K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
12 tháng 9 2021
\(\left(\dfrac{2}{\sqrt{6}-1}+\dfrac{3}{\sqrt{6}-2}-\dfrac{3}{3-\sqrt{6}}\right)\cdot\dfrac{5}{9\sqrt{6}+4}\)
\(=\left(\dfrac{2+2\sqrt{6}}{5}+\dfrac{6+3\sqrt{6}}{2}-3-\sqrt{6}\right)\cdot\dfrac{5}{9\sqrt{6}+4}\)
\(=\dfrac{4+4\sqrt{6}+30+15\sqrt{6}-30-10\sqrt{6}}{10}\cdot\dfrac{5}{9\sqrt{6}+4}\)
\(=\dfrac{1}{2}\)
AH
Akai Haruma
Giáo viên
26 tháng 8 2023
Lời giải:
a.
\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)
$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$
b.
$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$
$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$
$=|\sqrt{3}-3|+|\sqrt{3}+3|$
$=(3-\sqrt{3})+(\sqrt{3}+3)=6$
c.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
H9
HT.Phong (9A5)
CTVHS
14 tháng 7 2023
\(\dfrac{15}{\sqrt{6}-1}+\dfrac{8}{\sqrt{6}+2}+\dfrac{6}{3-\sqrt{6}}-9\sqrt{6}\)
\(=\dfrac{15\left(\sqrt{6}+1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{8\left(\sqrt{6}-2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}+\dfrac{6\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}-9\sqrt{6}\)
\(=\dfrac{15\left(\sqrt{6}+1\right)}{6-1}+\dfrac{8\left(\sqrt{6}-2\right)}{6-4}+\dfrac{6\left(3+\sqrt{6}\right)}{9-6}-9\sqrt{6}\)
\(=3\left(\sqrt{6}+1\right)+4\left(\sqrt{6}-2\right)+2\left(3+\sqrt{6}\right)-9\sqrt{6}\)
\(=3\sqrt{6}+3+4\sqrt{6}-8+6+2\sqrt{6}-9\sqrt{6}\)
\(=9\sqrt{6}+1-9\sqrt{6}\)
\(=1\)
H9
HT.Phong (9A5)
CTVHS
14 tháng 7 2023
\(\sqrt{\left(\sqrt{5}-1\right)\sqrt{14-6\sqrt{5}}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{9-6\sqrt{5}+5}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{3^2-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\sqrt{\left(3-\sqrt{5}\right)^2}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\left|3-\sqrt{5}\right|}\)
\(=\sqrt{\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}\)
\(=\sqrt{3\sqrt{5}-5-3+\sqrt{5}}\)
\(=\sqrt{4\sqrt{5}-8}\)
\(=\sqrt{4\left(\sqrt{5}-2\right)}\)
\(=2\sqrt{\sqrt{5}-2}\)
20 tháng 5 2022
Câu 1:
\(A=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
Câu 2:
\(\Leftrightarrow\left|2x-3\right|=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=2\sqrt{3}\\2x-3=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{3}+3}{2}\\x=\dfrac{-2\sqrt{3}+3}{2}\end{matrix}\right.\)
1
1
16 tháng 6 2021
\(D=\sqrt{9+6\sqrt{2}}-\sqrt{9-6\sqrt{2}}-\sqrt{21-12\sqrt{3}}\)
\(D=\sqrt{9+2.\sqrt{3}.\sqrt{3}.\sqrt{2}}-\sqrt{9-2\sqrt{3}.\sqrt{3}.\sqrt{2}}-\sqrt{21-2.2\sqrt{3}.3}\)
\(D=\sqrt{\left(\sqrt{6}\right)^2+2\sqrt{6}.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{6}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(-\sqrt{3^2-2.3.2\sqrt{3}+\left(2\sqrt{3}\right)^2}\)
\(D=\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}-\sqrt{\left(3-2\sqrt{3}\right)^2}\)
\(D=\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}-2\sqrt{3}+3=3\)
22 tháng 5 2018
a ) \(\sqrt{6+\sqrt{35}}.\sqrt{6-\sqrt{35}}=1\)
\(\Leftrightarrow VT=\sqrt{\left(6+\sqrt{35}\right)\left(6-\sqrt{35}\right)}\)
\(\Leftrightarrow VT=\sqrt{6^2-35}=\sqrt{1}=1=VP\)
b ) \(VT=\left(\sqrt{2}-1\right)^2=2+1-2\sqrt{2}=3-2\sqrt{2}\)
\(VP=\sqrt{9}-\sqrt{8}=3-2\sqrt{2}\)
=> \(VT=VP.\)
BK
1
PT
24 tháng 9 2017
\(B=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+....+\sqrt{6}}}}}\)
\(\Rightarrow B^2=6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+....+\sqrt{6}}}}}\)
\(\Rightarrow B^2=6+B\)
\(\Rightarrow B^2-B-6=0\)
\(\Rightarrow B^2-2B+3A-6=0\)
\(\Rightarrow B\left(B-2\right)+3\left(B-2\right)=0\)
\(\Rightarrow\left(B+3\right)\left(B-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}B+3=0\\B-2=0\end{matrix}\right.\)
\(\Rightarrow B=2\left(B\in Z\right)\)
\(\Rightarrow\) Đpcm.
\(\sqrt{6\sqrt{6+\sqrt{6}}}< \sqrt{6\sqrt{6+\sqrt{9}}}\)
\(=\sqrt{6\sqrt{6+3}}=\sqrt{6\sqrt{9}}\)
\(=\sqrt{6.3}< \sqrt{81}=9\)
Vậy \(\sqrt{6\sqrt{6+\sqrt{6}}}< 9\)