P=0.75=0.6+3/7+3/13 trên 2.75=2.2+11/7+11/13
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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)
\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(=\frac{3}{11}\)
\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}=\frac{3}{11}\)
\(A=\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}\)
\(=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)
\(=\frac{3.\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11.\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}\)
\(=\frac{3}{11}\)
Ta có:
\(A=\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}\)
\(A=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)
\(A=\frac{3\cdot\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\cdot\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}\)
\(A=\frac{3}{11}\)
Bài 1 :
0,7 + 0,6 - \(\frac{3}{7}\)- \(\frac{3}{13}\)
= 1,3 - \(\frac{3}{7}\)- \(\frac{3}{13}\)
= \(\frac{61}{70}\)- \(\frac{3}{13}\)
= \(\frac{583}{910}\)
2,75 + 2,2 - \(\frac{11}{7}\)- \(\frac{11}{13}\)
= 4,95 - \(\frac{11}{7}-\frac{11}{13}\)
= \(\frac{473}{140}-\frac{11}{13}\)
= \(\frac{4609}{1820}\)
=43/306:599/105x1473/1820
=1505/61098x1473/1820
=0.01993613035