`~Zurden~`
Câu 18: $\dfrac{x-1}{12}+ \dfrac{x-1}{20}+ \dfrac{x-1}{30}+ \dfrac{x-1}{42}+ \dfrac{x-1}{56}+ \dfrac{x-1}{72}=\dfrac{16}9$
$(x-1)\left(\dfrac1{12}+\dfrac1{20}+\dfrac1{30}+\dfrac1{42}+\dfrac1{56}+\dfrac1{72}\right)=\dfrac{16}9$
$(x-1)\left(\dfrac1{3.4}+\dfrac1{4.5}+\dfrac1{5.6}+\dfrac1{6.7}+\dfrac1{7.8}+\dfrac1{8.9}\right)=\dfrac{16}9$
$(x-1)\left(\dfrac13-\dfrac14+\dfrac15-\dfrac15+\dfrac16-\dfrac16+\dfrac17-\dfrac17+\dfrac18-\dfrac19\right)=\dfrac{16}9$
$(x-1)\left(\dfrac13-\dfrac19\right)=\dfrac{16}9$
$\dfrac29(x-1)=\dfrac{16}9$
$x-1=8$
$x=9$
Đáp án $C$
Câu 19: $\left(x-\dfrac12\right)\times \dfrac16+\left(x-\dfrac12\right)\times \dfrac1{12}+\left(x-\dfrac12\right)\times \dfrac1{20}+...+\left(x-\dfrac12\right)\times \dfrac1{90}=\dfrac{19}5$
$\left(x-\dfrac12\right)\left(\dfrac16+\dfrac1{12}+\dfrac1{20}+...+\dfrac1{90}\right)=\dfrac{19}5$
$\left(x-\dfrac12\right)\left(\dfrac1{2.3}+\dfrac1{3.4}+\dfrac1{4.5}+...+\dfrac1{9.10}\right)=\dfrac{19}5$
$\left(x-\dfrac12\right)\left(\dfrac12-\dfrac13+\dfrac13-\dfrac14+\dfrac14-\dfrac15+...+\dfrac19-\dfrac1{10}\right)=\dfrac{19}5$
$\left(x-\dfrac12\right)\left(\dfrac12-\dfrac1{10}\right)=\dfrac{19}5$
$\left(x-\dfrac12\right)\dfrac25=\dfrac{19}5$
$x-\dfrac12=\dfrac{19}2$
$x=10$
Đáp án $C$
Câu 20: $\dfrac15+\dfrac1{45}+\dfrac1{117}+\dfrac1{221}+...+\dfrac1{x(x+4)}=\dfrac{53}{216}$
$\dfrac14 \left(\dfrac45+\dfrac4{45}+\dfrac4{117}+...+\dfrac1{x(x+4)}\right)=\dfrac{53}{216}$
$\dfrac14 \left(1-\dfrac15+\dfrac15-\dfrac19+\dfrac19-\dfrac1{13}+...+\dfrac{1}{x}-\dfrac1{x+4}\right)=\dfrac{53}{216}$
$\dfrac14 \left(1-\dfrac1{x+4}\right)=\dfrac{53}{216}$
$1-\dfrac1{x+4}=\dfrac{53}{54}$
$\dfrac1{x+4}=\dfrac1{54}$
$x+4=54$
$x=50$
Đáp án $C$
