\(\frac{x}{5}\)-\(\frac{5}{2}\) =-31. Tim x
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b,5x+2=625
=>5x+2=54
=>x+2=4
=>x=4-2
=>x=2
b,\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}...\frac{30}{62}.\frac{31}{64}=\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}...\frac{30}{2.31}.\frac{31}{2.32}=\frac{1.2.3...30.31}{2.2.2.3.2.4...2.31.2.32}=\frac{1}{2.2.2...2.2}=\frac{1}{2^{31}}\)
Kết quả ko thì nhanh thôi !
\(\left(a\right)\frac{x}{5}-\frac{5}{2}=31\)
\(x=167\frac{1}{2}\)
\(\left(b\right)\frac{x}{5}-\frac{5}{2}=-31\)
\(x=-142\frac{1}{2}\)
k nhé !
Đặt A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{5}{31}\)
2A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
2A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{\left(2x+1\right)}-\frac{1}{2x+3}=\frac{10}{31}\)
2A = \(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
Ta có : \(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
=> 2x + 3 = 93
2x = 90
x = 45
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
-5.(x+1/5) -1/2.(x-2/3)=3/2x-5/6
-5x + (-1) -1/2x -1/3=3/2x-5/6
-5x-1/2x-3/2x=1+1/3-5/6
x.(-5-1/2-3/2)= 6/6+2/6+(-5/6)
x.(-10/2+(-1/2)+(-3/2))=3/6
x.6/2=1/2
x=1/2:6/2
x=1/6
Vậy x = 1/6
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
\(X-\left(\frac{31}{5}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\right)=\frac{9}{13}\)
\(X-\left(\frac{31}{5}+\frac{31}{3\cdot5}+\frac{31}{5\cdot7}+\frac{31}{7\cdot9}+\frac{31}{9\cdot11}+\frac{31}{11\cdot13}\right)=\frac{9}{13}\)
\(X-\left[\frac{31}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)+\frac{31}{5}\right]=\frac{9}{13}\)
\(X-\left[\frac{31}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)+\frac{31}{5}\right]=\frac{9}{13}\)
\(X-\left[\frac{31}{2}\cdot\frac{10}{39}+\frac{31}{5}\right]=\frac{9}{13}\)
\(X-\frac{1984}{195}=\frac{9}{13}\)
\(\Rightarrow X=\frac{9}{13}+\frac{1984}{195}=\frac{163}{15}\)
Ta có ;
\(\frac{x}{5}-\frac{5}{2}=-31\)
\(\Rightarrow\frac{2x-25}{10}=-31\)
\(\Rightarrow2x-25=-310\)
\(\Rightarrow2x=-285\)
\(\Rightarrow x=-\frac{285}{2}\)
Hé hé,bài con gà