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Bài 1:
\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)
\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)
\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)
Bài 2:
a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)
\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)
b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
Bài 3:
\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)
\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)
Bài 4:
\(\dfrac{3}{4}-x=1\)
\(\Rightarrow-x=1-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
\(x+4=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-4\)
\(\Rightarrow x=-\dfrac{19}{5}\)
Vậy: \(x=-\dfrac{19}{5}\)
\(x-\dfrac{1}{5}=2\)
\(\Rightarrow x=2+\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{11}{5}\)
Vậy: \(x=\dfrac{11}{5}\)
\(x+\dfrac{5}{3}=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)
\(\Rightarrow x=-\dfrac{134}{81}\)
Vậy: \(x=-\dfrac{134}{81}\)
\(\left(\frac{3}{4}+\frac{2}{11}-\frac{8}{13}\right)+\left(\frac{1}{4}-\frac{5}{13}\right)\)
\(=\frac{3}{4}+\frac{2}{11}-\frac{8}{13}+\frac{1}{4}-\frac{5}{13}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{2}{11}-\left(\frac{8}{13}+\frac{5}{13}\right)\)
\(=1+\frac{2}{11}-1\)
\(=\frac{2}{11}\)
\(\left(\frac{9}{53}+\frac{21}{31}\right)+\left(\frac{44}{53}-\frac{16}{7}\right)+\frac{10}{31}\)
\(=\frac{9}{53}+\frac{21}{31}+\frac{44}{53}-\frac{16}{7}+\frac{10}{31}\)
\(=\left(\frac{9}{53}+\frac{44}{53}\right)+\left(\frac{21}{31}+\frac{10}{31}\right)-\frac{16}{7}\)
\(=1+1-\frac{16}{7}\)
\(=2-\frac{16}{7}\)
\(=\frac{14}{7}-\frac{16}{7}\)
\(=\frac{-2}{7}\)
\(\left(\dfrac{21}{31}+\dfrac{-16}{7}\right)+\left(\dfrac{44}{53}+\dfrac{10}{31}\right)+\dfrac{9}{53}.\)
\(\dfrac{21}{31}-\dfrac{16}{7}+\dfrac{44}{53}+\dfrac{10}{31}+\dfrac{9}{53}=\left(\dfrac{21}{31}+\dfrac{10}{31}\right)+\left(\dfrac{44}{53}+\dfrac{9}{53}\right)-\dfrac{16}{7}=1+1-\dfrac{16}{7}=\dfrac{-2}{7}.\)
\(\left(\frac{21}{31}+\frac{2013}{6039}\right)-\left(\frac{44}{53}-\frac{10}{31}\right)-\frac{9}{53}\)
\(=\frac{21}{31}+\frac{1}{3}-\frac{44}{53}+\frac{10}{31}-\frac{9}{53}\)
\(=\left(\frac{21}{31}+\frac{10}{31}\right)+\left(\frac{-44}{53}-\frac{9}{53}\right)+\frac{1}{3}\)
\(=1+\left(-1\right)+\frac{1}{3}\)
\(=\frac{1}{3}\)
a/ \(=\frac{21}{23}+\frac{125}{143}-\frac{101.21}{101.23}-\frac{1001.125}{1001.143}=0\)
b/ \(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}-\frac{10}{21}+\frac{3}{20}=\frac{7}{20}-\frac{1}{5}=\frac{4}{20}\)
c/ \(\frac{C}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(\frac{C}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{21-20}{20.21}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{C}{2}=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\Rightarrow C=\frac{19}{21}\)
Câu 1:
\(=\dfrac{21}{31}+\dfrac{1}{3}-\dfrac{44}{53}+\dfrac{10}{31}-\dfrac{9}{53}=\dfrac{1}{3}\)
Câu 2:
1: =>1,3x=-1,31
=>130x=-131
=>x=-131/130
2: =>(4,5-2x)=11/14:(-11/7)=-7/14=-1/2=-0,5
=>2x=5
=>x=5/2
Bài 1: Ta chỉ cần bỏ ngoặc rồi cộng hai phân số để ra kết quả là số tự nhiên là xong
Bài 2:
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+............+\frac{1}{2003.2004}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.............-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}\)
A = \(1-\frac{1}{2004}\)
A = \(\frac{2004}{2004}-\frac{1}{2004}=\frac{2003}{2004}\)
Mk làm mẫu 1 bài nha
Bài 1 :
a, = (1/4+3/4) - (5/13+8/13)+2/11
= 1 - 1 + 2/11
= 2/11
Tk mk nha
\(A=\frac{21}{31}+\frac{-16}{7}+\frac{44}{53}+\frac{10}{21}+\frac{9}{53} \)
\(A=\left(\frac{16}{7}+\frac{10}{21}\right)+\left(\frac{44}{53}+\frac{9}{53}\right)+\frac{21}{31}\)
\(A=\frac{58}{21}+1+\frac{21}{31}\)
\(A=\frac{100}{21}\)
\(B=6\left(\frac{1}{15.18}+\frac{1}{18.21}+...+\frac{1}{87.90}\right)\)
\(B=6\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=6\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=6.\frac{1}{18}\)
\(B=\frac{1}{3}\)
\(\frac{42}{46}\)\(+\frac{250}{286}+\frac{-125125}{143143}\)
\(=\frac{21}{23}+\frac{125}{143}+\frac{-125}{143}\)
\(=\frac{21}{23}+\left(\frac{125}{143}+\frac{-125}{143}\right)\)
\(=\frac{21}{23}+0\)
\(=\frac{21}{23}\)
k cho mik nha✔
(\(\frac{21}{31}+\frac{-16}{7}\)) \(+\) (\(\frac{44}{53}+\frac{10}{31}\)) \(+\frac{9}{53}\)
\(=\frac{21}{31}+\frac{-16}{7}+\frac{44}{53}+\frac{10}{31}+\frac{9}{53}\)
\(=\)(\(\frac{21}{31}+\frac{10}{31}\)) \(+\)( \(\frac{44}{53}+\frac{9}{53}\))\(+\frac{-16}{7}\)
\(=1+1+\frac{-16}{7}\)
\(=2\:+\frac{-16}{7}\)
\(=\frac{-2}{7}\)