BC
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22 tháng 5 2022
a: \(=347\cdot4\cdot9\cdot400:8=347\cdot36\cdot50=624600\)
c: \(=16:\left\{400:\left[200-37-138\right]\right\}\)
\(=16:\left\{400:25\right\}=16:16=1\)
e: \(=46-\left[300:15\right]-2=46-20-2=24\)
15 tháng 12 2017
a) \(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}\)
\(=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)
\(=1+\left(-1\right)\)
\(=0\)
b) \(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(20.5\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
c) \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{12}.\left(2^{18}+2^8\right)}{2^{12}.\left(1+2^{10}\right)}=\frac{2^{18}+2^8}{1+2^{10}}=256\)
9 tháng 8 2017
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)
\(A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(A=1-\dfrac{1}{99}\)
\(A=\dfrac{98}{99}\)
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\)
\(B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(B=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(B=\dfrac{1}{2}-\dfrac{1}{90}\)
\(B=\dfrac{22}{45}\)
9 tháng 8 2017
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+.........+\dfrac{2}{97.99}\)
\(\Leftrightarrow A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{97}-\dfrac{1}{99}\)
\(\Leftrightarrow A=1-\dfrac{1}{99}=\dfrac{98}{99}\)
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+.......+\dfrac{2}{8.9.10}\)
\(\Leftrightarrow B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+......+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(\Leftrightarrow B=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(\Leftrightarrow B=\dfrac{1}{2}-\dfrac{1}{90}\)
\(\Leftrightarrow B=\dfrac{22}{45}\)
TH
20 tháng 6 2018
a/Ta có: \(\dfrac{4}{3}-\left[\left(\dfrac{-11}{6}\right)-\left(\dfrac{2}{9}+\dfrac{5}{3}\right)\right]\)
\(=\) \(\dfrac{4}{3}-\left[\dfrac{-11}{6}-\dfrac{2}{9}-\dfrac{5}{3}\right]\)
\(=\) \(\dfrac{4}{3}+\dfrac{11}{6}+\dfrac{2}{9}+\dfrac{5}{3}\)
\(=\) \(\dfrac{24}{18}+\dfrac{33}{18}+\dfrac{4}{18}+\dfrac{30}{18}\)
\(=\) \(\dfrac{91}{18}\)
b/Ta có: \(\left(8-\dfrac{9}{4}+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+\dfrac{5}{4}\right)-\left(3+\dfrac{2}{4}-\dfrac{9}{7}\right)\)
\(=\) \(8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)
\(=\) \(8+6-3-\dfrac{9}{4}-\dfrac{5}{4}-\dfrac{2}{4}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\)
\(=\) \(11-\dfrac{2}{4}+\dfrac{14}{7}\)
\(=\) \(11-\dfrac{1}{2}+2\)
\(=\) \(9-\dfrac{1}{2}\)
\(=\) \(\dfrac{17}{2}\)
Chúc bn học tốt!!!
7 tháng 7 2018
a) \(\left(-\dfrac{1}{2}\right)-\left(-\dfrac{3}{5}\right)+\left(-\dfrac{1}{5}\right)+\dfrac{1}{131}-\left(-\dfrac{2}{7}\right)+\dfrac{4}{35}-\dfrac{7}{18}\\ =-\dfrac{1}{2}+\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{1}{131}+\dfrac{2}{7}+\dfrac{4}{35}-\dfrac{7}{18}\\ =\left(\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)-\left(\dfrac{1}{2}+\dfrac{7}{18}\right)+\dfrac{1}{131}\\ =\dfrac{4}{5}-\dfrac{8}{9}+\dfrac{1}{131}=-\dfrac{4}{45}+\dfrac{1}{131}\\ =-\dfrac{479}{5895}\)
7 tháng 7 2017
C= 1/100-(1/1.2+1/2.3+...+1/97.98+1/98.99+1/99.100)
C=1/100-(1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)
C=1/100-(1-1/100)
C=1/100-99/100
C=-98/100=-49/50
7 tháng 7 2017
\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)+\dfrac{1}{100}\)
\(=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=\left(-1\right)+\dfrac{1}{50}=-\dfrac{49}{50}\)
27 tháng 6 2017
\(A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(\dfrac{-6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)
\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)
\(A=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)-\left(\dfrac{7}{13}+\dfrac{6}{13}\right)+\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\)
\(A=1-1+1=1\)
\(B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)
\(B=\dfrac{-3}{2}:\dfrac{3}{-4}.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=2.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(=-9-\dfrac{1}{4}=\dfrac{-37}{4}\)
27 tháng 6 2017
\(a,A=\left(\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}\right)+\left(-\dfrac{6}{13}+\dfrac{1}{2}+1\dfrac{1}{3}\right)\)
\(A=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}+\dfrac{-6}{13}+\dfrac{1}{2}+\dfrac{4}{3}\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{7}{13}-\dfrac{6}{13}\right)+\left(-\dfrac{1}{3}+\dfrac{4}{3}\right)\)
\(A=-1+1=0\)
\(b,B=\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right)\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}\)
\(B=\left(-\dfrac{3}{2}.\dfrac{-4}{3}\right).\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=8.\dfrac{-9}{2}-\dfrac{1}{4}\)
\(B=-36-\dfrac{1}{4}\)
B = \(-\dfrac{145}{4}\)
Ta có:
\( \dfrac{7256.4375-725}{3650+4375.7255}\\ =\dfrac{7255.4375+4375-7255}{3650+4375.7255}\\ =\dfrac{7255.4375+\left(4375-725\right)}{3650+4375.7255}\\ =\dfrac{7255.4375+3650}{3650+4375.7255}\\ =1\)
Vậy giá trị của \(\dfrac{7256.4375-725}{3650+4375.7255}=1\)
7256.4375-725=(7255+1).4375-725=7255.4375+4375-725=7255.4375+3650
⇒(7256.4375−725)/(3650+4375.7255)=(7255.4375+3650)/(3650+4375.7255)=1