\(\dfrac{50}{\sqrt{2}}\)=?
K
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22 tháng 9 2021
\(6\sqrt{\dfrac{1}{2}}+\dfrac{2}{\sqrt{2}}-\sqrt{50}=3\sqrt{2}+\sqrt{2}-5\sqrt{2}=-\sqrt{2}\\ \dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}=\dfrac{\left(2+\sqrt{3}\right)^2-\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\\ =\dfrac{7+4\sqrt{3}-\left(7-4\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\dfrac{8\sqrt{3}}{1}=8\sqrt{3}\)
29 tháng 8 2021
Ta có: \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\cdot\sqrt{4.5}+\dfrac{2}{5}\sqrt{50}\right):\dfrac{4}{15}\sqrt{\dfrac{1}{8}}\)
\(=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{9}{4}\sqrt{2}+2\sqrt{2}\right):\dfrac{\sqrt{2}}{15}\)
\(=0\)
29 tháng 6 2021
`c)1/(2sqrt2)-3/2sqrt{4,5}+2/5sqrt{50}`
`=1/(2sqrt2)-3/2sqrt{9/2}+2/5sqrt{25.2}`
`=1/(2sqrt2)-9/(2sqrt2)+2sqrt2`
`=2sqrt2-8/(2sqrt2)`
`=2sqrt2-sqrt2=sqrt2`
`d)4/(3+sqrt5)-8/(1+sqrt5)+15/sqrt5`
`=(4(3-sqrt5))/(9-5)-(8(sqrt5-1))/(5-1)+3sqrt5`
`=3-sqrt5-2(sqrt5-1)+3sqrt5`
`=3+3sqrt5-3sqrt5+2=5`
27 tháng 7 2021
1) \(\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}\)
2) \(\dfrac{\sqrt{12.5}}{0.5}=\sqrt{\dfrac{12.5}{0.25}}=5\sqrt{2}\)
3) \(\sqrt{\dfrac{25}{64}}=\dfrac{5}{8}\)
4) \(\dfrac{\sqrt{230}}{\sqrt{2.3}}=\sqrt{\dfrac{230}{2.3}}=\sqrt{100}=10\)
5) \(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=0\cdot\sqrt{6}=0\)
H9
HT.Phong (9A5)
CTVHS
14 tháng 8 2023
\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}+\dfrac{\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}}{2}+\dfrac{\sqrt{5}}{2}\)
\(=\dfrac{3\sqrt{5}}{2}\)
H9
HT.Phong (9A5)
CTVHS
14 tháng 8 2023
\(\left(\sqrt{3}+1\right)\cdot\dfrac{\sqrt{3}-3}{2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\cdot\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\cdot\dfrac{1-\sqrt{3}}{2}\)
\(=\dfrac{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}{2}\)
\(=\dfrac{1-3}{2}\)
\(=-1\)
HM
1
Y
23 tháng 6 2023
\(\left(\sqrt{\dfrac{8}{2}}\sqrt{6}+\sqrt{\dfrac{50}{3}}\right)\sqrt{6}\)
\(=\left(\dfrac{\sqrt{8}}{\sqrt{2}}.\sqrt{2}.\sqrt{3}+\dfrac{\sqrt{50}}{\sqrt{3}}\right)\sqrt{6}\)
\(=\left(2\sqrt{2}.\sqrt{3}+\dfrac{5\sqrt{2}}{\sqrt{3}}\right)\sqrt{6}\)
\(=\dfrac{2\sqrt{6}.\sqrt{3}+5\sqrt{2}}{\sqrt{3}}.\sqrt{6}\)
\(=\dfrac{6\sqrt{2}+5\sqrt{2}}{\sqrt{3}}.\sqrt{3}.\sqrt{2}\)
\(=11\sqrt{2}.\sqrt{2}\)
\(=11.2=22\)
16 tháng 9 2021
\(=\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)+2\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\\ =\dfrac{x\sqrt{x}+2x+2x-50+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\\ =\dfrac{x\sqrt{x}-5\sqrt{x}+4x}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}\left(x+4\sqrt{x}-5\right)}{2\sqrt{x}\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-1}{2}\)
16 tháng 9 2021
\(\dfrac{x+2\sqrt{x}}{2\sqrt{x}+10}+\dfrac{\sqrt{x}-5}{\sqrt{x}}+\dfrac{50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\left(đk:x>0\right)\)
\(=\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)+2\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\)
\(=\dfrac{x\sqrt{x}+2x+2x-50+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{x\sqrt{x}+4x-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-1}{2}\)
NT
1
2 tháng 8 2018
Vời mọi k > 0 , ta có:
\(\dfrac{1}{\sqrt{k}}=\dfrac{2}{\sqrt{k}+\sqrt{k}}< \dfrac{2}{\sqrt{k}+\sqrt{k-1}}=2\left(\sqrt{k}-\sqrt{k-1}\right)\)
\(A=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{50}}\)
\(\Rightarrow A< 2\left[\left(\sqrt{50}-\sqrt{49}\right)+...+\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{1}-0\right)\right]\)
\(\Rightarrow A< 2\left(\sqrt{50}-0\right)=2\sqrt{50}=10\sqrt{2}\)(đpcm)
5 tháng 10 2021
a: Ta có: \(\left(4\sqrt{2}-\dfrac{11}{2}\sqrt{8}-\dfrac{1}{3}\sqrt{288}+\sqrt{50}\right)\cdot\left(\dfrac{1}{2}\sqrt{2}\right)\)
\(=\dfrac{1}{2}\sqrt{2}\cdot\left(4\sqrt{2}-11\sqrt{2}-4\sqrt{2}+5\sqrt{2}\right)\)
\(=\dfrac{1}{2}\sqrt{2}\cdot6\sqrt{2}=3\)
đây mà là toán lớp 1:V
\(\dfrac{50}{\sqrt{2}}=25\sqrt{2}\)