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1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)

2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)

3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)

4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)

hôi lì sít

\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=2\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)

\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=-\dfrac{49}{8}\)

\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

a)$\left(\genfrac{}{}{0ex}{}{1}{2}+1,5\right)\cdot x=\genfrac{}{}{0ex}{}{1}{5}$

$2\cdot x=\genfrac{}{}{0ex}{}{1}{5}$

$x=\genfrac{}{}{0ex}{}{1}{5}:2$

$x=\genfrac{}{}{0ex}{}{1}{10}$
b) $\left(-1\genfrac{}{}{0ex}{}{3}{5}+x\right):\genfrac{}{}{0ex}{}{12}{13}=2\genfrac{}{}{0ex}{}{1}{6}$

$-1\genfrac{}{}{0ex}{}{3}{5}+x=\genfrac{}{}{0ex}{}{13}{6}\cdot \genfrac{}{}{0ex}{}{12}{13}$
$x=2+1\genfrac{}{}{0ex}{}{3}{5}$

$x=3\genfrac{}{}{0ex}{}{3}{5}$
c) $\left(x:2\genfrac{}{}{0ex}{}{1}{3}\right)\cdot \genfrac{}{}{0ex}{}{1}{7}=\genfrac{}{}{0ex}{}{-3}{8}$

$x\cdot \genfrac{}{}{0ex}{}{3}{7}\cdot \genfrac{}{}{0ex}{}{1}{7}=\genfrac{}{}{0ex}{}{-3}{8}$

$x=\genfrac{}{}{0ex}{}{-3}{8}:\genfrac{}{}{0ex}{}{3}{49}$
$x=\genfrac{}{}{0ex}{}{-49}{8}=-6\genfrac{}{}{0ex}{}{1}{8}$
d) $\genfrac{}{}{0ex}{}{-4}{7}\cdot x+\genfrac{}{}{0ex}{}{7}{5}=\genfrac{}{}{0ex}{}{1}{8}:\left(-1\genfrac{}{}{0ex}{}{2}{3}\right)$

$\genfrac{}{}{0ex}{}{-4}{7}x+\genfrac{}{}{0ex}{}{7}{5}=\genfrac{}{}{0ex}{}{1}{8}\cdot \genfrac{}{}{0ex}{}{-3}{5}$
$-\genfrac{}{}{0ex}{}{4}{7}x=\genfrac{}{}{0ex}{}{-3}{40}-\genfrac{}{}{0ex}{}{7}{5}x=\genfrac{}{}{0ex}{}{-59}{40}:\genfrac{}{}{0ex}{}{-4}{7}=\genfrac{}{}{0ex}{}{413}{160}=2\genfrac{}{}{0ex}{}{93}{160}$
 

\(\dfrac{-5}{9}+1\dfrac{5}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{15}{20}-\dfrac{8}{20}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\dfrac{7}{20}:49\\ =\dfrac{-5}{9}+\dfrac{49}{90}:49\\ =\dfrac{-5}{9}+\dfrac{1}{90}\\ =\dfrac{-50}{90}+\dfrac{1}{90}\\ =\dfrac{-49}{90}\)

\(1\dfrac{13}{15}.0,75-\left(\dfrac{104}{195}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\\ =\dfrac{7}{5}-\left(\dfrac{32}{60}+\dfrac{15}{60}\right).\dfrac{24}{47}-\dfrac{51}{13}.\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =1-\dfrac{17}{13}\\ =\dfrac{13}{13}-\dfrac{17}{13}\\ =\dfrac{-4}{13}\)

1) \(A=1+2+2^2+2^3+......+2^{2015}\)

\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)

\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)

\(\Leftrightarrow A=2^{2016}-1\)

Vậy \(A=2^{2016}-1\)

6)Ta có: \(13+23+33+43+.......+103=3025\)

\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)

\(\Leftrightarrow26+46+66+86+.......+206=6050\)

\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)

\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)

\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)

\(\Leftrightarrow23+43+63+83+.......+203+=6020\)

Vậy S=6020

b, B có 19 thừa số

=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)

<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)

<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)

<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)

<=>\(B=\frac{-21}{40} \)

a) $A=\frac{3}{5}.\frac{6}{7}+\frac{3}{7}.\frac{3}{5}-\frac{2}{7}.\frac{3}{5}$
$=\frac{3}{5}\cdot \left(\frac{6}{7}+\frac{3}{7}-\frac{2}{7}\right)=\frac{3}{5}$
b) $B=\left(-13\cdot \frac{2}{5}+\frac{-2}{9}\cdot \frac{2}{5}+\frac{2}{5}\cdot \frac{11}{9}\right)\cdot \frac{5}{2}$
$=\left(-13-\frac{2}{9}+\frac{11}{9}\right)\cdot \frac{2}{5}\cdot \frac{5}{2}=-13+\left(\frac{11}{9}-\frac{2}{9}\right)=-12.$
c) $C=\left(\frac{-4}{5}+\frac{5}{7}\right)\cdot \frac{3}{2}+\left(\frac{-1}{5}+\frac{2}{7}\right)\cdot \frac{3}{2}=\left(\frac{-4}{5}+\frac{5}{7}+\frac{-1}{5}+\frac{2}{7}\right)\cdot \frac{3}{2}=\left(\left(\frac{-4}{5}+\frac{-1}{5}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\right)\cdot \frac{3}{2}=0.$
d) $D=\frac{4}{9}:\left(\frac{1}{15}-\frac{10}{15}\right)+\frac{4}{9}:\left(\frac{2}{22}-\frac{5}{22}\right)$
$=\frac{4}{9}:\frac{-3}{5}+\frac{4}{9}:\frac{-3}{22}=\frac{4}{9}\cdot \frac{-5}{3}+\frac{4}{9}.\frac{-22}{3}$
$=\frac{4}{9}\cdot \left(\frac{-5}{3}+\frac{-22}{3}\right)=\frac{4}{9}.\frac{-27}{3}=-4.$

a) $\mathrm{A}=\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{6}{7}+\genfrac{}{}{0ex}{}{3}{7}.\genfrac{}{}{0ex}{}{3}{5}-\genfrac{}{}{0ex}{}{2}{7}.\genfrac{}{}{0ex}{}{3}{5}$

b) $\mathrm{B}=\left(-13\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{-2}{9}\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{5}{2}$
$=\left(-13-\genfrac{}{}{0ex}{}{2}{9}+\genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{5}{2}=-13+\left(\genfrac{}{}{0ex}{}{11}{9}-\genfrac{}{}{0ex}{}{2}{9}\right)=-12.$
c) $\mathrm{C}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}+\left(\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{-1}{5}\right)+\left(\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{2}{7}\right)\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=0.$
d) $\mathrm{D}=\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{1}{15}-\genfrac{}{}{0ex}{}{10}{15}\right)+\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{2}{22}-\genfrac{}{}{0ex}{}{5}{22}\right)$

\(a,=\dfrac{3^6\cdot5^4\cdot9^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+9^6\cdot5^6}=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\\ =\dfrac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

\(b,=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}\cdot\dfrac{16}{3}\right)^3}{2^7\cdot5^2+2^9}=\dfrac{2^7+12^3}{2^7\left(5^2+2^2\right)}=\dfrac{2^7+4^3\cdot3^3}{2^7\cdot29}=\dfrac{2^6\left(2+3^3\right)}{2^7\cdot29}=\dfrac{1}{2}\)

Giải:

a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)

\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)

\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)

\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)

\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)

\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)

\(\Leftrightarrow x=-\dfrac{79}{210}\)

Vậy \(x=-\dfrac{79}{210}\).

b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)

\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)

\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)

\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)

\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)

\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)

\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)

\(\Leftrightarrow3x=\dfrac{67}{75}\)

\(\Leftrightarrow x=\dfrac{67}{75}:3\)

\(\Leftrightarrow x=\dfrac{67}{225}\)

Vậy \(x=\dfrac{67}{225}\).

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