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a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
\(\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{10}{7}+\frac{10}{3}\right):\left(\frac{37}{3}-\frac{100}{7}\right)}=\frac{\left(\frac{4293}{324}-\frac{708}{324}-\frac{3510}{324}\right).\frac{5751}{25}+\frac{187}{4}}{\left(\frac{30}{21}+\frac{70}{21}\right):\left(\frac{259}{21}-\frac{300}{21}\right)}=\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:\left(-\frac{41}{21}\right)}\)=\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}=100:\left(-\frac{100}{4}\right)=-4\)
\(30+\frac{14}{5}:\left(\frac{24}{150}-\frac{270}{150}-\frac{25}{150}\right)=30+\frac{14}{5}:\left(-\frac{271}{150}\right)=30+\left(-\frac{420}{271}\right)=\frac{7710}{271}\)
1) tự làm (thực hiện từ dưới lên)
2) B = \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{4}\right)^5.3}{\frac{\frac{1}{1024}.1}{3}-\left(\frac{1}{2}\right)^{11}}\)
= \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{2}\right)^{10}.3}{\left(\frac{1}{2}\right)^{10}.\frac{1}{3}-\left(\frac{1}{2}\right)^{10}.\frac{1}{2}}\)
= \(\frac{\left(\frac{1}{2}\right)^{10}.\left(5-3\right)}{\left(\frac{1}{2}\right)^{10}.\left(\frac{1}{3}-\frac{1}{2}\right)}\)
= \(\frac{2}{-\frac{1}{6}}\)= 2 . (-6) = -12
1) \(5+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{4}}}}=5+\frac{15}{7}=\frac{5}{1}+\frac{15}{7}=\frac{50}{7}\)
\(\left(\frac{1}{216}-\frac{1}{1^3}\right)\) Ko phải là \(\left(\frac{1}{216}-\frac{1}{13}\right)\) nha
Trong tích trên có thừa số \(\frac{1}{216}-\frac{1}{6^3}=\frac{1}{216}-\frac{1}{216}=0\)
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