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a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)
a) \(\frac{1.3+3.5+5.7+7.9}{3.6+9.10+15.14+21.18}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.2.3+3.5.2.3+5.7.2.3+7.9.2.3}\)
= \(\frac{1.3+3.5+5.7+7.9}{1.3.6+3.5.6+5.7.6+7.9.6}\)
= \(\frac{1.3+3.5+5.7+7.9}{6.\left(1.3+3.5+5.7+7.9\right)}=\frac{1}{6}\)
Dấu "." là dấu nhân cấp 2
b) \(\frac{1.2+2.3+3.4+4.5}{3.6+6.9+9.12+12.15}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.3.3+2.3.3.3+3.4.3.3+4.5.3.3}\)
= \(\frac{1.2+2.3+3.4+4.5}{1.2.9+2.3.9+3.4.9+4.5.9}\)
= \(\frac{1.2+2.3+3.4+4.5}{9.\left(1.2+2.3+3.4+4.5\right)}=\frac{1}{9}\)
Dấu "." là dấu nhân cấp 2
c) \(\frac{0,3+\frac{3}{7}+\frac{3}{11}}{0,4+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{\frac{3}{10}+\frac{3}{7}+\frac{3}{11}}{\frac{4}{10}+\frac{4}{7}+\frac{4}{11}}\)= \(\frac{3.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}{4.\left(\frac{1}{10}+\frac{1}{7}+\frac{1}{11}\right)}=\frac{3}{4}\)
a) 8/15 + 1 = 23/15
b) 3/14 + 11/14 - 2/3 = 1- 2/3 = 1/3
c) 7/15 * (5/3+2/3) = 7/15 * 7/3= 49/45
d) 6/11 + 5/11 - 3/5 = 1 - 3/5 = 2/5
a) \(85\times16+16\times15\)
\(=16\times\left(85+15\right)\)
\(=16\times100=1600\)
b) \(\left(\dfrac{3}{7}\times\dfrac{11}{15}+\dfrac{3}{7}\times\dfrac{4}{15}\right):\left(4+2\dfrac{6}{11}-\dfrac{6}{11}\right)\)
\(=\left[\dfrac{3}{7}\times\left(\dfrac{11}{15}+\dfrac{4}{15}\right)\right]:\left(4+\dfrac{28}{11}-\dfrac{6}{11}\right)\)
\(=\dfrac{3}{7}\times1:\left(4+2\right)\)
\(=\dfrac{3}{7}:6\)
\(=\dfrac{3}{7}\times\dfrac{1}{6}=\dfrac{3}{42}=\dfrac{1}{14}\)
c) \(x+\left(x+3\right)+\left(x+6\right)+...+\left(x+102\right)=1855\)
\(\Rightarrow35x+\left(3+6+...+102\right)=1855\)
\(\Rightarrow35x+\left(102+3\right)\times34:2=1855\)
\(\Rightarrow35x+1785=1855\)
\(\Rightarrow35x=1855-1785\)
\(\Rightarrow35x=70\)
\(\Rightarrow x=70:35\)
\(\Rightarrow x=2\)
Vậy...
\(#Tmiamm\)
\(\frac{7}{3}+\frac{13}{3}=\frac{7+13}{3}=\frac{20}{3}\)
a) 3/2:9/4+21/8
= 3/2 x 4/9 + 21/8
=12/18 + 21/8
= 2/3 + 21/8 = 16/24 + 63/24 = 79/24
b) 3/10+5/2x1/3
=3/10 + 5/6
= 9/30 + 25/30 = 34/30 = 17/15
c) 8/15:3/11+2/15:2/11
= 8/15x11/3 + 2/15x11/2
= 88/45 + 22/30
= 88/45 + 11/15 = 88/45 + 33/45 = 121/45
d) 19/5-11/4:9/8
= 19/5 - 11/4x8/9
= 19/5 - 88/36
= 19/5 - 22/9 = 171/45 - 110/45 = 61/45
e) 33/7:11/6+37/15x21/74
= 33/7x6/11 + 7/10
= 198/77 + 7/10 = 1980/770 + 1386/770 = 153/35
a) 3/2:9/4+21/8
= 3/2 x 4/9 + 21/8
=12/18 + 21/8
= 2/3 + 21/8 = 16/24 + 63/24 = 79/24
b) 3/10+5/2x1/3
=3/10 + 5/6
= 9/30 + 25/30 = 34/30 = 17/15
c) 8/15:3/11+2/15:2/11
= 8/15x11/3 + 2/15x11/2
= 88/45 + 22/30
= 88/45 + 11/15 = 88/45 + 33/45 = 121/45
d) 19/5-11/4:9/8
= 19/5 - 11/4x8/9
= 19/5 - 88/36
= 19/5 - 22/9 = 171/45 - 110/45 = 61/45
e) 33/7:11/6+37/15x21/74
= 33/7x6/11 + 7/10
= 198/77 + 7/10 = 1980/770 + 1386/770 = 153/35
23/12
11/5 - 3/4 + 7/15
= 29/20 + 7/15
= 23/12