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1) Tìm x
\(\frac{11}{2}.x+\frac{1}{3}.x=1\)
\(\Rightarrow x\left(\frac{11}{2}+\frac{1}{3}\right)=1\)
\(\Rightarrow x\left(\frac{33}{6}+\frac{2}{6}\right)=1\)
\(\Rightarrow x.\frac{35}{6}=1\)
\(\Rightarrow x=\frac{6}{35}\)
2) So sánh
\(\frac{59}{40}< \frac{50}{31}\)( cái này bạn quy đồng là ra, mik chỉ ghi kq, bạn tự tính )
3)\(\frac{1}{3}+\frac{4}{7}-\frac{5}{14}-\frac{1}{2}-\frac{2}{3}\)
\(=\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{4}{7}-\frac{5}{14}\right)-\frac{1}{2}\)
\(=-\frac{1}{3}+\frac{3}{14}-\frac{1}{2}\)
\(=-\frac{13}{21}\)
1)\(\frac{11}{2}.x+\frac{1}{3}.x=1\)
\(x.\left(\frac{11}{2}+\frac{1}{3}=1\right)\)
\(x.\frac{35}{6}=1\)
\(x=1:\frac{35}{6}\)
\(x=\frac{6}{35}\)
2) Ta có:
\(\frac{59}{40}=\frac{1829}{1240}\)
\(\frac{50}{31}=\frac{2000}{1240}\)
Vì \(2000>1829\Rightarrow\frac{2000}{1240}>\frac{1829}{1240}\Rightarrow\frac{50}{31}>\frac{59}{40}\)
3)\(\frac{1}{3}+\frac{4}{7}-\frac{5}{14}-\frac{1}{2}-\frac{2}{3}\)
\(=\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{4}{7}-\frac{5}{14}-\frac{1}{2}\right)\)
\(=-\frac{1}{3}+\left(\frac{8}{14}-\frac{5}{14}-\frac{7}{14}\right)\)
\(=\frac{-1}{3}+\frac{-4}{14}\)
\(=\frac{-1}{3}+\frac{-2}{7}\)
\(=\frac{-7}{21}+\frac{-6}{21}\)
\(=\frac{-13}{21}\)
A = \(\frac{1}{3}-\frac{3}{4}-\frac{-3}{5}+\frac{1}{73}-\frac{1}{36}+\frac{1}{15}+\frac{-2}{9}\)
A = \(\left(\frac{1}{3}-\frac{2}{9}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)+\frac{1}{73}\)
A = \(\left(\frac{3-2}{9}\right)-\left(\frac{27+1}{36}\right)+\left(\frac{9+1}{15}\right)+\frac{1}{73}\)
A = \(\frac{1}{9}-\frac{7}{9}+\frac{6}{9}+\frac{1}{73}\)
A = \(0+\frac{1}{73}=\frac{1}{73}\)
Làm
B = 1/3 - 3/4 - (-3)/5 + 1/73 - 1/36 + 1/15 + -2/9
B = 1/3 -3/4 + 3/5 +1/73 - 1/36 + 1/15 -2/9
B = [ 1/3 + 3/5 + 1/15 ] + [ -3/4 - 1/36 -2/9] + 1/73
B = [ 5/15 + 9/15 + 1/15 ] + [ -27/36 - 8/36 - 1/36 ] + 1/73
B = 1 + (-1) + 1/73
B = 1/73
HỌC TỐT Ạ
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)