Tính a + b + c , biết :
\(\frac{a}{30}+\frac{b}{35}+\frac{c}{42}=4\); \(\frac{a}{42}+\frac{b}{35}+\frac{c}{30}=6\)
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Ta có \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Ta có \(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=\frac{1}{2}-\frac{1}{7}\)
\(=\frac{5}{14}\)
Ta có \(C=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{1}{6}-\frac{1}{22}\)
\(=\frac{4}{33}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(B=\frac{1}{2}-\frac{1}{7}\)
\(B=\frac{5}{14}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(C=\frac{1}{6}-\frac{1}{22}=\frac{4}{33}\)
\(\frac{1}{3}a=\frac{3}{4}b=\frac{4}{5}c\)
\(\frac{a}{3}=\frac{3b}{4}=\frac{4c}{5}\)
\(\Rightarrow\frac{a}{3.12}=\frac{3b}{4.12}=\frac{4c}{5.12}\)
hay \(\frac{a}{36}=\frac{b}{16}=\frac{c}{15}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\frac{a}{36}=\frac{b}{16}=\frac{c}{15}=\frac{a-c}{36-15}=\frac{42}{21}=2\)
\(\Rightarrow a=2.36=72;b=2.16=32;c=2.15=30\)
Vậy ...
Bài 2: Mình nghĩ câu a là a+2b-3c=-20
a) Ta có: a/2 = b/3 = c/4 = 2b/6 = 3c/12 = a + 2b - 3c/ 2 + 6 - 12 = -20/-4 = 5
a/2 = 5 => a = 2 . 5 = 10
b/3 = 5 => b = 5 . 3 = 15
c/4 = 5 => c = 5 . 4 = 20
Vậy a = 10; b = 15; c = 20
b) Ta có: a/2 = b/3 => a/10 = b/15
b/5 = c/4 => b/15 = c/12
=> a/10 = b/15 = c/12 = a - b + c / 10 - 15 + 12 = -49/7 = -7
a/10 = -7 => a = -7 . 10 = -70
b/15 = -7 => b = -7 . 15 = -105
c/12 = -7 => c = -7 . 12 = -84
Vậy a = -70; b = -105; c = -84.
a)
\(A=\left(\frac{1}{9}-\frac{1}{10}\right)-\left(\frac{1}{8}-\frac{1}{9}\right)-....-\left(1-\frac{1}{2}\right)=\frac{1}{9}-\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-....-1+\frac{1}{2}\)
\(A=-\left(\frac{1}{10}+1\right)=-\frac{11}{10}\)
a)\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\\ \Rightarrow A=-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\\ \Rightarrow A=-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)Đặt \(B=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}+\frac{1}{90}\)
\(\Rightarrow B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow B=1-\frac{1}{10}=\frac{9}{10}\)
Ta có : \(A=-B\)
\(\Rightarrow A=-\frac{9}{10}\)
Mình cũng nghĩ như bạn vậy. Thay vì 30/43, bạn ấy lại ghi là 30/42.
\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
Vậy a = 1; b = 2 ; c = 3 ; d = 4
a)\(\frac{77}{24}\)=\(3\frac{5}{24}\)
b)=10
c)\(\frac{24}{35}\)