1+1=?
3+9=?
1/3 + 5/9 1/9 x 9/3
1/7 - 1/9 1/3 : 1/7
3 : 5/9 9 + 9/3
3 x 5/9 4 - 2/4
\(\dfrac{1}{3}+\dfrac{5}{9}=\dfrac{6}{18}+\dfrac{10}{18}=\dfrac{16}{18}=\dfrac{8}{9}\)
\(\dfrac{1}{7}-\dfrac{1}{9}=\dfrac{9}{63}-\dfrac{7}{63}=\dfrac{2}{63}\)
\(3:\dfrac{5}{9}=3.\dfrac{9}{5}=\dfrac{27}{5}\)
\(3.\dfrac{5}{9}=\dfrac{15}{9}=\dfrac{5}{3}\)
\(\dfrac{1}{9}.\dfrac{9}{3}=\dfrac{1}{3}\)
\(\dfrac{1}{3}:\dfrac{1}{7}=\dfrac{7}{3}\)
\(9+\dfrac{9}{3}=9+3=12\)
\(4-\dfrac{2}{4}=4-\dfrac{1}{2}=\dfrac{7}{2}\)
\(\dfrac{1}{3}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{9}\) \(+\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3+5}{9}\) \(=\) \(\dfrac{8}{9}\)
\(\dfrac{1}{7}\) \(-\) \(\dfrac{1}{9}\) \(=\) \(\dfrac{9}{63}\) \(-\) \(\dfrac{7}{63}\) \(=\) \(\dfrac{9-7}{63}\) \(=\) \(\dfrac{2}{63}\)
\(\dfrac{1}{9}\) \(\times\) \(\dfrac{9}{3}\) \(=\) \(\dfrac{1\times9}{9\times3}\) \(=\) \(\dfrac{1}{3}\)
\(\dfrac{1}{3}\) \(\div\) \(\dfrac{1}{7}\) \(=\) \(\dfrac{1}{3}\) \(\times\) \(\dfrac{7}{1}\) \(=\) \(\dfrac{1\times7}{3\times1}\) \(=\) \(\dfrac{7}{3}\)
\(3\) \(\div\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{3}{1}\) \(\div\) \(\dfrac{5}{9}\) \(=\dfrac{3}{1}\times\dfrac{9}{5}=\dfrac{3\times9}{1\times5}=\dfrac{27}{5}\)
\(3\times\dfrac{5}{9}=\dfrac{3}{1}\times\dfrac{5}{9}=\dfrac{3\times5}{1\times9}=\dfrac{5}{3}\)
\(9+\dfrac{9}{3}=\dfrac{9}{1}+\dfrac{9}{3}=\dfrac{27}{3}+\dfrac{9}{3}=\dfrac{27+9}{3}=\dfrac{36}{3}=12\)
\(4\) \(-\dfrac{2}{4}=\dfrac{4}{1}-\dfrac{2}{4}=\dfrac{16}{4}-\dfrac{2}{4}=\dfrac{14}{4}=\dfrac{7}{2}\)
a,1/1*2+1/2*3+1/3*4+1/4*5+......+1/9*10
b,2/1*3+2/3*5+2/5*7+2/7*9+2/9*11
c,3/1*3+3/3*5+3/5*7+3/7*9+3/9*11
d,5/1*3+5/3*5+5/5*7+5/7*9+5/9*11
Giúp mình với
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
= \(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(\frac{10}{11}\)
A= \(\frac{10}{11}:\frac{2}{3}\)
A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)
d) giả tương tự câu c kết quả \(\frac{25}{11}\)
tổng đặc biệt đó bạn
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(1-\frac{1}{10}=\frac{9}{10}\)
những câu sau cũng áp dụng như vậy nhé
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{3}{2}.\left(1-\frac{1}{11}\right)=\frac{3}{2}.\frac{10}{11}=\frac{15}{11}\)
Khi bỏ dấu ngoặc biểu thức số : 2 - ( 9 – 1 + 3) ta được:
A. ) 2 + 9 – 1 -3 B) 2 - 9 + 1 - 3 C) 2 - 9 - 1 – 3 D) 2 - 9 + 1 +3
B)2-9+1-3
.vì bỏ ngoặc trước nó là dấu trừ thì ta đổi dấu các số hạng trong ngoặc
so sánh
P=\(\dfrac{1+7^2+7^3+...+7^{100}}{1+7^2+7^3+...+7^{99}}\)
Q=\(\dfrac{1+9^2+9^3+...+9^{100}}{1+9^2+9^3+...+9^{99}}\)
1+1+1+1+1+2+3+4+5+6+7+8+9+10 x 1+1+1+1+1+2+3+4+5+6+7+8+9+10 +1+1+1+1+1+2+3+4+5+6+7+8+9+10 - 10 x 9 - 9 + 45 +46 +47 =
a,1/3 -3/4 - (-3/5) + 1/72-2/9-1/36 +1/15
b, 1/5 - 3/7 + 5/9 -1/11 + 7/13 - 9/16 - 7/13 + 2/11 -5/9 + 3/7 -1/5
help me pls
a) \(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{72}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{72}=1-1+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(=\dfrac{1}{5}-\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{5}{9}-\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{7}{13}-\dfrac{7}{13}-\dfrac{9}{16}\)
\(=\dfrac{9}{16}\)