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\(1,A=\dfrac{1}{21}\\ 2,B=\dfrac{101}{200}\\ 3,a\in\left\{-14;-13;-12;-11;-10\right\}\\ 4,D=\dfrac{48}{7}\\ 5,E=-\dfrac{1}{3}\\ 6,F=2-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Câu 8:
Ta có: \(A=2+2^2+2^3+2^4+...+2^{50}\)
\(\Leftrightarrow2\cdot A=2^2+2^3+...+2^{51}\)
\(\Leftrightarrow A=2^{51}-2\)
\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{19}\right)\left(1-\dfrac{1}{20}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{18}{19}.\dfrac{19}{20}=\dfrac{1}{20}>\dfrac{1}{21}\)
Lời giải:
a)
\(\frac{\frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\frac{2}{11}}{\frac{8}{3}-\frac{8}{5}+\frac{8}{7}-\frac{8}{9}+\frac{8}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}{8\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}\) \(=\frac{2}{8}=\frac{1}{4}\)
b)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{50}-1\right)\left(\frac{1}{51}-1\right)\)
\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}....\frac{1-50}{50}.\frac{1-51}{2}=\frac{(-1)(-2)(-3)...(-49)(-50)}{2.3.4....50.51}\)
\(=\frac{(-1)^{50}.1.2.3....49.50}{2.3.4...50.51}=\frac{1}{51}\)
\(a,=\dfrac{13}{50}\cdot\dfrac{50}{13}\cdot\left(-\dfrac{31}{2}\right)\cdot\dfrac{169}{2}=-\dfrac{5239}{2}\\ b,=\dfrac{-\dfrac{49}{100}\cdot\left(-125\right)}{-\dfrac{343}{27}\cdot\dfrac{81}{16}\cdot\left(-1\right)}=\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{245}{4}\cdot\dfrac{16}{1029}=\dfrac{20}{21}\)
a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}=\dfrac{13}{50}.-75:\dfrac{13}{50}.\dfrac{169}{2}=-\dfrac{75.169}{2}=-\dfrac{12675}{2}\)
b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}=\dfrac{0,49.\left(-125\right)}{-\dfrac{343}{27}.\dfrac{81}{16}.\left(-1\right)}=-\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{20}{21}\)
\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\left(-1\right)+1=\dfrac{1}{3}+0=\dfrac{1}{3}\)
\(\left(\dfrac{15}{4}-5x\right).\left(9x^2-4\right)=0\)
\(\left[{}\begin{matrix}\dfrac{15}{4}-5x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}5x=\dfrac{15}{4}\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)
a) \(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)
\(A=\left(-0,75-0,25\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right)\cdot\dfrac{-1}{3}\)
\(A=\left(-1\right):\left(-5\right)+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{1}{5}+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{119}{720}\)
b) \(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)
\(B=\left(0,24-1,24\right):\dfrac{3}{7}:\left[\left(\dfrac{7}{2}-\dfrac{11}{3}\right):\dfrac{1}{14}\right]\)
\(B=-1:\dfrac{3}{7}:\left(-\dfrac{1}{6}:\dfrac{1}{14}\right)\)
\(B=-\dfrac{7}{3}:-\dfrac{7}{3}\)
\(B=1\)
a, A = (-0,75 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{48}\) - (- \(\dfrac{1}{6}\)) : (-3)
A = -(0,75 + 0,25): (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = -1 : (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{1}{5}\) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{53}{240}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{119}{720}\)
b, B = (\(\dfrac{6}{25}\) - 1,24): \(\dfrac{3}{7}\): [(3\(\dfrac{1}{2}\) - 3\(\dfrac{2}{3}\)): \(\dfrac{1}{14}\)]
B = (0,24 - 1,24): \(\dfrac{3}{7}\):[(\(\dfrac{7}{2}\)-\(\dfrac{11}{3}\)): \(\dfrac{1}{14}\)]
B = -1: \(\dfrac{3}{7}\):[ (-\(\dfrac{1}{6}\) : \(\dfrac{1}{14}\))]
B = -1: \(\dfrac{3}{7}\): (- \(\dfrac{7}{3}\))
B = 1 \(\times\) \(\dfrac{7}{3}\) \(\times\) \(\dfrac{3}{7}\)
B = 1