Mn ơi giải cho mình bài 16 câu 3,4,6 với ạ October 18, 2020 by Nick Mn ơi giải cho mình bài 16 câu 3,4,6 với ạ
Đáp án: 6) 2 Giải thích các bước giải: \(\begin{array}{l}3)\dfrac{{\left( {9\sqrt 7 – 19} \right)\left( {3 + \sqrt 7 } \right)}}{{9 – 7}} + \dfrac{{\left( {\sqrt 7 + 1} \right)\left( {\sqrt 7 – 3} \right)}}{{7 – 9}} – \dfrac{{\left( {33 – 9\sqrt 7 } \right)\left( {\sqrt 7 + 2} \right)}}{{7 – 4}}\\ = \dfrac{{27\sqrt 7 + 63 – 57 – 19\sqrt 7 }}{2} + \dfrac{{7 – 3\sqrt 7 + \sqrt 7 – 3}}{{ – 2}} – \dfrac{{33\sqrt 7 + 66 – 63 – 18\sqrt 7 }}{3}\\ = \dfrac{{8\sqrt 7 + 6}}{2} – \dfrac{{4 – 2\sqrt 7 }}{2} – \dfrac{{15\sqrt 7 + 3}}{3}\\ = 4\sqrt 7 + 3 – 2 + \sqrt 7 – 5\sqrt 7 – 1 = 0\\4)\dfrac{{\left( {5 + \sqrt 5 } \right)\left( {\sqrt 5 – 2} \right)}}{{5 – 4}} + \dfrac{{\sqrt 5 \left( {\sqrt 5 + 1} \right)}}{{5 – 1}} – \dfrac{{3\sqrt 5 \left( {3 – \sqrt 5 } \right)}}{{9 – 5}}\\ = \dfrac{{5\sqrt 5 – 10 + 5 – 2\sqrt 5 }}{1} + \dfrac{{5 + \sqrt 5 }}{4} – \dfrac{{9\sqrt 5 – 15}}{4}\\ = \dfrac{{4\left( {3\sqrt 5 – 5} \right) + 5 + \sqrt 5 – 9\sqrt 5 + 15}}{4}\\ = \dfrac{{12\sqrt 5 – 20 + 5 + \sqrt 5 – 9\sqrt 5 + 15}}{4}\\ = \dfrac{{4\sqrt 5 }}{4} = \sqrt 5 \\6)\dfrac{1}{{\sqrt {2 – 2\sqrt 2 .1 + 1} }} – \dfrac{1}{{\sqrt {2 + 2\sqrt 2 .1 + 1} }}\\ = \dfrac{1}{{\sqrt {{{\left( {\sqrt 2 – 1} \right)}^2}} }} – \dfrac{1}{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} }}\\ = \dfrac{1}{{\sqrt 2 – 1}} – \dfrac{1}{{\sqrt 2 + 1}}\\ = \dfrac{{\sqrt 2 + 1 – \sqrt 2 + 1}}{{2 – 1}} = 2\end{array}\) Reply
Đáp án:
6) 2
Giải thích các bước giải:
\(\begin{array}{l}
3)\dfrac{{\left( {9\sqrt 7 – 19} \right)\left( {3 + \sqrt 7 } \right)}}{{9 – 7}} + \dfrac{{\left( {\sqrt 7 + 1} \right)\left( {\sqrt 7 – 3} \right)}}{{7 – 9}} – \dfrac{{\left( {33 – 9\sqrt 7 } \right)\left( {\sqrt 7 + 2} \right)}}{{7 – 4}}\\
= \dfrac{{27\sqrt 7 + 63 – 57 – 19\sqrt 7 }}{2} + \dfrac{{7 – 3\sqrt 7 + \sqrt 7 – 3}}{{ – 2}} – \dfrac{{33\sqrt 7 + 66 – 63 – 18\sqrt 7 }}{3}\\
= \dfrac{{8\sqrt 7 + 6}}{2} – \dfrac{{4 – 2\sqrt 7 }}{2} – \dfrac{{15\sqrt 7 + 3}}{3}\\
= 4\sqrt 7 + 3 – 2 + \sqrt 7 – 5\sqrt 7 – 1 = 0\\
4)\dfrac{{\left( {5 + \sqrt 5 } \right)\left( {\sqrt 5 – 2} \right)}}{{5 – 4}} + \dfrac{{\sqrt 5 \left( {\sqrt 5 + 1} \right)}}{{5 – 1}} – \dfrac{{3\sqrt 5 \left( {3 – \sqrt 5 } \right)}}{{9 – 5}}\\
= \dfrac{{5\sqrt 5 – 10 + 5 – 2\sqrt 5 }}{1} + \dfrac{{5 + \sqrt 5 }}{4} – \dfrac{{9\sqrt 5 – 15}}{4}\\
= \dfrac{{4\left( {3\sqrt 5 – 5} \right) + 5 + \sqrt 5 – 9\sqrt 5 + 15}}{4}\\
= \dfrac{{12\sqrt 5 – 20 + 5 + \sqrt 5 – 9\sqrt 5 + 15}}{4}\\
= \dfrac{{4\sqrt 5 }}{4} = \sqrt 5 \\
6)\dfrac{1}{{\sqrt {2 – 2\sqrt 2 .1 + 1} }} – \dfrac{1}{{\sqrt {2 + 2\sqrt 2 .1 + 1} }}\\
= \dfrac{1}{{\sqrt {{{\left( {\sqrt 2 – 1} \right)}^2}} }} – \dfrac{1}{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} }}\\
= \dfrac{1}{{\sqrt 2 – 1}} – \dfrac{1}{{\sqrt 2 + 1}}\\
= \dfrac{{\sqrt 2 + 1 – \sqrt 2 + 1}}{{2 – 1}} = 2
\end{array}\)