Tính
(\(\sqrt{20}\)-\(\sqrt{45}\)+\(\sqrt{5}\)):\(\sqrt{5}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
H
18 tháng 8 2023
\(=\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}\\ =\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\\ =\dfrac{2\sqrt{5}}{\sqrt{6}}\\ =\dfrac{\sqrt{2}.\sqrt{2}.\sqrt{5}}{\sqrt{2}.\sqrt{3}}\\ =\dfrac{\sqrt{10}}{\sqrt{3}}=\dfrac{\sqrt{30}}{3}\)
H
18 tháng 8 2023
\(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\\ =\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}\\ =\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\\ =2\sqrt{5}:\sqrt{6}\\ =2\sqrt{5}.\dfrac{1}{\sqrt{6}}\\ =\dfrac{2\sqrt{5}}{\sqrt{6}}\\ =\dfrac{2\sqrt{5}.\sqrt{6}}{6}\\ =\dfrac{1}{3}.\sqrt{30}\\ =\dfrac{\sqrt{30}}{3}\)
10 tháng 7 2021
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
10 tháng 7 2021
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
10 tháng 7 2021
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
10 tháng 7 2021
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
HT
1
6 tháng 11 2021
\(=\sqrt{5}\left(2\sqrt{5}-3\right)+3\sqrt{5}=10-3\sqrt{5}+3\sqrt{5}=10\)
11 tháng 5 2022
a: \(=\left(2\sqrt{3}-12\sqrt{3}+15\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)
b: \(=\left(6\sqrt{2}-16\sqrt{2}+15\sqrt{2}\right):5=\sqrt{2}\)
c: \(=\dfrac{\left(2\sqrt{5}-6\sqrt{5}+15\sqrt{5}\right)}{\sqrt{5}}=17-6=11\)
4 tháng 10 2021
1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)
2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)
3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)
4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)
PD
17 tháng 8 2020
\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}=6\sqrt{5}-6\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)
\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{2}.\sqrt{7}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)
\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}=6\sqrt{3}-12\sqrt{3}+20\sqrt{3}=14\sqrt{3}\)
câu tiếp tương tự câu thứ 2 nha
ID
16 tháng 4 2022
\(M=10-3\sqrt{5}+\sqrt{45}\)
\(M=10-3\sqrt{5}+3\sqrt{5}\)
\(M=10\)
6 tháng 12 2021
\(=5\sqrt{2}-9\sqrt{5}-6\sqrt{2}+10\sqrt{5}=\sqrt{5}-\sqrt{2}\)
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
a) \(\sqrt{200}-\sqrt{32}+\sqrt{72}\)
\(=\sqrt{10^2\cdot2}-\sqrt{4^2\cdot2}+\sqrt{6^2\cdot2}\)
\(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}\)
\(=\left(10-4+6\right)\sqrt{2}\)
\(=12\sqrt{2}\)
b) \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}\)
\(=4\cdot2\sqrt{5}-3\cdot5\sqrt{5}+5\cdot3\sqrt{5}-3\sqrt{5}\)
\(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}\)
\(=\left(8-15+15-3\right)\sqrt{5}\)
\(=5\sqrt{5}\)
c) \(\left(2\sqrt{8}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\sqrt{20}-2\sqrt{2}\right)\)
\(=\left(2\cdot2\sqrt{2}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\cdot2\sqrt{5}-2\sqrt{2}\right)\)
\(=\left(3\sqrt{5}-3\sqrt{2}\right)\left(72-10\sqrt{5}-2\sqrt{2}\right)\)
\(\left(\sqrt{20}-\sqrt{45}+\sqrt{5}\right):\sqrt{5}\)
\(=\sqrt{20}:\sqrt{5}-\sqrt{45}:\sqrt{5}+\sqrt{5}:\sqrt{5}\)
\(=2-3+1\)
\(=0\)