Đáp án: $x=\dfrac{1+\sqrt{5}}{2}$

Giải thích các bước giải:

a) Mình chịu nhé

$b)x=\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}\\\Leftrightarrow x^{2}=x-\dfrac{1}{x}+2\sqrt{\left ( x-\dfrac{1}{x} \right )\left ( 1-\dfrac{1}{x} \right )}+1-\dfrac{1}{x}\\\Leftrightarrow x^{2}=x-\dfrac{1}{x}+2\sqrt{\dfrac{x^{2}-1}{x}.\dfrac{x-1}{x}}+1-\dfrac{1}{x}\\\Leftrightarrow x^{2}=x-\dfrac{1}{x}+2\sqrt{\dfrac{(x^{2}-1)(x-1)}{x^{2}}}+1-\dfrac{1}{x}\\\Leftrightarrow x^{2}=x-\dfrac{1}{x}+2\sqrt{\dfrac{x^{3}-x^{2}-x+1}{x^{2}}}+1-\dfrac{1}{x}\\\Leftrightarrow x^{2}=x-\dfrac{1}{x}+2.\dfrac{\sqrt{x^{3}-x^{2}-x+1}}{|x|}+1-\dfrac{1}{x}\\\Leftrightarrow x^{2}=x-\dfrac{1}{x}+\dfrac{2\sqrt{x^{3}-x^{2}-x+1}}{|x|}+1-\dfrac{1}{x}\\\Leftrightarrow x^{2}=x-\dfrac{2}{x}+\dfrac{2\sqrt{x^{3}-x^{2}-x+1}}{|x|}+1\\\Leftrightarrow x^{2}-x+\dfrac{2}{x}-\frac{2\sqrt{x^{3}-x^{2}-x+1}}{|x|}=1\\\Leftrightarrow x^{2}-x+\dfrac{2}{x}-\dfrac{2\sqrt{x^{3}-x^{2}-x+1}}{±x}=1$
Vậy $x=\dfrac{1+\sqrt{5}}{2}$

Đáp án:Tham khảo

Giải thích các bước giải:

A) $\text{Ta có}$:x=$\dfrac{(\sqrt{5}+2)\sqrt[3]{17}-\sqrt{5}-38}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}$ 

=$\dfrac{(\sqrt{5}+2).\sqrt[3]{(\sqrt{5}-2)³}}{\sqrt{5}+\sqrt{(3-\sqrt{5})²}}$ 

$=\dfrac{(\sqrt{5}+2)(\sqrt{5}-2)}{\sqrt{5}+3-\sqrt{5}}$

$=\dfrac{1}{3}$
$\text{Thay x=$\dfrac{1}{3}$ vào biểu thức $B=(3²+8x²-2)^{2018}$ ta được}$

$B=[3.(\dfrac{1}{3})²+8(\dfrac{1}{3})²-2]^{2018}$
$=[3.\dfrac{1}{27}+8.\dfrac{1}{9}-2]^{2018}$
$=(-1)^{2018}$
$=1$

$\text{Vậy$ B=1 $tại x=$\dfrac{(\sqrt{5}+2)\sqrt[3]{17}-\sqrt{5}-38}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}$}$