I) Hình bạn tự vẽ nha 

Ta có DY//BH ; YH//DB 

=> DYHB hình bình hành => DY = HB 

Tương tự được ZE = FC

mà \(\frac{BH}{BC}=1-\frac{HC}{BC}=1-\frac{1}{\sqrt{2}}\)\(\left(\Delta HIC\approx\Delta BAC;\frac{AB}{IH}=\sqrt{2}\right)\)(1)

Tương tự được \(\frac{FC}{BC}=1-\frac{BF}{BC}=1-\frac{1}{\sqrt{2}}\)(2) 

Từ (1) ; (2) => BH = FC hay DY = ZE 

Ta có x⁵+1=(x+1)(x⁴-x³+x²-x+1)+7

Vậy x⁵+8 chia cho x+1 dư 7

Khá phổ biến!

\(\sqrt{1+2016^2+\dfrac{2016^2}{2017^2}}+\dfrac{2016}{2017}=\sqrt{\left(2016+1\right)^2-2.2016+\dfrac{2016^2}{2017^2}}+\dfrac{2016}{2017}\) \(=\sqrt{2017^2-2.2016+\dfrac{2016^2}{2017^2}}+\dfrac{2016}{2017}=\sqrt{\left(2017-\dfrac{2016}{2017}\right)^2}+\dfrac{2016}{2017}\)

\(=2017-\dfrac{2016}{2017}+\dfrac{2016}{2017}=2017\)

@Nguyễn Thị Thu Sương :

\(\frac{\sqrt{3+\sqrt{15}}}{\sqrt{2}}=\sqrt{\frac{3+\sqrt{15}}{2}}\)

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

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

\(=\sqrt{\frac{\sqrt{3}}{\sqrt{5}-\sqrt{3}}}\)

a) \(\left(\sqrt{12}-\sqrt{27}+\sqrt{3}\right):\sqrt{3}\)

\(=\left(2\sqrt{3}-3\sqrt{3}+\sqrt{3}\right):\sqrt{3}\)

\(=\sqrt{3}\left(2-3+1\right):\sqrt{3}\)

\(=0:\sqrt{3}=0\)

b) \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)

\(=\frac{5\sqrt{3}}{\sqrt{15}}+\frac{3\sqrt{5}}{\sqrt{15}}\)

\(=\frac{5\sqrt{3}}{\sqrt{3}\cdot\sqrt{5}}+\frac{3\sqrt{5}}{\sqrt{3}\cdot\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{3}\)

\(\sqrt{50}-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)

\(=5\sqrt{2}-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{12}\)

\(=-15\sqrt{2}\)

\(\dfrac{10\sqrt{6}-12}{\sqrt{6}-5}-3\sqrt{\dfrac{2}{3}}+\dfrac{15}{\sqrt{6}-1}\)
\(=\dfrac{2\sqrt{6}\left(5-\sqrt{6}\right)}{\sqrt{6}-5}-3.\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{15\left(\sqrt{6}+1\right)}{\left(\sqrt{6}-1\right)\left(\sqrt{6}+1\right)}\)
\(=-2\sqrt{6}-\sqrt{3}.\sqrt{2}+\dfrac{15\left(\sqrt{6}+1\right)}{6-1}\)
\(=-2\sqrt{6}-\sqrt{6}+3\left(\sqrt{6}+1\right)\)
\(=3\).

em cảm ơn

c) \(\sqrt{6-4\sqrt{2}}+\sqrt{19-6\sqrt{2}}\) = \(\dfrac{\sqrt{12-8\sqrt{2}}}{\sqrt{2}}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)

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

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

d )Đặt A = \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)

\(\Leftrightarrow A^2=\left(\sqrt{12-3\sqrt{7}}\right)^2-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}+\left(\sqrt{12+3\sqrt{7}}\right)^2\)

\(\Leftrightarrow A^2=12-3\sqrt{7}-2\sqrt{144-63}+12+3\sqrt{7}\)

\(\Leftrightarrow A^2=24-2\sqrt{81}\)

\(\Leftrightarrow A^2=24-18=6\)

=> A = \(\sqrt{6}\)

Vậy \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}=\sqrt{6}\)