\(\sqrt[]{\dfrac{1}{125}}.\sqrt[]{\dfrac{32}{35}}:\sqrt[]{\dfrac{56}{225}}\)
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TN
3
BL
11 tháng 7 2017
\(=\sqrt{\dfrac{1.32.225}{125.35.56}}\\ =\sqrt{\dfrac{2^5.25.9}{25.5.35.8.7}}\\ =\sqrt{\dfrac{2^2.9}{5.35.7}}\\ =\sqrt{\dfrac{36}{1225}}\\ =\dfrac{6}{35}\)
DH
11 tháng 7 2017
\(\sqrt{\dfrac{1}{125}}.\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}\)
\(=\sqrt{\dfrac{1}{125}}.\sqrt{\dfrac{32}{35}}.\sqrt{\dfrac{225}{56}}\)
\(=\sqrt{\dfrac{1.32.225}{125.35.56}}=\sqrt{\dfrac{2^5.5^2.3^2}{5^3.5.7.7.2^3}}\)
\(=\sqrt{\dfrac{2^2.3^2}{5^2.7^2}}=\dfrac{6}{35}\)
Chúc bạn học tốt!!!
12 tháng 7 2021
Bài 1:
a) \(\sqrt{72}:\sqrt{8}=\sqrt{72:8}=3\)
b) \(\left(\sqrt{28}-\sqrt{7}+\sqrt{112}\right):\sqrt{7}=5\sqrt{7}:\sqrt{7}=5\)
Bài 2:
a) \(\sqrt{\dfrac{49}{8}}:\sqrt{3\dfrac{1}{8}}=\sqrt{\dfrac{49}{8}:\dfrac{25}{8}}=\sqrt{\dfrac{49}{25}}=\dfrac{7}{5}\)
b) \(\sqrt{54x}:\sqrt{6x}=\sqrt{54x:6x}=\sqrt{9}=3\)
c) \(\sqrt{\dfrac{1}{125}}\cdot\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}\)
\(=\dfrac{\sqrt{5}}{25}\cdot\dfrac{4\sqrt{2}}{\sqrt{35}}:\dfrac{2\sqrt{14}}{15}\)
\(=\dfrac{\sqrt{5}\cdot4\sqrt{2}\cdot15}{25\cdot\sqrt{35}\cdot\sqrt{14}\cdot2}\)
\(=\dfrac{6}{35}\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
\(\dfrac{\sqrt{8}+3}{\sqrt{17-3\sqrt{32}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{17-12\sqrt{2}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{3^2-2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{3-2\sqrt{5}}{\sqrt{3^2-2\cdot3\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\cdot\sqrt{7}+\left(\sqrt{7}\right)^2}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{\left(2\sqrt{2}-3\right)^2}}-\dfrac{3-2\sqrt{5}}{\sqrt{\left(3-2\sqrt{5}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{5}+\sqrt{7}\right)^2}}\)
\(=\dfrac{2\sqrt{2}+3}{2\sqrt{2}-3}+\dfrac{3-2\sqrt{5}}{3-2\sqrt{5}}-\dfrac{1}{\sqrt{5}+\sqrt{7}}\)
\(=\dfrac{\left(2\sqrt{2}+3\right)^2}{\left(2\sqrt{2}+3\right)\left(2\sqrt{2}-3\right)}+1-\dfrac{\sqrt{5}-\sqrt{7}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}\)
\(=17-12\sqrt{2}+1-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)
\(=\dfrac{2\cdot\left(18-12\sqrt{2}\right)}{2}-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)
\(=\dfrac{36-24\sqrt{2}-\sqrt{5}+\sqrt{7}}{2}\)
NN
12 tháng 7 2017
Giải:
Ta có tính chất tổng quát:
\(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\)
\(=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Áp dụng vào biểu thức
\(\Rightarrow A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)
\(=1-\frac{1}{\sqrt{225}}\)
15 tháng 8 2021
đoạn cuối thiếu dấu"+"
\(A=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5}-\sqrt{6}}{5-6}+....+\dfrac{\sqrt{34}-\sqrt{35}}{34-35}+\dfrac{\sqrt{35}-\sqrt{36}}{335-36}\)
\(A=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+....+\sqrt{35}-\sqrt{36}}{-1}=\dfrac{\sqrt{4}-\sqrt{36}}{-1}\)
\(A=\sqrt{36}-\sqrt{4}=6-2=4\)
\(\sqrt{\dfrac{1}{125}}.\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}=\sqrt{\dfrac{1}{125}.\dfrac{32}{35}:\dfrac{56}{225}}=\sqrt{\dfrac{36}{1225}}=\dfrac{6}{35}\)