Đáp án là A

Đáp án là D

Đáp án là A

Đáp án cần chọn là: A

$\begin{array}{l}x:\frac{5}{8}=\frac{-14}{35}\cdot \frac{15}{-42}\\ x:\frac{5}{8}=\frac{-2}{5}.\frac{5}{-14}\\ x:\frac{5}{8}=\frac{1}{7}\\ x=\frac{1}{7}.\frac{5}{8}\\ x=\frac{5}{56}\end{array}$

Đáp án cần chọn là: C

$\begin{array}{l}\frac{-3}{7}-x:\frac{4}{7}=-2\\ x:\frac{4}{7}=\frac{-3}{7}-(-2)\\ x:\frac{4}{7}=\frac{11}{7}\\ x=\frac{4}{7}.\frac{11}{7}\\ x=\frac{44}{49}\end{array}$

Đáp án cần chọn là: C

$\begin{array}{l}\left(\frac{7}{6}+x\right):\frac{16}{25}=\frac{-5}{4}\\ \frac{7}{6}+x=\frac{-5}{4}.\frac{16}{25}\\ \frac{7}{6}+x=\frac{-1}{1}.\frac{4}{5}\\ \frac{7}{6}+x=\frac{-4}{5}\\ x=\frac{-4}{5}-\frac{7}{6}\\ x=\frac{-59}{30}\end{array}$

Đáp án cần chọn là: D

$\begin{array}{l}\left(x-\frac{11}{25}\right):\frac{3}{25}=-\frac{7}{15}-\frac{8}{6}\\ \left(x-\frac{11}{25}\right):\frac{3}{25}=-\frac{9}{5}\\ x-\frac{11}{25}=-\frac{9}{5}.\frac{3}{25}\\ x-\frac{11}{25}=-\frac{27}{125}\\ x=\frac{28}{125}\end{array}$

Đáp án cần chọn là: B

$\begin{array}{l}M=\frac{17}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.{\left(\frac{-1}{2}\right)}^3\\ M=\frac{17.\left(-31\right)\mathrm{.1.10.}{\left(-1\right)}^3}{{\mathrm{5.125.2.17.2}}^3}\\ M=\frac{-31.\left(-1\right)}{{125.2}^3}\\ M=\frac{31}{1000}\end{array}$

$\begin{array}{l}N=\left(\frac{17}{28}+\frac{28}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\\ N=\left(\frac{17}{28}+\frac{28}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{3}{12}+\frac{2}{12}\right)\\ N=\left(\frac{17}{28}+\frac{28}{29}-\frac{19}{30}-\frac{20}{31}\right).0\\ N=0\end{array}$

Vậy $M+N=\frac{31}{1000}+0=\frac{31}{1000}$

Đáp án cần chọn là: C

$\begin{array}{l}F=\left(-\frac{9}{25}\right).\frac{53}{3}-{\left(-\frac{3}{5}\right)}^2.\frac{22}{3}\\ =-\frac{9}{25}.\frac{53}{3}-\frac{9}{25}.\frac{22}{3}\\ =-\frac{9}{25}.\left(25\right)\\ =-9\end{array}$

$\begin{array}{l}\begin{array}{l}P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)\mathrm{...}\\ \mathrm{...}\left(1-\frac{1}{999}\right).\left(1-\frac{1}{1000}\right)\end{array}\\ =\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\mathrm{.....}\frac{998}{999}.\frac{999}{1000}\\ =\frac{\mathrm{1.2.3.....999}}{\mathrm{2.3.4...999.1000}}=\frac{1}{1000}\end{array}$

$F+P=\frac{1}{1000}+(-9)=-\frac{8999}{1000}$

Đáp án cần chọn là: A

$\begin{array}{l}B=\frac{2^2}{3}\cdot \frac{3^2}{8}\cdot \frac{4^2}{15}\cdot \frac{5^2}{24}\cdot \frac{6^2}{35}\cdot \frac{7^2}{48}\cdot \frac{8^2}{63}\cdot \frac{9^2}{80}\\ =\frac{2.2}{1.3}\cdot \frac{3.3}{2.4}\cdot \frac{4.4}{3.5}\cdot \frac{5.5}{4.6}\cdot \frac{6.6}{5.7}\cdot \frac{7.7}{6.8}\cdot \frac{8.8}{7.9}\cdot \frac{9.9}{8.10}\\ =\frac{\mathrm{2.3.4.5.6.7.8.9}}{\mathrm{1.2.3.4.5.6.7.8.}}\cdot \frac{\mathrm{2.3.4.5.6.7.8.9}}{\mathrm{3.4.5.6.7.8.9.10}}\\ =\frac{9}{1}\cdot \frac{2}{10}=\frac{9.2}{1.10}=\frac{9}{5}\end{array}$

Đáp án cần chọn là: B

$\begin{array}{l}M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}\mathrm{.....}\frac{{99}^2}{99.100}.\frac{{100}^2}{101}\\ =\frac{1^2{.2}^2{.3}^2{\mathrm{....99}}^2{.100}^2}{\left(1.2\right).\left(2.3\right).\left(3.4\right)\mathrm{....}\left(99.100\right).\left(100.101\right)}\\ =\frac{\left(\mathrm{1.2.3.....99.100}\right).\left(\mathrm{1.2.3....99.100}\right)}{\left(\mathrm{1.2.3.....99.100}\right).\left(\mathrm{2.3....99.100}\right).101}\\ =\frac{1}{101}\end{array}$

Đáp án cần chọn là: A

$\begin{array}{l}A=\left(\frac{11}{4}.\frac{-5}{9}-\frac{4}{9}.\frac{11}{4}\right).\frac{8}{33}\\ =\frac{11}{4}.\left(\frac{-5}{9}-\frac{4}{9}\right).\frac{8}{33}=\frac{11}{4}.\frac{-9}{9}.\frac{8}{33}\\ =\frac{-11}{4}.\frac{8}{33}=\frac{-2}{3}\end{array}$

Đáp án cần chọn là: B

$\begin{array}{l}G=\left(\frac{1}{57}-\frac{1}{5757}+\frac{1}{35}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\\ =\left(\frac{1}{57}-\frac{1}{5757}+\frac{1}{35}\right).0\\ =0\end{array}$

Đáp án cần chọn là: C

$\begin{array}{l}\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\mathrm{...}+\frac{1}{x(x+1):2}=\frac{2019}{2021}\\ 2.[\frac{1}{2.3}+\frac{1}{3.4}+\mathrm{...}+\frac{1}{x(x+1)}]=\frac{2019}{2021}\\ 2.(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\mathrm{...}+\frac{1}{x}-\frac{1}{x+1})=\frac{2019}{2021}\\ 2.(\frac{1}{2}-\frac{1}{x+1})=\frac{2019}{2021}\\ 1-\frac{2}{x+1}=\frac{2019}{2021}\\ \frac{2}{x+1}=1-\frac{2019}{2021}\\ \frac{2}{x+1}=\frac{2}{2021}\\ x+1=2021\\ x=2020\end{array}$

Đáp án cần chọn là: D

$\begin{array}{l}M=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{...}+\frac{1}{2^{99}}+\frac{1}{2^{100}}\\ 2M=2.\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{...}+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)\\ =2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{...}+\frac{1}{2^{99}}\end{array}$

Ta có : 

$\begin{array}{l}\\ \begin{array}{l}M=2M-M\\ =\left(2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{...}+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\\ -\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\mathrm{...}+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)\\ =2-\frac{1}{2^{100}}\\ =\frac{2^{101}-1}{2^{100}}\end{array}\end{array}$