\(\sqrt{28+10\sqrt{3}}\)=??
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12 tháng 8 2020
1)
BT = \(\sqrt{\left(5+\sqrt{3}\right)^2}-\sqrt{\left(5-\sqrt{3}\right)^2}\)
= \(\left(5+\sqrt{3}\right)-\left(5-\sqrt{3}\right)\)
= \(2\sqrt{3}\)
2:
BT = \(\frac{\sqrt{2}.\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\frac{5\left(\sqrt{6}-1\right)}{\left(1+\sqrt{6}\right)\left(\sqrt{6}-1\right)}\)
= \(\sqrt{6}-\frac{5\left(\sqrt{6}-1\right)}{5}\) = \(\sqrt{6}-\left(\sqrt{6}-1\right)\) =1
L
1
4 tháng 9 2023
\(\sqrt{28-10\sqrt{3}}\\ =\sqrt{3-10\sqrt{3}+25}\\ =\sqrt{\left(\sqrt{3}-5\right)^2}\\ =\left|\sqrt{3}-5\right|\\ =5-\sqrt{3}\)
\(\sqrt{41+12\sqrt{5}}\\ =\sqrt{5+12\sqrt{5}+36}\\ =\sqrt{\left(\sqrt{5}+6\right)}\\ =\left|\sqrt{5}+6\right|\\ =\sqrt{5}+6\)
\(\sqrt{32-10\sqrt{7}}\\ =\sqrt{7-10\sqrt{7}+25}\\ =\sqrt{\left(\sqrt{7}-5\right)^2}\\ =\left|\sqrt{7}-5\right|\\ =5-\sqrt{7}\)
\(\sqrt{11-4\sqrt{7}}\\ =\sqrt{7-4\sqrt{7}+4}\\ =\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|\\ =\sqrt{7}-2\)
LD
1
HN
19 tháng 10 2021
Ta có:
\(\sqrt[3]{7}< \sqrt[3]{8}=2\) và \(\sqrt{15}< \sqrt{16}=4\), suy ra \(\sqrt[3]{7}+\sqrt{15}< 6\).
\(\sqrt{10}>\sqrt{9}=3\) và \(\sqrt[3]{28}>\sqrt[3]{27}=3\), suy ra \(\sqrt{10}+\sqrt[3]{28}>6\).
Vậy \(\sqrt[3]{7}+\sqrt{15}< \sqrt{10}+\sqrt[3]{28}\).
1 tháng 10 2021
a: Ta có: \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}-\sqrt{20}\)
\(=4\sqrt{7}+6\sqrt{7}-15\sqrt{7}+4\sqrt{7}-2\sqrt{5}\)
\(=-\sqrt{7}-2\sqrt{5}\)
GH
22 tháng 6 2023
a)
\(7\sqrt{2}=\sqrt{49.2}=\sqrt{98}\\ 2\sqrt{8}=\sqrt{4.8}=\sqrt{32}\\ 5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)
Do 98 > 50 > 32 > 28 nên \(\sqrt{98}>\sqrt{50}>\sqrt{32}>\sqrt{28}\)
=> \(7\sqrt{2}>5\sqrt{2}>2\sqrt{8}>\sqrt{28}\)
b)
\(3\sqrt{10}=\sqrt{9.10}=\sqrt{90}\\ 5\sqrt{3}=\sqrt{25.3}=\sqrt{75}\)
\(\dfrac{20}{\sqrt{5}}=\dfrac{20\sqrt{5}}{5}=4\sqrt{5}=\sqrt{16.5}=\sqrt{80}\)
\(12\sqrt{\dfrac{2}{3}}=\sqrt{144.\dfrac{2}{3}}=\sqrt{96}\)
Do 96 > 90 > 80 > 75 => \(\sqrt{96}>\sqrt{90}>\sqrt{80}>\sqrt{75}\)
=> \(12\sqrt{\dfrac{2}{3}}>3\sqrt{10}>\dfrac{20}{\sqrt{5}}>5\sqrt{3}\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 7 2023
\(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6-\sqrt{6}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}-1}+\dfrac{\sqrt{6}\cdot\sqrt{6}-\sqrt{6}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}}\)
\(=\dfrac{\sqrt{6}}{1}+\dfrac{\sqrt{6}-1}{1}\)
\(=\sqrt{6}+\sqrt{6}-1\)
\(=2\sqrt{6}-1\)
=======================
\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\cdot\sqrt{3}+\sqrt{6}\cdot\sqrt{2}}\)
\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)-3\left(\sqrt{2}-\sqrt{3}\right)}{-\sqrt{6}}\)
\(=\dfrac{2\sqrt{3}+3\sqrt{2}-3\sqrt{2}+3\sqrt{3}}{-\sqrt{6}}\)
\(=\dfrac{5\sqrt{3}}{-\sqrt{6}}=-\dfrac{5}{\sqrt{2}}\)
17 tháng 7 2019
\(\left(\frac{15}{3-\sqrt{3}}-\frac{2}{1-\sqrt{3}}+\frac{3}{\sqrt{3}-2}\right):\sqrt{28+10\sqrt{3}}\)
\(=\left(\frac{15}{3-\sqrt{3}}-\frac{2}{1-\sqrt{3}}+\frac{3}{\sqrt{3}-2}\right):\sqrt{3}+5\)
\(=\left(\frac{5\sqrt{3}}{\sqrt{3}-1}-\frac{2}{1-\sqrt{3}}+\frac{3}{\sqrt{3}-2}\right):\sqrt{3}+5\)
\(=-\frac{5\sqrt{3}-8}{5-3\sqrt{3}}:\sqrt{3}+5\)
\(=-\frac{5\sqrt{3}-8}{\left(5-3\sqrt{3}\right)\left(5+3\sqrt{3}\right)}\)
\(=-\frac{5\sqrt{3}-8}{16-10\sqrt{3}}\)
= -(-1/2) = 1/2
17 tháng 12 2023
a: \(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)
\(=2\sqrt{4\cdot2\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\cdot2\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-\sqrt{2\sqrt{3}}-3\sqrt{2\sqrt{3}}\)
=0
b: \(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|2-\sqrt{3}\right|\)
\(=\sqrt{3}+2-\sqrt{3}\)
=2
c: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)
\(=\left|\sqrt{7}-4\right|-2\sqrt{7}+3\sqrt{7}\)
\(=4-\sqrt{7}+\sqrt{7}\)
=4
d: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(=\dfrac{\sqrt{10}\left(15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\right)}{\sqrt{10}}\)
\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=15\sqrt{5}+5\cdot2\sqrt{5}-3\cdot3\sqrt{5}\)
\(=16\sqrt{5}\)
e: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)
\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)
\(=-16\sqrt{3}\)
DD
0
1231+23456=
\(\sqrt{28+10\sqrt{3}}\)=\(\sqrt{\left(\sqrt{3}\right)^2+2.5.\sqrt{3}+\left(\sqrt{25}\right)^2}\)
=\(\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{25}.\sqrt{3}+\left(\sqrt{25}\right)^2}\)
=\(\sqrt{\left(\sqrt{3}+\sqrt{25}\right)^2}\)
=\(\sqrt{\left(\sqrt{3}+5\right)^2}\)
=\(|\sqrt{3}+5|\)
=\(\sqrt{3}+5\)(vì \(\sqrt{3}+5\)\(\ge0\))