{\sqrt{4x^2+5x+1} + 2.\sqrt{x^2-x+1}} = 9(3x-1) \\ \\ \frac{3x-1}{\sqrt{4x^2+5x+1} + 2.\sqrt{x^2-x+1}} = 3x-1 \Rightarrow x = \frac{1}{3} \\ \\ \frac{1}{\sqrt{4x^2+5x+1} + 2.\sqrt{x^2-x+1}} = 1 $

Ta có: $x^2-x+1 = (x-\frac{1}{2})^2 + \frac{3}{4} \Rightarrow 2.\sqrt{x^2-x+1} \geq 2\frac{\sqrt{3}}{2} > 1 \\ \\ \Rightarrow \sqrt{4x^2+5x+1} + 2.\sqrt{x^2-x+1} > 1 \\ \\ \Rightarrow \frac{1}{\sqrt{4x^2+5x+1} + 2.\sqrt{x^2-x+1}} < 1$

vậy pt có 1 nghiệm duy nhất $x = \frac{1}{3}$
 
B

braga

$$pt\iff \sqrt{4x^2+5x+1} - \sqrt{4x^2-4x+4}=9x-3 \\ \text{Đặt: } \sqrt{4x^2+5x+1}=a \ ; \ \sqrt{4x^2-4x+4}=b \\ pt\iff a-b=a^2-b^2\iff (a-b)(a+b-1)=0$$
 
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