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giải phương trình

1, $\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}= x$

2, $x-2\sqrt{x-1}-(x-1)\sqrt{x}+\sqrt{x^{2}-x}= 0$

tran85295 · Answer 1 · 6/5/2016 11:07:06 AM

Điều kiện $x \ge 1$

$pt\Leftrightarrow \biggl[(x-1)-2\sqrt{x-1}+1 \bigg]-\sqrt{x(x-1)} \biggl(\sqrt{x-1}-1 \biggr)=0$

$\Leftrightarrow (\sqrt{x-1}-1)^2-\sqrt{x(x-1)} \biggl(\sqrt{x-1}-1 \biggr)=0$

$\Leftrightarrow \biggl(\sqrt{x-1}-1\biggr)\underset{<0 \forall x \ge 1}{\underbrace{\biggl(\sqrt{x-1}-1-\sqrt{x(x-1)} \biggr)}}=0$

$\Leftrightarrow \boxed{x=2}$

Trả lời 05-06-16 11:07 AM

tran85295
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tran85295 · Answer 2 · 6/5/2016 10:49:56 AM

Điều kiện $x \ge 1$

Áp dụng bdt cosi, ta có :

$VT=\sqrt{1.\left(x-\frac 1x \right)}+\sqrt{\biggl( x-1 \biggr).\frac 1x} \le \frac 12 \Biggl(1+x-\frac 1x+\frac 1x+x-1\Biggr)=x$

Đẳng thức xảy ra $\Leftrightarrow x=\frac{\sqrt 5+1}{2}$

Vậy $x=\frac{\sqrt 5+1}{2}$ là nghiệm duy nhất của pt

Trả lời 05-06-16 10:49 AM

tran85295
1

10M 559K

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