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8 tháng 11 2016
Ta có:
\(A=\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+...+\frac{1}{\sqrt{25}+\sqrt{24}}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{25}-\sqrt{24}}{25-24}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+...+\frac{\sqrt{25}-\sqrt{24}}{1}\)
\(=5-1=4\)
10 tháng 10 2020
1) Đặt \(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3D=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3D-D=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(\Leftrightarrow2D=1-\frac{1}{3^{100}}\)
\(\Leftrightarrow D=\frac{3^{100}-1}{2\cdot3^{100}}\)
Vậy \(D=\frac{3^{100}-1}{2\cdot3^{100}}\)
2) Ta có: \(\frac{49}{58}\cdot\frac{2^5}{4^2}-\frac{7^2}{-58}\cdot3\)
\(=\frac{49}{58}\cdot2-\frac{49}{58}\cdot3\)
\(=-1\cdot\frac{49}{58}\)
\(=-\frac{49}{58}\)
1 tháng 1 2020
Bài 1
\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)
\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)
\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)
\(=\frac{9}{25}+\frac{8}{9}-1\)
\(=\frac{56}{225}\)
\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)
\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)
\(=1:\frac{4}{3}=\frac{3}{4}\)
1 tháng 1 2020
Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v
\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)
\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)
\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)
\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)
\(=-\frac{1}{2}\)
SL
9
KN
20 tháng 11 2019
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{24}+\sqrt{25}}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{1}+\sqrt{2}}+\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}}\)
\(+...+\frac{\left(\sqrt{25}-\sqrt{24}\right)\left(\sqrt{25}+\sqrt{24}\right)}{\sqrt{24}+\sqrt{25}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{25}-\sqrt{24}\)
\(=\sqrt{25}-1=5-1=4\)
30 tháng 10 2016
\(\frac{1}{\sqrt{1}\sqrt{2}}+\frac{1}{\sqrt{2}\sqrt{3}}+...+\frac{1}{\sqrt{24}\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}\)
17 tháng 10 2018
C = \(\frac{2}{3}\sqrt{144}-\left(-\frac{3}{4}\right)\div\sqrt{\frac{225}{144}}\)
C = \(\frac{2}{3}.12+\frac{3}{4}\div\frac{5}{4}\)
C = \(8+\frac{3}{5}\)
C = \(8\frac{3}{5}\)
D = \(\frac{4^6.25^5-2^{12}.25^4}{2^{12}.5^8-10^8.64}\)
D = \(\frac{\left(2^2\right)^6.\left(5^2\right)^5-2^{12}.\left(5^2\right)^4}{2^{12}.5^8-\left(2.5\right)^8.2^6}\)
D = \(\frac{2^{12}.5^{10}-2^{12}.5^8}{2^{12}.5^8-2^8.5^8.2^6}\)
D = \(\frac{2^{12}.5^8.\left(25-1\right)}{2^{12}.5^8.\left(1-2^2\right)}\)
D = \(\frac{24}{-3}\)
D = \(-8\)
T
18 tháng 10 2018
\(C=\frac{2}{3}\sqrt{144}-\left(\frac{-3}{4}\right):\sqrt{\frac{225}{144}}\)
\(=\frac{2}{3}\cdot12+\frac{3}{4}:\frac{5}{4}\)
\(=8+\frac{3}{4}\cdot\frac{4}{5}\)
\(=8+\frac{3}{5}\)
\(=\frac{40}{5}+\frac{3}{4}=\frac{43}{5}\)
\(D=\frac{4^6\cdot25^5-2^{12}\cdot25^4}{2^{12}\cdot5^8-10^8\cdot64}=\frac{\left(2^2\right)^6\cdot\left(5^2\right)^5-2^{12}\cdot\left(5^2\right)^4}{2^{12}\cdot5^8-\left(2\cdot5\right)^8\cdot2^6}\)
\(=\frac{2^{12}\cdot5^{10}-2^{12}\cdot5^8}{2^{12}\cdot5^8-2^{14}\cdot5^8}=\frac{5^8\left(2^{12}\cdot5^2-2^{12}\right)}{5^8\left(2^{12}-2^{14}\right)}\)
\(=\frac{2^{12}\cdot5^2-2^{12}}{2^{12}-2^{14}}=\frac{2^{12}\left(5^2-1\right)}{2^{12}\left(1-2^2\right)}=\frac{24}{-3}=-8\)
T
29 tháng 10 2018
\(\sqrt{32}\cdot18+2\cdot\sqrt{25}+\left|\frac{-1}{3}\right|\cdot\left|-6\right|-2^2\)
\(=4\cdot\sqrt{2}\cdot18+2\cdot5+\frac{1}{3}\cdot6-4\)
\(=72\cdot\sqrt{2}+\left(10+2-4\right)\)
\(=72\cdot\sqrt{2}+8\)
\(=8+72\sqrt{2}\)
T
29 tháng 10 2018
\(\left(x^2-4\right)\cdot\sqrt{x}=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x^2-4\right)=0\\\sqrt{x}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0+4\\x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=4\\x=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-2\\x=2\\x=0\end{cases}}\)
NN
3
NR
4 tháng 4 2016
theo công thức, ta tính đc:
A = 1- 1/3 + 1/3 - 1/5 + 1/5 -1/7 +..... + 1/49 - 1/51
=> A bằng 1- 1/51 ( các cặp phân số đối nhau thì lược bỏ như - 1/3 và + 1/3 )
VT
0