1=\(\frac{6}{ }\)
\(A=\frac{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}{\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)-\frac{1}{4}.\frac{1}{5}.\frac{1}{6}}\)
Ta có \(A=\frac{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}{\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)-\frac{1}{4.5.6}}=\frac{\frac{\left(15+12+10\right)}{60}}{\left(\frac{15+12-10}{60}\right)-\frac{1}{120}}\)\(=\frac{\frac{37}{60}}{\frac{27}{60}-\frac{1}{120}}=\frac{37}{60}:\frac{54}{120}=\frac{37}{60}.\frac{120}{54}=\frac{37.2.60}{60.2.37}=1\)
\(\frac{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}{(\frac{1}{4}+\frac{1}{5}-\frac{1}{6})-\frac{1}{4}.\frac{1}{5}.\frac{1}{6}}\)=\(\frac{\frac{37}{60}}{\frac{17}{60}-\frac{1}{120}}\)=\(\frac{\frac{37}{60}}{\frac{33}{120}}\)=\(\frac{37}{60}:\frac{33}{120}=\frac{37}{60}.\frac{120}{33}\)=\(\frac{74}{33}\)
Anime Tổng hợp ơi,\((12+15-10)=17\) chứ ko phải =27
So sánh:\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac{5}{6}}{\frac{7}{8}}}+\frac{\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{4}{3}}{\frac{2}{1}}}\) và\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}+\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{5}{6}}{\frac{7}{8}}+\frac{\frac{4}{3}}{\frac{2}{1}}}\)và \(\frac{\frac{\frac{1}{2}+\frac{8}{7}}{\frac{3}{4}+\frac{6}{5}}}{\frac{\frac{5}{6}+\frac{4}{3}}{\frac{7}{8}+\frac{2}{1}}}\)và\(\frac{\frac{\frac{1+8}{2+7}}{\frac{3+6}{4+5}}}{\frac{5+4}{\frac{6+3}{2+1}}}\)
bài 1: tính A:=\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{2}{3}-\frac{1}{2}\)
Bài 2: Cho B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{49}-\frac{1}{50}\)
Chứng minh rằng: \(\frac{7}{12}< A< \frac{5}{6}\)
A = \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}-1\right)\times\left(1-\frac{8}{1}-\frac{4}{1}-\frac{2}{1}\right)\)
B = \(\frac{\frac{3}{1}-\frac{6}{3}-\frac{9}{6}-\frac{369}{1}}{\frac{1}{3}+\frac{3}{6}+\frac{6}{9}-\frac{1}{963}}\)
C = \(\frac{1}{1}-\frac{1}{2}+\frac{3}{1}-\frac{1}{4}+\frac{5}{1}-\frac{1}{6}+\frac{7}{1}-\frac{1}{8}+\frac{9}{1}-\frac{1}{10}\)
so sánh các số trên ( A , B , C )
a= 1/2 + 1/4 + 1/8 - 1 x 1 + 8/1 - 4/1 - 2/1=\(1\frac{7}{8}\)=1,875
b=3/1 - 6/3 - 9/6 - 369/1 : 1/3 + 3/6 + 6/9 - 1/963 \(\approx\)186,665628245067
c=1/1 - 1/2 + 3/1 - 1/4 + 5/1 - 1/6 + 7/1 - 1/8 + 9/1 - 1/10=\(\approx\)23,8583333333333
vậy a>b>c
**************************l i k e***********************************8
A = \(\left(-\frac{1}{8}\right)\times\left(-13\right)=\frac{13}{8}\) => 0 < A < 2
B: Tử âm ; mẫu dương => B < 0
C = \(\left(\frac{1}{1}+\frac{3}{1}+\frac{5}{1}+\frac{7}{1}+\frac{9}{1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
= 25 \(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
Dễ có: B < A < C
\(\frac{\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\div\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)}{\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)\div\left(\frac{1}{4}-\frac{1}{6}\right)}\)
Tính
P= \(\frac{\frac{6}{8}-\frac{6}{10}+\frac{6}{14}+\frac{6}{26}}{\frac{11}{4}+\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
Q=\(\frac{\frac{9}{1}+\frac{8}{2}+.....+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{10}}\)
Ta có :
\(P=\frac{\frac{6}{8}+\frac{6}{10}+\frac{6}{14}+\frac{6}{26}}{\frac{11}{4}+\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(\Rightarrow P=\frac{\frac{3}{4}+\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3}{11}\)
Vậy \(P=\frac{3}{11}\)
\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}=\frac{3}{11}\)
đề bài của bn sai nên mk sửa luôn nha
Tính nhanh E=\(\frac{6}{8+1}.\frac{6}{18+1}.\frac{6}{30+1}...\frac{6}{10700+1}\)
e,\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{42}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}=4-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=4-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1}-\frac{1}{7}\right)=4-\frac{6}{7}=3\frac{1}{7}\)
BN mún hỏi j vậy, đây k phải câu hỏi, mà có thì phải là toán lớp 6
Tính giá trị của biểu thức:
\(\frac{\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right):\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)}{\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right):\left(\frac{1}{4}-\frac{1}{6}\right)}\)
\(\frac{6-\frac{1}{\frac{1}{2}-\frac{1}{3}}}{6+\frac{1}{\frac{1}{2}-\frac{1}{3}}}\)
\(\frac{6-\frac{1}{\frac{1}{2}-\frac{1}{3}}}{6+\frac{1}{\frac{1}{2}-\frac{1}{3}}}=0\) NHA Nguyễn Giang !