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\(-\frac{5}{6}x-1\frac{13}{19}:3\frac{7}{19}-\left(-0,12\right)=\left(-\frac{1}{2}\right)^2\)
\(\Rightarrow-\frac{5}{6}x-\frac{32}{19}:\frac{64}{19}+0,12=\frac{1}{4}\)
\(\Rightarrow-\frac{5}{6}x-\frac{1}{2}=\frac{1}{4}-0,12=\frac{13}{100}\)
\(\Rightarrow-\frac{5}{6}x=\frac{63}{100}\)
\(\Rightarrow x=-\frac{189}{250}\)
-5/6x-32/19:64/19+3/25=1/4
-5/6x-32/19×19/64+3/25=1/4
-5/6x-1/2+3/25=1/4
-5/6x-31/50=1/4
-5/6x=1/4+31/50
-5/6x=87/100
x=87/100:-5/6
x=87/100×-6/5
x=-261/250
vậy x=-261/250
mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn
a, [ 6 + 1/8 - 1/2 ]: 3/12=
45/8 : 3/12=45/2
b,4/9.(58/3 - 118/3)=
4/9. (-20) =-80/9
c,1/4:1/4-2.(-1/2)=
1 - (-1) = 2
[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)
\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)
Ê hỗn số \(1\frac{19}{19}\)là sai vì 19 vẫn chia típ đc cho 19
\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=2\)
\(\Rightarrow2\left(1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=2-\frac{2}{3}\)
\(\Rightarrow2\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}+\frac{1}{x+1}\right)=\frac{4}{3}\)
\(\Rightarrow2\left(1+\frac{1}{2}-\frac{1}{x+1}\right)=\frac{4}{3}\)
\(\Rightarrow3+\frac{1}{x+1}=\frac{4}{3}\)
\(\Rightarrow\frac{1}{x+1}=\frac{4}{3}-3\)
\(\Rightarrow\frac{1}{x+1}=\frac{-5}{3}\)
\(\Rightarrow x=\frac{-8}{5}\)
Mà \(x\in N\Rightarrow\)đề sai :D