\(\sqrt{5^2-4^2}-\sqrt{0,25}\)+ \(\sqrt{\frac{9}{4}}\)
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TM
17 tháng 7 2019
1. a) 3+2=5
b) 0,5-0,1=0,4
c) 4/5-1/9=31/45
d) 2-0,6=1,4
2. a) 8-4+3=7
b) 11+5-3=13
c) 3/2-4/6-7-37/6
d) 4+5-6=3
26 tháng 11 2021
a) \(\sqrt{25-9}\) = \(\sqrt{16}\) = 4
b) \(\sqrt{0,01}-\sqrt{0,25}\) = 0,1 - 0,5 = -0,4
c)\(\sqrt{2.2^2+4^2}+5^2\) = \(\sqrt{2.4+16+25}\) = \(\sqrt{8+16+25}\) = \(\sqrt{49}\) = 7
12 tháng 6 2017
\(\sqrt{\frac{4}{9-4\sqrt{5}}}+\sqrt{\frac{9}{9+4\sqrt{5}}}=\sqrt{\frac{4}{4-4\sqrt{5}+5}}+\sqrt{\frac{9}{4+4\sqrt{5}+5}}\)
\(=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}+\sqrt{\frac{9}{\left(2+\sqrt{5}\right)^2}}=\frac{2}{\sqrt{5}-2}+\frac{3}{2+\sqrt{5}}\)
\(=\frac{2\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\frac{3\left(2-\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}=\frac{2\sqrt{5}+4}{5-4}+\frac{6-3\sqrt{5}}{4-5}\)
\(=2\sqrt{5}+4+3\sqrt{5}-6=5\sqrt{5}-2\)
b) \(\left(5-4\sqrt{3}\right):\frac{2+\sqrt{3}}{2-\sqrt{3}}=\left(5-4\sqrt{3}\right).\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\left(5-4\sqrt{3}\right).\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\left(5-4\sqrt{3}\right).\frac{4-4\sqrt{3}+3}{4-3}\)
\(=\left(5-4\sqrt{3}\right)\left(7-4\sqrt{3}\right)=35-28\sqrt{3}-20\sqrt{3}+48\)
\(=73-48\sqrt{3}\)
Mình chịu câu c nha
28 tháng 2 2022
d: \(D=-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-8\cdot\dfrac{1}{2}:\dfrac{27-14}{12}\)
\(=-4:\dfrac{13}{12}\)
\(=-4\cdot\dfrac{12}{13}=-\dfrac{48}{13}\)
e: \(E=5\cdot4-4\cdot3+5-0.3\cdot20\)
=20-12+5-6
=8+5-6
=13-6=7
f: \(F=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{2}:6\)
\(=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{12}\)
\(=\dfrac{27}{12}+\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{34}{12}=\dfrac{17}{6}\)
2 tháng 10 2018
\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\)
\(\Rightarrow A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{2}}\right)\left(4-\sqrt{10+2\sqrt{2}}\right)}+4-\sqrt{10+2\sqrt{5}}\)
\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{5-2\sqrt{5.1}+1}=8+2\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
\(B=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+...+\frac{1}{\sqrt{2001}+\sqrt{2005}}\)
\(=\frac{1-\sqrt{5}}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+\frac{\sqrt{5}-\sqrt{9}}{\left(\sqrt{5}+\sqrt{9}\right)\left(\sqrt{5}-\sqrt{9}\right)}+...+\frac{\sqrt{2001}-\sqrt{2005}}{\left(\sqrt{2001}+\sqrt{2005}\right)\left(\sqrt{2001}-\sqrt{2005}\right)}\)
\(=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(=-\frac{1}{4}\left(1-\sqrt{5}+\sqrt{5}-\sqrt{9}+....+\sqrt{2001}-\sqrt{2005}\right)\)
\(=-\frac{1}{4}\left(1-\sqrt{2005}\right)\)
\(=10,94430659\)
\(\text{Lm hơi vắn tắt thông cảm nha!!}\)
PN
19 tháng 8 2020
\(a,\frac{2}{\sqrt{2}-1}-\frac{2}{\sqrt{2}+1}=\frac{2\left(\sqrt{2}+1\right)-2\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(=\frac{2\sqrt{2}+2-2\sqrt{2}+2}{\sqrt{2}^2-1^2}=\frac{4}{2-1}=4\)
\(b,\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{4+2.2.\sqrt{2}+2}+\sqrt{4-2.2.\sqrt{2}+2}\)
\(=\sqrt{2^2+2.2.\sqrt{2}+\sqrt{2}^2}+\sqrt{2^2-2.2.\sqrt{2}+\sqrt{2}^2}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=|2+\sqrt{2}|+|2-\sqrt{2}|=2+2=4\)
\(c,\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{4+2.2.\sqrt{5}+5}+\sqrt{4-2.2.\sqrt{5}+5}\)
\(=\sqrt{2^2+2.2.\sqrt{5}+\sqrt{5}^2}+\sqrt{2^2-2.2.\sqrt{5}+\sqrt{5}^2}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|+|2-\sqrt{5}|=2+\sqrt{5}+\sqrt{5}-2=2\sqrt{5}\)
câu d bạn cứ nhân bình thường
O
26 tháng 7 2019
a)\(\sqrt{0,09}\)+2.\(\sqrt{0,25}\)=0,3+2.0,5
=0,3+1
=1,3
b)0,5.\(\sqrt{100}\)-\(\sqrt{\frac{4}{25}}\)=0,5.10-0,4
=5-0,4
=4,6
c)(\(\sqrt{1\frac{9}{16}}\) -\(\sqrt{\frac{9}{16}}\)):5=(1,25-0,75):5
=0,5:5
=0,1
d)3.\(\sqrt{1\frac{17}{64}}\) -2.\(\sqrt{0,0625}\)=1,125-2.0,25
=1,125-0,5
=0,625
\(\sqrt{5^2-4^2}-\sqrt{0,25}+\sqrt{\frac{9}{4}}\)
\(=\sqrt{3^2}-\sqrt{\left(0,5\right)^2}+\sqrt{\left(\frac{3}{2}\right)^2}\)
\(=3-\left(0,5\right)+\frac{3}{2}\)
\(=4\)
\(=\sqrt{25-16}-0,5+\frac{3}{2}\)
\(=\sqrt{9}-0,5+1,5\)
\(=3-0,5+1,5\)
\(=4\)