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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)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)
\(\dfrac{3}{10}-x=\dfrac{7}{10}\)
x = \(\dfrac{3}{10}-\dfrac{7}{10}\)
x=\(\dfrac{-4}{10}\)
b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)
\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)
\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)
\(x=\dfrac{-64}{9}\)
c)=>2.18=(x-3).(x-3)
=>36=(x-3)\(^2\)
=>6\(^2\)=(x-3)\(^2\)
6= x-3
x=6+3=9
a)\(=\dfrac{211}{180}\)
b)\(=\dfrac{5}{39}\)
c)=\(=-\dfrac{65}{168}\)
a: \(=\dfrac{6}{18}+\dfrac{3}{18}+\dfrac{2}{18}+\dfrac{1}{18}=\dfrac{15}{18}=\dfrac{5}{6}\)
b: \(=\dfrac{7}{4}-\dfrac{5}{4}-\dfrac{2}{4}=0\)
c: \(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
=1/2-1/10=4/10=2/5
Bài 2:
a: \(=44\cdot82-400+18\cdot44\)
\(=44\cdot100-400=4400-400=4000\)
b: \(=6^2:\left\{780:\left[390-125\cdot49+65\right]\right\}\)
\(=36:\left\{780:\left[-5670\right]\right\}\)
\(=36:\dfrac{-26}{189}=\dfrac{-3402}{13}\)
\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)=-\dfrac{1}{6}\)
\(\left(\dfrac{1}{5}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\dfrac{1}{5^{12}}.\dfrac{1}{4^{20}}=5^{-12}.4^{-20}=125^{-4}.1024^{-4}=\left(125.1024\right)^{-4}=128000^{-4}\)
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.2^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\dfrac{2.6}{3.7}=\dfrac{4}{7}\)
\(a.\dfrac{3}{5}-\dfrac{-7}{10}-\dfrac{13}{-20}=\dfrac{12}{20}-\dfrac{-14}{20}-\dfrac{-13}{20}=\dfrac{12-\left(-14\right)-\left(-13\right)}{20}=\dfrac{39}{20}\)
\(b.\dfrac{3}{4}+\dfrac{-1}{3}-\dfrac{5}{18}=\dfrac{3}{4}+\left(\dfrac{-6}{18}-\dfrac{5}{18}\right)=\dfrac{3}{4}+\dfrac{-11}{18}=\dfrac{27}{36}-\dfrac{-22}{36}=\dfrac{49}{36}\)
\(c.\dfrac{3}{13}-\dfrac{5}{-8}+\dfrac{-1}{2}=\dfrac{3}{13}-\left(\dfrac{5}{-8}+\dfrac{-4}{8}\right)=\dfrac{3}{13}-\dfrac{1}{8}=\dfrac{24}{104}-\dfrac{13}{104}=\dfrac{11}{104}\)
\(d.\dfrac{1}{2}+\dfrac{1}{-3}=\dfrac{3}{6}+\dfrac{-2}{6}=\dfrac{1}{6}\)
\(a,\dfrac{3}{5}-\dfrac{-7}{10}-\dfrac{13}{-20}\)
\(=\dfrac{12}{20}+\dfrac{14}{20}+\dfrac{13}{20}\)
\(=\dfrac{12+14+13}{20}\)
\(=\dfrac{39}{20}\)
\(b,\dfrac{3}{4}+\dfrac{-1}{3}-\dfrac{5}{18}\)
\(=\dfrac{27}{36}+\dfrac{-12}{36}-\dfrac{10}{36}\)
\(=\dfrac{27+\left(-12\right)-10}{36}\)
\(=\dfrac{5}{36}\)
\(c,\dfrac{3}{13}-\dfrac{5}{-8}+\dfrac{-1}{2}\)
\(=\dfrac{24}{104}-\dfrac{-65}{104}+\dfrac{-52}{104}\)
\(=\dfrac{24-\left(-65\right)+\left(-52\right)}{104}\)
\(=\dfrac{37}{104}\)
\(d,\dfrac{1}{2}+\dfrac{1}{-3}\)
\(=\dfrac{3}{6}+\dfrac{-2}{6}\)
\(=\dfrac{3+\left(-2\right)}{6}\)
\(=\dfrac{1}{6}\)
Vì 18/91 < 18/90 =1/5
23/114>23115=1/5
vậy 18/91<1/5<23/114
suy ra 18/91<23/114
vì 21/52=210/520
Mà 210/520=1-310/520
213/523=1-310/523
310/520>310/523
vậy 210/520<213/523
suy ra 21/52<213/523
Ta quy về dạng tổng quát xét cho dễ nhé.
\(\dfrac{1}{x\cdot\left(x+2\right)}=\dfrac{1}{2}.\dfrac{2}{x.\left(x+2\right)}=\dfrac{1}{2}.\left(\dfrac{1}{x}-\dfrac{1}{x-2}\right)\)
Từ đó áp dụng dạng tổng quát để rút gọn là ra.
Chúc em học tốt!