Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có :
\(0,\left(27\right)+0,\left(72\right)==\dfrac{27}{99}+\dfrac{72}{99}=\dfrac{99}{99}=1\)
\(\Rightarrow0,\left(27\right)+0,\left(72\right)=1\rightarrowđpcm\)
b) Ta có :
\(0,\left(22\right).\dfrac{9}{2}=\dfrac{2}{9}.\dfrac{9}{2}=\dfrac{18}{18}=1\)
\(\Rightarrow0,22.\dfrac{9}{2}=1\rightarrowđpcm\)
c) Ta có :
\(\left[0,\left(11\right).9\right]^{2003}=\left[\dfrac{1}{9}.9\right]^{2003}=\left[\dfrac{9}{9}\right]^{2003}=1^{2003}=1\)
\(\Rightarrow\left[0,\left(11\right).9\right]^{2003}=1\rightarrowđpcm\)
a. \(\dfrac{\left(x+1\right)}{10}+\dfrac{\left(x+1\right)}{11}+\dfrac{\left(x+1\right)}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)
\(\Rightarrow x+1=0\)
\(x=-1\)
b, \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\\ \left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\\ \dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\\ x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)\)
vì \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\\ \Rightarrow x+2004=0\\ x=-2004\)
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Vì \(10< 11< 12< 13< 14\) nên \(\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}>\dfrac{1}{14}\)
\(\Rightarrow\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}>0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
Vậy.................
b, \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Vì \(2000< 2001< 2002< 2003\) nên \(\dfrac{1}{2000}>\dfrac{1}{2001}>\dfrac{1}{2002}>\dfrac{1}{2003}\)
\(\Rightarrow\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}>0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
Vậy.................
Chúc bạn học tốt!!!
a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)
= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)
= \(\dfrac{-1}{24}-\dfrac{5}{8}\)
= \(\dfrac{-2}{3}\)
b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)
= \(8\dfrac{5}{8}\)
c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)
= \(-49\)
d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)
= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)
= 0
\(\dfrac{4}{9}:\left(\dfrac{-1}{7}\right)+6\dfrac{5}{9}.\left(\dfrac{2}{3}\right)\)
\(=\dfrac{4}{9}.\left(-7\right)+\dfrac{59}{9}.\dfrac{2}{3}\)
\(=\dfrac{2}{9}.\left(-14\right)+\dfrac{2}{9}.\dfrac{59}{3}\)
\(=\dfrac{2}{9}.\left(-14+\dfrac{59}{3}\right)\)
\(=\dfrac{2}{9}.\dfrac{17}{3}\)
\(=\dfrac{34}{27}\)
\(\left(\dfrac{-1}{3}\right)^2.\dfrac{4}{11}+\dfrac{7}{11}.\left(\dfrac{-1}{3}\right)^2\)
\(=\dfrac{1}{9}.\dfrac{4}{11}+\dfrac{7}{11}.\dfrac{1}{9}\)
\(=\dfrac{1}{9}.\left(\dfrac{4}{11}+\dfrac{7}{11}\right)\)
\(=\dfrac{1}{9}.1=\dfrac{1}{9}\)
~ \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}.\dfrac{2}{3}\)
\(=\dfrac{4}{9}.\left(-7\right)+\dfrac{59}{9}.\dfrac{2}{3}\)
\(=-\dfrac{28}{9}+\dfrac{118}{27}\)
\(=-\dfrac{84}{27}+\dfrac{118}{27}\)
\(=\dfrac{34}{27}\)
~ \(\left(-\dfrac{1}{3}\right)^2.\dfrac{4}{11}+\dfrac{7}{11}.\left(-\dfrac{1}{3}\right)^2\)
\(=\left(-\dfrac{1}{3}\right)^2.\left(\dfrac{4}{11}+\dfrac{7}{11}\right)\)
\(=\dfrac{1}{9}.\dfrac{11}{11}\)
\(=\dfrac{1}{9}.1\)
\(=\dfrac{1}{9}\)
\(\left(\dfrac{4}{9}-\dfrac{5}{11}\right):\dfrac{3}{10}+\left(\dfrac{3}{9}-\dfrac{9}{11}\right):\dfrac{3}{10}-\left(\dfrac{2}{9}-\dfrac{8}{11}\right)\cdot\left(-\dfrac{10}{3}\right)\\ =\left(\dfrac{4}{9}-\dfrac{5}{11}\right)\cdot\dfrac{10}{3}+\left(\dfrac{3}{9}-\dfrac{9}{11}\right)\cdot\dfrac{10}{3}-\left(\dfrac{2}{9}-\dfrac{8}{11}\right)\cdot\left(-1\right)\cdot\dfrac{10}{3}\\ =\left(\dfrac{4}{9}-\dfrac{5}{11}\right)\cdot\dfrac{10}{3}+\left(\dfrac{3}{9}-\dfrac{9}{11}\right)\cdot\dfrac{10}{3}+\left(\dfrac{2}{9}-\dfrac{8}{11}\right)\cdot\dfrac{10}{3}=\dfrac{10}{3}\cdot\left(\dfrac{4}{9}-\dfrac{5}{11}+\dfrac{3}{9}-\dfrac{9}{11}+\dfrac{2}{9}-\dfrac{8}{11}\right)\\ =\dfrac{10}{3}\cdot\left(-1\right)\\ =-\dfrac{10}{3}\)
Ta có:(4/9-5/11):3/10+(3/9-9/11):3/10-(2/9-8/11).(-10/3)
=[(4/9-5/11)+(3/9-9/11)]:3/10+(-2/9+8/11).(-10/3)
=[(4/9+3/9)+(-5/11-9/11)]:3/10+(-2/9+8/11):(-3/10)
=(7/9-14/11):3/10+(2/9-8/11):3/10 (nhân chuyển dấu)
=[(7/9-14/11)+(2/9-8/11)]:3/10
=(1-2):3/10
=-1.10/3
=-10/3.
*Trả lời :
a) \(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)
= \(-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\dfrac{-68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)
=\(\dfrac{3}{4}.\cdot\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)
=\(\dfrac{3}{4}.\left(-8\right)\)
= \(-6\)
b)\(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{41}{9}-\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
=\(\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\left(-\dfrac{5}{7}\right)\)
=\(\dfrac{90}{9}:\left(-\dfrac{5}{7}\right)\)
=\(10:\left(-\dfrac{5}{7}\right)\)
=\(-14\)
c)\(\left(-\dfrac{3}{5}+\dfrac{4}{9}\right):\dfrac{7}{11}+\left(-\dfrac{2}{5}+\dfrac{5}{9}\right):\dfrac{7}{11}\)
=\(\left(-\dfrac{3}{5}\right)+\dfrac{4}{9}:\dfrac{7}{11}+\left(-\dfrac{2}{5}\right)+\dfrac{5}{9}:\dfrac{7}{11}\)(áp dụng tính chất phá ngoặc )
=\(\left\{\left[-\dfrac{3}{5}+\left(-\dfrac{2}{5}\right)\right]+\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\right\}:\dfrac{7}{11}\)
=\(\left(-\dfrac{5}{5}+\dfrac{9}{9}\right):\dfrac{7}{11}\)
=\(\left(-1+1\right):\dfrac{7}{11}\)
\(=0:\dfrac{7}{11}\)
=0.
d)\(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}:\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)
=\(\dfrac{6}{7}:\left[\dfrac{3}{26}+\left(-\dfrac{6}{26}\right)\right]+\dfrac{6}{7}:\left[\dfrac{1}{10}+\left(-\dfrac{16}{10}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{3}{26}\right)+\dfrac{6}{7}:\left(-\dfrac{3}{2}\right)\)
=\(\dfrac{6}{7}:\left[\left(-\dfrac{3}{26}\right)+\left(-\dfrac{39}{26}\right)\right]\)
=\(\dfrac{6}{7}:\left(-\dfrac{21}{13}\right)\)
=\(-\dfrac{26}{49}\)
