48 - 3 - 4 = ?
K
Khách
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NN
3
8 tháng 3 2017
\(A=\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{47\cdot48\cdot49}+\frac{2}{48\cdot49\cdot50}\)
\(A=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{47\cdot48}-\frac{1}{48\cdot49}+\frac{1}{48\cdot49}-\frac{1}{49\cdot50}\)
\(A=\frac{1}{2\cdot3}-\frac{1}{49\cdot50}\)
\(A=\frac{1}{6}-\frac{1}{2450}\)
\(A=\frac{611}{3675}\)
mong giúp đc bn.thk cho mk
TT
22 tháng 1 2019
a) 13.(23+22)-3.(17+28)
= 13.45 - 3.45
= 45(13-3)
= 45.10 =450
b) -48+48.(-78)+48(-21)
= -48 + 48(-78 - 21)
= -48 + 48.(-99)
= -48 + ( -4752)
= -4800
c) 135.(171-123)-171.(135-123)
= 135.171 - 135.123 - 171.135 + 171.123
= (135.171 - 171.135) - (135.123 - 171.123)
= - 123(135 - 171) = (-123).(-36)= 4428
AY
7 tháng 7 2016
Tính nhanh :
\(a.1+2+3+...+98+99+100\)
\(=\left(100+1\right).100:2\)
\(=101.100:2\)
\(=10100:2\)
\(=5050\)
\(b.16.48+8.48+16.28\)
\(=48.\left(16+8\right)+16.28\)
\(=48.24+16.28\)
\(=1152+448\)
\(=1600\)
AT
12 tháng 7 2021
\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)
\(=\sqrt{12}+1=2\sqrt{3}+1\)
\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}-1\)
\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
2) biến đổi khúc sau như câu 1:
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
AT
12 tháng 7 2021
1) Ta có: \(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{13+\sqrt{4.12}}}=\sqrt{5-\sqrt{13+2\sqrt{12}}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{12}\right)^2+2.\sqrt{12}+1^2}}=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}=\sqrt{5-\left|\sqrt{4.3}+1\right|}\)
\(=\sqrt{5-\left(2\sqrt{3}+1\right)}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=2\sqrt{3+\sqrt{3}-1}=2\sqrt{2+\sqrt{3}}\)
\(=2\sqrt{\dfrac{4+2\sqrt{3}}{2}}=2\sqrt{\dfrac{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}{2}}=2\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}\)
\(=2.\dfrac{\left|\sqrt{3}+1\right|}{\sqrt{2}}=\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{6}+\sqrt{2}\)
2) Ta có: \(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{3}-1\) (như trên)
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
6 
2 = 4
6 
2 = 16
9 
3 = 9
9 
3 = 1
CM
10 tháng 2 2018
a) 48 : 6 : 2 = 4
48 : 6 x 2 = 16
b) 27 : 9 x 3 = 9
27 : 9 : 3 = 1
DB
1
NM
1
MV
23 tháng 5 2017
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\\ A=1-\dfrac{1}{2^{100}}\)
MV
23 tháng 5 2017
\(E=\dfrac{3^2}{2\cdot4}+\dfrac{3^2}{4\cdot6}+...+\dfrac{3^2}{198\cdot200}\\ =3^2\cdot\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{198\cdot200}\right)\\ =9\cdot\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{198\cdot200}\right)\\ =\dfrac{9}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{198}-\dfrac{1}{200}\right)\\ =\dfrac{9}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{200}\right)\\ =\dfrac{9}{2}\cdot\dfrac{99}{200}\\ =\dfrac{891}{400}\)
48-3-4=41
41 nha bạn