A:-1/4

B:0

C:0

D-11/17

bạn có cần cả bài giải ko?

ĐS. a) b) ;

c) ; d)

\(\left(\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{30}\right).\left(\dfrac{21}{22}+\dfrac{22}{23}+...+\dfrac{102}{103}\right)\)

\(=0.\left(\dfrac{21}{22}+\dfrac{22}{23}+...+\dfrac{102}{103}\right)\)

\(=0\)

vì \(\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{30}\)=0 nên

(15−16−13015−16−130 ).(2122+2223+......+102103)=0

a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)

\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)

\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)

\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)

\(=5\dfrac{4}{23}.23\)

\(=\dfrac{119}{23}.23\)

\(=\dfrac{119}{23}\)

b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)

\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)

\(=\dfrac{-4}{6}+\dfrac{3}{2}\)

\(=\dfrac{-2}{3}+\dfrac{3}{2}\)

\(=\dfrac{-4}{6}+\dfrac{9}{6}\)

\(=\dfrac{5}{6}\)

c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)

\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)

\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)

\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)

\(=1-1\)

\(=0\)

d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)

(đợi đã, mình chưa tìm được hướng làm...)

quy đồng lên

d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B

cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A

Suy ra B>A(chuc ban hoc goi nhe)

a)

\(\dfrac{-2}{3}\cdot\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\cdot\left(2x-1\right)\\ \dfrac{-2}{3}x-\left(\dfrac{-1}{6}\right)=\dfrac{2}{3}x-\dfrac{1}{3}\\ \dfrac{-2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\\ \dfrac{-2}{3}x-\dfrac{2}{3}x=\dfrac{-1}{3}-\dfrac{1}{6}\\ -x\cdot\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{-2}{6}-\dfrac{1}{6}\\ \dfrac{-4}{3}x=\dfrac{-1}{2}\\ x=\dfrac{-1}{2}:\dfrac{-4}{3}\\ x=\dfrac{-1}{2}\cdot\dfrac{-3}{4}\\ x=\dfrac{3}{8}\)

a, \(2-\dfrac{14}{x}=\dfrac{-22}{3}\)

\(\dfrac{14}{x}=2-\dfrac{-22}{3}=\dfrac{28}{3}\)

\(\dfrac{14}{x}=\dfrac{28}{3}\)

=> \(x.28=14.3\)

\(x.28=42\)

\(x=42:28\)

\(x=\dfrac{3}{2}=1,5\)

b, \(\left(\dfrac{2x}{5}+1\right):\left(-7\right)=\dfrac{1}{35}\)

\(\dfrac{2x}{5}+1=\dfrac{1}{35}.\left(-7\right)=-\dfrac{1}{5}\)

\(\dfrac{2x}{5}=-\dfrac{1}{5}-1=\dfrac{-6}{5}\)

\(\dfrac{2x}{5}=\dfrac{-6}{5}\)

=> \(2x=-6\)

\(x=-6:2=-3\)

a)

\(2-\dfrac{14}{x}=-\dfrac{22}{3}\)

\(\Rightarrow\dfrac{14}{x}=2-\dfrac{-22}{3}=\dfrac{28}{3}\)

\(\Rightarrow x=\dfrac{14.3}{28}=\dfrac{3}{2}=1,5\)

b)

\(\left(\dfrac{2x}{5}+1\right):\left(-7\right)=\dfrac{1}{35}\)

\(\Rightarrow\dfrac{2x}{5}+1=\dfrac{1}{35}.\left(-7\right)\)

\(\Rightarrow\dfrac{2x}{5}+1=-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{2x}{5}=-\dfrac{1}{5}-1=-\dfrac{6}{5}\)

Hay \(\dfrac{2x}{5}=-\dfrac{6}{5}\)

\(\Rightarrow2x=-6\)

\(\Rightarrow x=-\dfrac{6}{2}=-3\)

Chúc bạn học tốt!