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\(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2008}-1\right).\left(\dfrac{1}{2009}-1\right)\)
\(=-\dfrac{1}{2}.-\dfrac{2}{3}.-\dfrac{3}{4}...-\dfrac{2007}{2008}.-\dfrac{2008}{2009}\)
\(=-\left(\dfrac{1.2.3...2007.2008}{2.3.4...2008.2009}\right)\)
\(=-\dfrac{1}{2009}\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\Leftrightarrow x=-2004\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
=>\(\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
=>\(\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
=>\(\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
=>x+2004=0
=>x=-2004
Người thợ săn trộn số lít sơn xanh và trắng là :
3,8 + 0,7 = 4,5 ( l )
Người thợ sơn còn lại số lít sơn xanh nhạt là :
4,5 - 1,25 = 3,25 ( l )
Đáp số : 3,25l sơn xanh nhạt
a) \(\dfrac{-3}{5}+\dfrac{28}{5}.\left(\dfrac{13}{56}-\dfrac{5}{24}+\dfrac{1}{7}\right)\\ =-\dfrac{3}{5}+\dfrac{28}{5}.\dfrac{13}{56}-\dfrac{28}{5}.\dfrac{5}{24}+\dfrac{28}{5}.\dfrac{1}{7}\\ =-\dfrac{3}{5}+\dfrac{28.13}{5.2.28}-\dfrac{4.7.5}{5.4.6}+\dfrac{4.7.1}{5.7}\\ =-\dfrac{3}{5}+\dfrac{13}{10}-\dfrac{7}{6}+\dfrac{4}{5}\\ =\dfrac{1}{5}+\dfrac{13}{10}-\dfrac{7}{6}\\ =\dfrac{6}{30}+\dfrac{39}{30}-\dfrac{35}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)
b) \(\dfrac{5}{7}.\dfrac{11}{18}+\dfrac{3}{7}.\dfrac{5}{8}+\dfrac{4}{9}\\ =\dfrac{5}{7}.\dfrac{11}{18}+\dfrac{5}{7}.\dfrac{3}{8}+\dfrac{4}{9}\\ =\dfrac{5}{7}.\left(\dfrac{11}{18}+\dfrac{3}{8}\right)+\dfrac{4}{9}\\ =\dfrac{5}{7}.\dfrac{7}{4}+\dfrac{4}{9}\\ =\dfrac{5}{4}+\dfrac{4}{9}=\dfrac{61}{36}\)
c) \(4\dfrac{5}{9}:\left(\dfrac{-5}{7}\right)+\dfrac{49}{9}:\left(\dfrac{-5}{7}\right)\\=\left(4\dfrac{5}{9}+\dfrac{49}{9}\right):\dfrac{-5}{7}\\ =\left(\dfrac{41}{9}+\dfrac{49}{9}\right).\dfrac{-7}{5}\\ =10.\dfrac{-7}{5}=-14\)
b)Ta có: \(x-y=7\Rightarrow x=y+7\)
\(\dfrac{2x+7}{3x-y}=\dfrac{2\left(y+7\right)+7}{3\left(y+7\right)-y}=\dfrac{2y+21}{2y+21}=1\)
\(\dfrac{2y-7}{3y-x}=\dfrac{2y-7}{3y-\left(y+7\right)}=\dfrac{2y-7}{2y-7}=1\)
\(\Rightarrow C=\dfrac{2x+7}{3x-y}+\dfrac{3y-7}{3y-x}=1+1=2\)
\(\left(\dfrac{1}{4}-\dfrac{2}{3}\right):\dfrac{3x}{5}=\dfrac{5}{2}\Leftrightarrow\dfrac{-5}{12}:\dfrac{3x}{5}=\dfrac{5}{2}\Leftrightarrow\dfrac{3x}{5}=-\dfrac{1}{6}\Leftrightarrow x=-\dfrac{5}{18}\)
5A:
\(x-128=\left(4\dfrac{20}{21}-5\right):\left(\dfrac{4141}{4242}-1\right):\left(\dfrac{636364}{646464}-1\right)\)
\(\Rightarrow x-128=\left(\dfrac{104}{21}-5\right):\left(\dfrac{41}{42}-1\right):\left(\dfrac{63}{64}-1\right)\)
\(\Rightarrow x-128=-\dfrac{1}{21}:\left(-\dfrac{1}{42}\right):\left(-\dfrac{1}{64}\right)\)
\(\Rightarrow x-128=-\dfrac{1}{21}.\left(-42\right).\left(-64\right)\)
\(\Rightarrow x-128=-128\)
\(\Rightarrow x=-128+128=0\)
6A:
Vì \(15558000< 21960000< 70560000< 85133000< 168723000\)
nên thứ tự tăng dần của các đại dương đó là:
Bắc Băng Dương; Nam Đại Dương; Ấn Độ Dương; Đại Tây Dương; Thái Bình Dương
4A:
\(\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}+\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{2}{193}.\dfrac{193}{17}-\dfrac{3}{386}.\dfrac{193}{17}+\dfrac{33}{34}\right):\left(\dfrac{7}{1931}.\dfrac{1931}{25}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right)\)
\(=\left(\dfrac{2}{17}-\dfrac{3}{34}+\dfrac{33}{34}\right):\left(\dfrac{7}{25}+\dfrac{11}{50}+\dfrac{9}{2}\right)\)
\(=\left(\dfrac{4}{34}+\dfrac{30}{34}\right):\left(\dfrac{14}{50}+\dfrac{11}{50}+\dfrac{9}{2}\right)\)
\(=\dfrac{34}{34}:\left(\dfrac{1}{2}+\dfrac{9}{2}\right)\)
\(=1:\dfrac{10}{2}=\dfrac{1}{5}\)
4B:
a, \(A=\dfrac{1,11+0,19-13.2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\)
\(=\dfrac{1,3-26}{2,6}-\dfrac{3}{4}:2\)
\(=\dfrac{-24,7}{2,6}-\dfrac{3}{8}=\dfrac{-19}{2}-\dfrac{3}{8}=-\dfrac{79}{8}\)
\(B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):2\dfrac{23}{26}\)
\(=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)
\(=\left(\dfrac{29}{8}-\dfrac{1}{2}\right).\dfrac{26}{75}\)
\(=\dfrac{25}{8}.\dfrac{26}{75}=\dfrac{13}{12}\)
b, Để \(A< x< B\) thì \(-\dfrac{79}{8}< x< \dfrac{13}{12}\)
\(\Rightarrow x\in\left\{-9;-8;-7;...;1\right\}\) (vì \(x\in\mathbb{Z}\))
$Toru$
a)
\(\dfrac{4}{5}-\left(\dfrac{-2}{3}\right)-\dfrac{1}{10}-\dfrac{2}{3}\\ =\dfrac{4}{5}+\dfrac{2}{3}-\dfrac{1}{10}-\dfrac{2}{3}\\ =\left(\dfrac{4}{5}-\dfrac{1}{10}\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)\\ =\dfrac{7}{10}+0\\ =\dfrac{7}{10}\)
b)
\(\dfrac{1}{3}-\dfrac{-1}{2}+\dfrac{1}{13}-\dfrac{5}{6}\\ =\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\dfrac{1}{13}-\dfrac{5}{6}\\ =\dfrac{5}{6}+\dfrac{1}{13}-\dfrac{5}{6}\\ =\dfrac{1}{13}\)
c)
\(\dfrac{-5}{12}-\left(\dfrac{-5}{6}-\dfrac{5}{12}\right)\\ =\dfrac{-5}{12}+\dfrac{5}{6}+\dfrac{5}{12}\\ =\left(-\dfrac{5}{12}+\dfrac{5}{12}\right)+\dfrac{5}{6}\\ =\dfrac{5}{6}\)