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Câu 1;
\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)
Câu 2:
\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{3}}=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{3}}=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}=\frac{3}{11}\)
Câu 1;
13 −17 −113 23 −27 −213 ·34 −316 −364 −3256 1−14 −116 −164 +58
=13 −17 −113 2(13 −17 −113 ) ·3(14 −116 −164 −1256 )4(14 −116 −164 −1256 ) +58
=12 ·34 +58 =38 +58 =1
Câu 2:
1; = ( -4/10 + 3/10 ) : ( -2/5 + 2/3 ) = -1/10 : ( -6/15 + 10/15 ) = -1/10 : 4/15 = -1/10 . 15/4 = -15/40 = -3/8
2; = 25/2 . -5/7 + 39/4 + -3/2 . 5/7 = -125/14 + 39/4 + -15/14 = ( -125/14 + -15/14 ) + 39/4 = -10 + 39/4 = -40/4 + 39/4 = -1/4
3; = 5/52 + 35/52 + 40/52 = 40/52 + 40/52 = 80/52 = 20/13
4; = ( -39/52 + 20/52 ) . 7/2 - ( 117/52 + 32/52 ) . 7/2 = -19/52 . 7/2 - 149/52 . 7/2 = ( -19/52 + -149/52 ) . 7/2 = -168/52 .7/2 = -147/13
5; = ( 36/12 + -9/12 + 8/12 ) - ( -12/6 + -8/6 + -9/6 ) - ( 6/6 - 14/6 - 27/6 ) = 35/12 + 10/12 + 70/12 = 115/12
6; = -1/3 + -8/35 +-2/9 + -1/135 +4/5 +-4/9 +3/7 = (-1/3 + -2/9 + -4/9 ) + ( -8/35 + 4/5 + 3/7 ) + -1/135 = ( -1/3 + -2/3 ) + ( -8/35 + 28/35 + 15/35 ) + -1/135 = -1 + 1 + -1/135 = -1/135
A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)
\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)
\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)
=1-82/84
=2/84=1/42
\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)
\(D=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
a. \(A=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}=\dfrac{3\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
Vậy \(A=\dfrac{3}{11}\)
b. \(B=\dfrac{2^{12}\cdot13+2^{12}\cdot65}{2^{10}\cdot104}+\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{2^{12}\left(13+65\right)}{2^{10}\cdot104}+\dfrac{3^{10}\left(11+5\right)}{3^9\cdot2^4}=\dfrac{2^{12}\cdot78}{2^{10}\cdot104}+\dfrac{3^{10}\cdot16}{3^9\cdot16}=\dfrac{2^2\cdot3}{1\cdot4}+3=\dfrac{12}{4}+3=3+3=6\)
Vậy \(B=6\)
Bài 1:
\(=\dfrac{-1}{2}+\dfrac{3}{5}-\dfrac{1}{9}+\dfrac{1}{131}+\dfrac{2}{7}+\dfrac{4}{35}-\dfrac{7}{18}\)
\(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{131}\)
\(=\dfrac{-9-2-7}{18}+\dfrac{21+10+4}{35}+\dfrac{1}{131}\)
=1/131
Bài 2:
b: \(B=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)
\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{99}-\dfrac{98}{99}=-\dfrac{97}{99}\)
Dấu "=" chính giữa mình ko biết là dấu + hay - nên mình sẽ làm cả hai TH nhé
TH1: \(P=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}+\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)
\(=\dfrac{3\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}+\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}+\dfrac{3}{11}=\dfrac{6}{11}\)
Tương tự TH2:
\(P=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}-\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}=\dfrac{3}{11}-\dfrac{3}{11}=0\)