NT
0
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
24 tháng 9 2021
a: Ta có: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)
\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)
\(=0\)
b: Ta có: \(\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}\)
\(=5+7-1\)
=11
H9
HT.Phong (9A5)
CTVHS
30 tháng 10 2023
\(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\dfrac{5}{\sqrt{5}}+\dfrac{1}{3}\cdot3\sqrt{5}+\left|2-\sqrt{5}\right|\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-2\)
\(=3\sqrt{5}-2\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 9 2023
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
19 tháng 10 2021
\(a,=\sqrt{5}\left(2\sqrt{5}-3\right)+3\sqrt{5}=10-3\sqrt{5}+3\sqrt{5}=10\\ b,=5-\sqrt{3}-\left(2-\sqrt{3}\right)=3\\ c,=\dfrac{2\left(\sqrt{5}-1\right)}{4}-\dfrac{2\left(3+\sqrt{5}\right)}{4}=\dfrac{2\sqrt{5}-2-6-2\sqrt{5}}{4}=\dfrac{-8}{4}=-2\)
T
thực hiện phép tính:\(\sqrt{\left(5-\sqrt{24}^{ }\right)^2}\)- \(\sqrt{\left(5+\sqrt{24}\right)^2}\)
2
18 tháng 5 2021
\(\sqrt{\left(5-\sqrt{24}\right)^2}-\sqrt{\left(5+\sqrt{24}\right)^2}\\ =\left|5-\sqrt{24}\right|-\left|5+\sqrt{24}\right|\\ =5-\sqrt{24}-5-\sqrt{24}\\ =-2\sqrt{24}=-4\sqrt{6}\)
TA
18 tháng 5 2021
`\sqrt((5-\sqrt24)^2) - \sqrt((5+\sqrt24)^2)`
`=|5-\sqrt24|-|5+\sqrt24|`
`=5-\sqrt24-5-\sqrt24`
`=-2\sqrt24`
`=-4\sqrt6`
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
21 tháng 9 2023
\(a,\left(2\sqrt{3}-\sqrt{2}\right)^2+2\sqrt{24}=\left[\left(2\sqrt{3}\right)^2-2.2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2\right]+2\sqrt{24}\\ =\left[12-4\sqrt{6}+2\right]+2\sqrt{24}=14-4\sqrt{6}+4\sqrt{6}=14\\ b,\left(3\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+2\sqrt{3}\right)-\sqrt{60}=3\sqrt{5}.\sqrt{5}-2\sqrt{3}.\sqrt{3}+3\sqrt{5}.2\sqrt{3}-\sqrt{3}.\sqrt{5}-\sqrt{60}\\ =15-6+6\sqrt{15}-\sqrt{15}-\sqrt{2^2.15}\\ =9+3\sqrt{15}\)
T
21 tháng 9 2023
\(a,\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(1+\sqrt{5}\right)^2}\)
\(=\left|2-\sqrt{5}\right|-\left|1+\sqrt{5}\right|\)
\(=\sqrt{5}-2-\left(1+\sqrt{5}\right)\)
\(=\sqrt{5}-2-1-\sqrt{5}\)
\(=-3\)
\(b,\dfrac{3-5\sqrt{3}}{\sqrt{3}-5}+6\sqrt{\dfrac{4}{3}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}-5\right)}{\sqrt{3}-5}+6\cdot\dfrac{\sqrt{4}}{\sqrt{3}}\)
\(=\sqrt{3}+\dfrac{12}{\sqrt{3}}\)
\(=\sqrt{3}+\dfrac{12\sqrt{3}}{3}\)
\(=\sqrt{3}+4\sqrt{3}\)
\(=5\sqrt{3}\)
#\(Toru\)
H
21 tháng 9 2023
\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(1+\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|-\left|1+\sqrt{5}\right|\\ =\sqrt{5}-2-1-\sqrt{5}\\ =-2-1\\ =-3\)
\(\dfrac{3-5\sqrt{3}}{\sqrt{3}-5}+6\sqrt{\dfrac{4}{3}}\\ =\dfrac{\sqrt{3}\left(\sqrt{3}-5\right)}{\sqrt{3}-5}+4\sqrt{3}\\ =\sqrt{3}+4\sqrt{3}\\ =5\sqrt{3}\)
24 tháng 11 2021
\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)
