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a. 857+3,15+127+4,35
=\(\dfrac{61}{7}+\dfrac{63}{20}+\dfrac{9}{7}+\dfrac{87}{20}\)
=\(\left(\dfrac{61}{7}+\dfrac{9}{7}\right)+\left(\dfrac{63}{20}+\dfrac{87}{20}\right)\)
=\(10+\dfrac{15}{2}\)
=\(\dfrac{35}{2}\)
b. (4523−225+7713)−(3523−6613)
=\(4\dfrac{5}{23}-2\dfrac{2}{5}+7\dfrac{7}{13}-3\dfrac{5}{23}+6\dfrac{6}{13}\)
=\(\left(4\dfrac{5}{23}-3\dfrac{5}{23}\right)+\left(7\dfrac{7}{13}+6\dfrac{6}{13}\right)-2\dfrac{2}{5}\)
=\(1+14-\dfrac{12}{5}\)
=15-\(\dfrac{12}{5}\)
=\(\dfrac{63}{5}\)
Câu C khó khó mình chưa giải được !!!
\(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
\(=\left(4\dfrac{3}{4}+\dfrac{1}{8}+3\dfrac{1}{12}\right)+\left[\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)\right]\)
\(=\left(\dfrac{19}{4}+\dfrac{1}{8}+\dfrac{37}{12}\right)+\left(-4,15\right)\)
\(=\dfrac{191}{24}-4,15\)
\(=\dfrac{457}{120}=3\dfrac{97}{120}\)
\(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)
\(=\dfrac{19}{4}+\left(-\dfrac{37}{100}\right)+\dfrac{1}{8}+\left(-\dfrac{32}{25}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{475}{100}+\left(-\dfrac{37}{100}\right)+\dfrac{1}{8}+\left(-\dfrac{32}{25}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{219}{50}+\dfrac{25}{200}+\left(-\dfrac{256}{200}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)
=\(\dfrac{219}{50}+\left(-\dfrac{231}{200}\right)+\left(-\dfrac{30}{12}+\dfrac{37}{12}\right)\)
=
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
\(a.\)
\(\dfrac{1}{3}\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+\dfrac{3}{5}=-5x+x\)
\(\Rightarrow-4x=-\dfrac{37}{30}\)
\(\Rightarrow4x=\dfrac{37}{30}\)
\(\Rightarrow x=\dfrac{37}{120}\)
\(b.\)
\(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{3}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{3}{2}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{2}\)
\(c.\)
\(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\)
\(\Rightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\)
\(\Rightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)
\(\Rightarrow x=\dfrac{19}{21}\)
\(d.\)
\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{39}{10}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{39}{10}+1\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{49}{10}\)
\(\Rightarrow x=\dfrac{49}{5}\)
\(e.\)
\(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=\dfrac{3}{5}\\4x+3=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{12}{5}\\4x=-\dfrac{18}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{9}{10}\end{matrix}\right.\)
a) \(\dfrac{1}{3}.\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\Leftrightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)
\(\Leftrightarrow5x-x=-\dfrac{3}{5}-\dfrac{1}{6}+2\Leftrightarrow4x=\dfrac{37}{30}\Leftrightarrow x=\dfrac{\dfrac{37}{30}}{4}=\dfrac{37}{120}\)
vậy \(x=\dfrac{37}{120}\)
b) \(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\Leftrightarrow\dfrac{3}{2}x-x=\dfrac{-3}{4}+1\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}x=\dfrac{-3}{4}+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.2=\dfrac{6}{4}=\dfrac{3}{2}\) vậy \(x=\dfrac{3}{2}\)
c) \(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\Leftrightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\Leftrightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)
\(\Leftrightarrow x=\dfrac{\dfrac{19}{12}}{\dfrac{7}{4}}=\dfrac{19}{21}\) vậy \(x=\dfrac{19}{21}\)
d) \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\Leftrightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)
\(\Leftrightarrow\dfrac{3}{2}x-x=1+\dfrac{5}{2}+\dfrac{7}{5}\Leftrightarrow\dfrac{1}{2}x=\dfrac{49}{10}\Leftrightarrow x=\dfrac{49}{10}.2=\dfrac{49}{5}\)
vậy \(x=\dfrac{49}{5}\)
e) \(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\Leftrightarrow\left|4x+3\right|=\dfrac{\dfrac{6}{15}}{\dfrac{2}{3}}=\dfrac{3}{5}\)
th1 : \(4x+3\ge0\Leftrightarrow4x\ge-3\Leftrightarrow x\ge\dfrac{-3}{4}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow4x+3=\dfrac{3}{5}\Leftrightarrow4x=\dfrac{3}{5}-3=\dfrac{-12}{5}\)
\(\Leftrightarrow x=\dfrac{\dfrac{-12}{5}}{4}=\dfrac{-3}{5}\left(tmđk\right)\)
th2: \(4x+3< 0\Leftrightarrow4x< -3\Leftrightarrow x< \dfrac{-3}{4}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow-\left(4x+3\right)=\dfrac{3}{5}\Leftrightarrow-4x-3=\dfrac{3}{5}\)
\(\Leftrightarrow4x=-3-\dfrac{3}{5}=\dfrac{-18}{5}\Leftrightarrow x=\dfrac{\dfrac{-18}{5}}{4}=\dfrac{-9}{10}\left(tmđk\right)\)
vậy \(x=\dfrac{-3}{5};x=\dfrac{-9}{10}\)
\(\dfrac{3}{-2}=\dfrac{-9}{6};\dfrac{-1}{-7}=\dfrac{1}{7}\)
Sắp xếp:
\(\dfrac{-1}{-7};\dfrac{0}{8};\dfrac{-7}{6};\dfrac{3}{-2}\)
b) \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
a) goi ƯCLN(n,n+1) là d
ta co : n ⋮ d ; n+1 ⋮d (1)
⇒ (n+1)-n ⋮ d
⇒1 ⋮ d (2)
Từ (1) và (2) ⇒ d = 1 hoac -1
Vậy \(\dfrac{n}{n+1}\) là phân số tối giản.
b) goi UCLN (n+1,2n+3)la d
=>(2n+3) - (n+1)⋮d
=>(2n+3) - [ 2(n+1)] ⋮ d
=>(2n+3)-(2n+2)⋮d
=>2n+3-2n-2 ⋮ d
=>1 ⋮ d => d=1
vay \(\dfrac{n+1}{2n+3}\) là phân số tối giản.
`(x-3/8) +1/4= 7/8`
`=> x-3/8 = 7/8-1/4`
`=> x-3/8 = 7/8-2/8`
`=> x-3/8 =5/8`
`=>x= 5/8 +3/8`
`=>x= 8/8=1`