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Câu a:
A = (-5/11).7/15.(-11/-5).(-30)
A = [(-5/11).(11/-5)].[7/15.(-30)]
A = 1 .[-14]
A = -14
\(\frac{-1}{6}.\frac{-15}{19}.\frac{38}{45}\)
\(\frac{\left(-1\right).\left(-15\right).2.19}{2.3.19.3.15}\)
\(=\frac{1}{3.3}=\frac{1}{9}\)
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
a, 6/7 + (2/11 - 6/7) - (13/11 + 1)
= 6/7 + 2/11 - 6/7 - 13/11 - 1
= (6/7 - 6/7) - (13/11 - 2/11) - 1
= 0 - 1 - 1
= -2



\(B=\left(\frac{-1}{6}\right).\left(\frac{-15}{19}\right).\left(\frac{38}{45}\right)\)
\(B=\frac{\left(-1\right).\left(-15\right).2.19}{2.3.19.15.3}\)
\(B=\frac{1.15.2.19}{2.3.19.15.3}\)
\(B=\frac{1}{9}\)
\(B=\left(\frac{-1}{6}\right).\left(\frac{-15}{19}\right).\left(\frac{38}{45}\right)\)
\(=\frac{570}{5130}=\frac{1}{9}\)
\(B=\frac{1\cdot15\cdot38}{6\cdot19\cdot45}\)
\(B=\frac{15\cdot2\cdot19}{6\cdot19\cdot3\cdot15}\)
\(B=\frac{15\cdot19\cdot2}{15\cdot19\cdot2\cdot3\cdot3}\)
\(B=\frac{1}{9}\)
B=(-1/6).(-15/19).(38/45)
B=-1/6.2/3=2/18
B=1/9