tính giá trị biểu thức
C=16/15.31+14/31.45+7/45.52+13/52.65+1/13.70
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= 1/15 - 1/31 + 1/31 - 1/45 + 1/45 - 1/52 + 1/52 - 1/65 + 1/13.70
= 1/15 - 1/65 + 1/13.70
= 2/39 + 1/910
= 11/210
Tk mk nha
. Tham khảo nha !!!
\(\frac{16}{15.31}+\frac{14}{31.45}+\frac{7}{45.52}+\frac{7}{52.65}+\frac{1}{13.70}\)
\(=\frac{16}{15.31}+\frac{14}{31.45}+\frac{7}{45.52}+\frac{7}{52.65}+\frac{5}{65.70}\)
\(=\frac{1}{15}-\frac{1}{31}+\frac{1}{31}-\frac{1}{45}+\frac{1}{45}-\frac{1}{52}+\frac{1}{52}-\frac{1}{65}+\frac{1}{65}-\frac{1}{70}\)
\(=\frac{1}{15}-\frac{1}{70}\)
\(=\frac{70}{1050}-\frac{15}{1050}\)
\(=\frac{55}{1050}\)
\(=\frac{11}{210}\)
Chúc bạn học tốt !!!
Lời giải:
$A=\frac{15-5}{5.15}+\frac{31-15}{15.31}+\frac{45-31}{31.45}+\frac{52-45}{45.52}+\frac{65-52}{52.65}+\frac{1}{13.70}+\frac{1}{70.15}$
$=\frac{1}{5}-\frac{1}{15}+\frac{1}{15}-\frac{1}{31}+\frac{1}{31}-\frac{1}{45}+\frac{1}{45}-\frac{1}{52}+\frac{1}{52}-\frac{1}{65}+\frac{1}{70}(\frac{1}{13}+\frac{1}{15})$
$=\frac{1}{5}-\frac{1}{65}+\frac{1}{70}.\frac{28}{195}$
$=\frac{12}{65}+\frac{2}{95}$
$=\frac{254}{1325}$
\(A=\dfrac{5}{11}.\dfrac{5}{7}+\dfrac{5}{11}.\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}.1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)
\(B=\dfrac{3}{13}.\dfrac{6}{11}+\dfrac{3}{13}.\dfrac{9}{11}-\dfrac{3}{13}.\dfrac{4}{11}=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}.1=\dfrac{3}{13}\)
\(C=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)=\left(\dfrac{12}{16}-\dfrac{31}{22}+\dfrac{14}{91}\right).0=0\)
Lời giải:
$=(\frac{3}{5}+\frac{2}{5})+(\frac{5}{11}+\frac{16}{11})+(\frac{7}{13}+\frac{19}{13})$
$=\frac{5}{5}+\frac{22}{11}=\frac{26}{13}=1+2+2=5$
\(\frac{16}{15.31}+\frac{14}{31.45}+\frac{7}{45.52}+\frac{13}{52.65}+\frac{1}{13.70}\)
\(=\frac{16}{15.31}+\frac{14}{31.45}+\frac{7}{45.52}+\frac{13}{52.65}+\frac{5}{65.70}\)
\(=\frac{1}{15}-\frac{1}{31}+\frac{1}{31}-\frac{1}{45}+\frac{1}{45}-\frac{1}{52}+\frac{1}{52}-\frac{1}{65}+\frac{1}{65}-\frac{1}{70}=\frac{1}{15}-\frac{1}{70}=\frac{11}{210}\)