\(\sqrt{x^2-6x+9}+x=11\); (ĐKXĐ\(\forall x\in R\))

<=> \(\sqrt{x^2-6x+9}=11-x\) 

<=> \(\sqrt{\left(x-3\right)^2}=11-x\)

<=> \(|x-3|=11-x\)

<=> \(\left[{}\begin{matrix}x-3=11-x\\x-3=-11+x\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}2x=14\\0x=-8\left(vô\right)lí\left(\right)\end{matrix}\right.\)

<=> x=7 (thỏa mãn ĐKXĐ)

Vậy phương trình có nghiệm là x=7

\(\sqrt{2-\sqrt{3}}\)

= \(\dfrac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}}\)

= \(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)

= \(\dfrac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}\)

= \(\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\)

= \(\dfrac{|\sqrt{3}-1|}{\sqrt{2}}\)

= \(\dfrac{\sqrt{3}-1}{\sqrt{2}}\)

= \(\dfrac{\sqrt{2}.\left(\sqrt{3}-1\right)}{2}\)

= \(\dfrac{\sqrt{6}-\sqrt{2}}{2}\)

 \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10.|2+\sqrt{3}|}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10.\left(2+\sqrt{3}\right)}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5.|5-\sqrt{3}|}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+5.\left(5-\sqrt{3}\right)}}\)

= \(\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)

= \(\sqrt{4+\sqrt{25}}\)

= \(\sqrt{4+5}\)

= \(\sqrt{9}\)

= \(3\)

Gọi tập hợp các số tự nhiên nhỏ hơn hoặc bằng 8 là Q

=> Q{0;1;2;3;4;5;6;7;8}

a, \(\dfrac{7}{12}=\dfrac{6+1}{12}=\dfrac{6}{12}+\dfrac{1}{12}=\dfrac{6}{6\times2}+\dfrac{1}{12}=\dfrac{1}{2}+\dfrac{1}{12}\)

b, \(\dfrac{13}{27}=\dfrac{12+1}{27}=\dfrac{12}{27}+\dfrac{1}{27}=\dfrac{3\times4}{3\times9}+\dfrac{1}{27}=\dfrac{4}{9}+\dfrac{1}{27}\)

đây là đáp án của mình ạ.