Tính giá trị biểu thức:1/11+1/12+1/13+...+1/14
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A = (1- 2) \(\times\) ( 4 - 3) \(\times\) (5 - 6) \(\times\) (8 - 7) \(\times\) (9 - 10) \(\times\) (12 - 11) \(\times\)(13 - 14)
A = (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1)
A = 1
Tính giá trị biểu thức\(1+\frac{1}{10}+1+\frac{1}{11}+1+\frac{1}{12}+1+\frac{1}{13}+1+\frac{1}{14}\)
\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{10}\)\(+\)\(\frac{1}{10}\)\(-\)\(\frac{1}{11}\)\(+\)\(\frac{1}{11}\)\(-\)\(\frac{1}{12}\)\(+\)\(\frac{1}{12}\)\(-\)\(\frac{1}{13}\)\(+\)\(\frac{1}{13}\)\(-\)\(\frac{1}{14}\)\(+\)\(\frac{1}{14}\)\(-\)\(\frac{1}{15}\)
\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{15}\)
\(A\)\(=\)\(\frac{2}{45}\)
\(A=\left(\frac{1}{9}.\frac{1}{10}+\frac{1}{10}.\frac{1}{11}\right)+\left(\frac{1}{11}.\frac{1}{12}+\frac{1}{12}.\frac{1}{13}\right)+\left(\frac{1}{13}.\frac{1}{14}+\frac{1}{14}.\frac{1}{15}\right)\)
Sau đó nhân phân phối ra là xong nhé bạn
\(A=\dfrac{5}{11}.\dfrac{5}{7}+\dfrac{5}{11}.\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}.1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)
\(B=\dfrac{3}{13}.\dfrac{6}{11}+\dfrac{3}{13}.\dfrac{9}{11}-\dfrac{3}{13}.\dfrac{4}{11}=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}.1=\dfrac{3}{13}\)
\(C=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right).0=0\)
Ta có: A\(=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}=\dfrac{2}{45}\)
\(A=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)
\(=\dfrac{1}{9}-\dfrac{1}{15}\)
\(=\dfrac{2}{45}\)
-Chúc bạn học tốt-
Câu a :
\(\dfrac{5}{12}+\dfrac{3}{4}+\dfrac{1}{3}\\ =\dfrac{5}{12}+\dfrac{9}{12}+\dfrac{4}{12}=\dfrac{3}{2}\)
câu b :
\(\dfrac{1}{4}+\dfrac{3}{7}+\dfrac{11}{14}\\ =\dfrac{7}{28}+\dfrac{12}{28}+\dfrac{22}{28}=\dfrac{41}{28}\)
câu c :
\(\dfrac{1}{4}+\dfrac{3}{12}+\dfrac{12}{36}\\ =\dfrac{3}{12}+\dfrac{3}{12}+\dfrac{4}{12}\\ =\dfrac{10}{12}=\dfrac{5}{6}\)
x=11 suy ra 12=x+1 thay vào A ta có:
A=x^17- (x+1)x^16 + (x+1)x^15 - (x+1)x^14 + .....- (x+1)x^2+(x+1)x -1
= x^17 - x^17 -x^16 + x^16 + x^15 - x^15 - x^14 +.....- x^3 -x^2 + x^2 +x -1
= x-1= 11-1=10
\(C=\left(1-2-3-4\right)+...+\left(197-198-199-200\right)\)
=-8x25=-200
\(D=-\left(11+13+...+99\right)+\left(10+12+...+98\right)\)
=(-1)+(-1)+...+(-1)
=-1x45=-45
mk làm lại:
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)
=\(\left(\frac{1}{11}-\frac{11}{11}\right)+\left(\frac{1}{12}-\frac{12}{12}\right)+\left(\frac{1}{13}-\frac{13}{13}\right)+...+\left(\frac{1}{20}-\frac{20}{20}\right)\)
=\(\frac{-10}{11}+\frac{-11}{12}+\frac{-12}{13}+...+\frac{-19}{20}\)
=\(\frac{-10}{20}\)
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{14}\)
=\(\left(\frac{1}{11}-\frac{11}{11}\right)+\left(\frac{1}{12}-\frac{12}{12}\right)+\left(\frac{1}{13}-\frac{13}{13}\right)+...+\left(\frac{1}{14}-\frac{14}{14}\right)\)
=\(\frac{-10}{11}+\frac{-11}{12}+\frac{-12}{13}+...+\frac{-13}{14}\)
=\(\frac{-10}{14}\)