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\(A=\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}\)
\(=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)
\(=\frac{3.\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11.\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}\)
\(=\frac{3}{11}\)
Ta có:
\(A=\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}\)
\(A=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)
\(A=\frac{3\cdot\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\cdot\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}\)
\(A=\frac{3}{11}\)
a) Ta có: \(\frac{0,4-\frac{2}{7}+\frac{2}{11}}{0,6-\frac{3}{7}+\frac{3}{11}}\)
\(=\frac{2.\left(0.2-\frac{1}{7}+\frac{1}{11}\right)}{3.\left(0,2-\frac{1}{7}+\frac{1}{11}\right)}\)
\(=\frac{2}{3}\)
b) Ta có: \(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{3.\left(0,25-0,2+\frac{1}{7}+\frac{1}{13}\right)}{11.\left(0,25-0,2+\frac{1}{7}+\frac{1}{13}\right)}\)
\(=\frac{3}{11}\)
Chuk pạn hok tốt!
a) \(\frac{0,4-\frac{2}{7}+\frac{2}{11}}{0,6-\frac{3}{7}+\frac{3}{11}}\\ =\frac{\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{3}{5}-\frac{3}{7}+\frac{3}{11}}\\ =\frac{2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{3\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}\\=\frac{2}{3}\)
b) \(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\\=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\\=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\\=\frac{3}{11}\)
k = ( 0,75 - 0, 6 + 3/7 + 3/3) : ( 11/7 + 11/3 + 2,75 + 2,2)
= ( 3/4 + 3/5 + 3/7 + 3/3) : ( 11/7 + 11/3 + +11/4 + 11/5 )
= 3 x (1/4 + 1/5 + 1/7 +1/3) : 11 x ( 1/7 + 1/3 + 1/4 + 1/5)
= 3/11
( Xem lại xem 3/3 hay 3/13)
\(\frac{\frac{3}{4}+\frac{3}{5}+\frac{3}{7}+\frac{3}{11}}{\frac{-13}{4}-\frac{13}{5}-\frac{13}{7}-\frac{13}{11}}\)
\(=\frac{3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}{-13\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}\)
\(=\frac{-3}{13}.\)
Ta có:\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{3}-\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)
\(\Rightarrow P=\frac{3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3}{11}\)
\(B=\frac{1}{3}-\frac{3}{4}+0,6+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}-\frac{48}{64}+\frac{9}{15}+\frac{1}{64}-\frac{8}{36}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}+\frac{9}{15}+\frac{1}{15}+\left(-\frac{48}{64}+\frac{1}{64}\right)+\left(-\frac{8}{36}-\frac{1}{36}\right)\)
\(\Rightarrow B=\frac{13}{15}-\frac{47}{64}-\frac{1}{4}\)
\(\Rightarrow B=-\frac{113}{960}\)
\(C=0\)
\(D=\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow D=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)
\(\Rightarrow D=1\)
D= \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}......-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
=\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{98}-\frac{1}{99}\right)\)
=\(\frac{1}{99}-\left[1-(\frac{1}{2}-\frac{1}{2}+......+\frac{1}{98}-\frac{1}{99})\right]\)
=\(\frac{1}{99}-\left(1-0-0-.....-0-\frac{1}{99}\right)\)
=\(\frac{1}{99}-1-\frac{1}{99}\)
=1
\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(=\frac{3}{11}\)
\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}=\frac{3}{11}\)