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Đáp án:

Giải thích các bước giải: ĐKXĐ: $x\ge\dfrac{19}{10}\Rightarrow 1-x<0$

Bất phương trình tương đương với

$x+3-\sqrt{14x-15}>\dfrac{1-\sqrt{10x-19}}{1-x}$

$\Leftrightarrow (x+3-\sqrt{14x-15})(1-x)<1-\sqrt{10x-19}$

$\Leftrightarrow x^2+2x-x\sqrt{14x-15}+\sqrt{14x-15}-\sqrt{10x-19}-2<0$

$\Leftrightarrow x(x+2-\sqrt{14x-15})-((x+2)-\sqrt{14x-15})+(x-\sqrt{10x-19})<0$

$\Leftrightarrow x.\dfrac{(x+2)^2-(14x-15)}{x+2+\sqrt{14x-15}}-\dfrac{(x+2)^2-(14x-15)}{x+2+\sqrt{14x-15}}+\dfrac{x^2-(10x-19)}{x+\sqrt{10x-19}}<0$

$\Leftrightarrow x.\dfrac{(x+2)^2-(14x-15)}{x+2+\sqrt{14x-15}}-\dfrac{(x+2)^2-(14x-15)}{x+2+\sqrt{14x-15}}+\dfrac{x^2-(10x-19)}{x+\sqrt{10x-19}}<0$

$\Leftrightarrow x.\dfrac{x^2-10x+19}{x+2+\sqrt{14x-15}}-\dfrac{x^2-10x+19}{x+2+\sqrt{14x-15}}+\dfrac{x^2-10x+19}{x+\sqrt{10x-19}}<0$

$\Leftrightarrow x^2-10x+19<0$ hoặc $\dfrac{x}{x+2+\sqrt{14x-15}}-\dfrac{1}{x+2+\sqrt{14x-15}}+\dfrac{1}{x+\sqrt{10x-19}}>0(1)$ hoặc $\Leftrightarrow x^2-10x+19>0$ và $\dfrac{x}{x+2+\sqrt{14x-15}}-\dfrac{1}{x+2+\sqrt{14x-15}}+\dfrac{1}{x+\sqrt{10x-19}}<0(1)$

Vì $x>\dfrac{19}{10}\to x+2+\sqrt{14x-15}>x+\sqrt{10x-19}$

nên (1)$\dfrac{x}{x+2+\sqrt{14x-15}}-\dfrac{1}{x+2+\sqrt{14x-15}}+\dfrac{1}{x+\sqrt{10x-19}}>0$

Vậy $x^2-10x+19<0\Leftrightarrow 5-\sqrt{6}<x<5+\sqrt{6}$