Giải phương trình:
Bạn nào giúp mình với, cảm ơn nhiều!!!!!!!

$\frac{1}{\sqrt{x}}$ +$\frac{1}{\sqrt{2x-3}}$=$\sqrt{3}$($\frac{1}{\sqrt{4x-3}}$+$\frac{1}{\sqrt{5x-6}}$)
Đk $x> \frac 65$
$pt\Leftrightarrow \frac 1{\sqrt{3x}}+\frac 1{\sqrt{6x-9}}=\frac 1{\sqrt{4x-3}}+\frac 1{\sqrt{5x-6}}$
$\Leftrightarrow \frac{1}{3x}+\frac{1}{6x-9}+\frac{2}{\sqrt{3x(6x-9)}}=\frac{1}{4x-3}+\frac{1}{5x-6}+\frac{2}{\sqrt{(4x-3)(5x-6)}}$
$\Leftrightarrow \frac{9(x-1)}{3x(6x-9)}+\frac{2}{\sqrt{3x(6x-9)}}=\frac{9(x-1)}{(4x-3)(5x-6)}+\frac{2}{\sqrt{(4x-3)(5x-6)}}$
Vì ta có $3x(6x-9) \le (4x-3)(5x-6)$
Do đó $\frac{2}{\sqrt{3x(6x-9)}} \ge \frac{2}{\sqrt{(4x-3)(5x-6)}}$
Với $x > \frac 65$ thì ta cũng có $\frac{9(x-1)}{3x(6x-9)} \ge \frac{9(x-1)}{(4x-3)(5x-6)}$

Do đó $VT \ge VP$, dấu bằng xảy ra khi $x=3$
Vậy $x=3$ là nghiệm duy nhất của pt

pp đánh giá nha –  tran85295 17-08-16 07:39 PM
Giải theo pp j đây –  pinocchio 17-08-16 05:44 PM

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