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\(\frac{28}{15}\times\frac{1}{4}\times3+\left(\frac{8}{15}-\frac{79}{60}\right)\times\frac{24}{47}\)
=\(\frac{7}{15}\times3+\left(\frac{32}{60}-\frac{79}{60}\right)\times\frac{24}{47}\)
=\(\frac{7}{5}+\frac{47}{60}\times\frac{24}{47}\)
=\(\frac{7}{5}+\frac{2}{5}\)
=\(\frac{5}{5}\)
=\(1\)
\(\frac{28}{15}\)x\(\frac{1}{4}\)x3+(\(\frac{8}{15}\)+\(\frac{-79}{60}\))x\(\frac{24}{47}\)=
=\(\frac{28}{15}\)x\(\frac{1}{4}\)x3+(\(\frac{32}{60}\)+\(\frac{-79}{60}\))x\(\frac{24}{47}\)
=\(\frac{28}{15}\)x\(\frac{1}{4}\)x3+\(\frac{-47}{60}\)x\(\frac{24}{47}\)
=\(\frac{7}{15}\)x3+\(\frac{-47}{60}\)x\(\frac{24}{47}\)
=\(\frac{7}{5}\)+\(\frac{-2}{5}\)
=\(\frac{5}{5}\)=1
mấy cái câu đầu thì tớ ko hiểu !(thông cảm)
17/60-5/18-64/90= -127/180
ko chắc là đúng^
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\frac{1}{2}\left(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{2}{9}\cdot\frac{1}{2}\)
\(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\)
\(x=18-1\)
\(x=17\)
sửa đề số cuối vế trái là \(\frac{1}{x\left(x+1\right)}\)
Đặt A là vế trái
\(\frac{1}{2}A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\)
\(=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}\)
\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\)
\(=\frac{1}{6}-\frac{1}{x+1}\)
\(\Rightarrow A=\frac{1}{3}-\frac{2}{x+1}=\frac{2}{9}\)
\(\frac{2}{x+1}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}=\frac{2}{18}\)
\(\Rightarrow x+1=18\Rightarrow x=17\)
Vậy x=17
ta có:
\(1+\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{x\left(x+3\right)}\)=1\(\frac{7}{24}\)\(=\frac{31}{24}\)
\(1+\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}=\frac{31}{24}\)
\(3+\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}=\frac{31}{3}\)
\(3+1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{31}{3}\)
\(3+1-\frac{1}{x+3}=\frac{31}{3}\)
\(4-\frac{1}{x+3}=\frac{31}{3}\)
\(\frac{1}{x+3}=4-\frac{31}{3}=\frac{-19}{3}\)
=>ko tìm được x
cậu tham khảo linh này nhé :
olm.vn/hoi-dap/question/976586.html
chúc cậu học tốt
60+24:(x+1)=28
84:(x+1)=28
x+1=84:28
x+1=3
x=3-1=2
60 + 24: [ x + 1 ] = 28
24 : [ x + 1 ] = 28 - 60
24 : [ x + 1 ] = - 32
x + 1 = 24 : ( - 32 )
x + 1 = \(\frac{-3}{4}\)
x = \(\frac{-3}{4}\)+ 1
x = \(\frac{1}{4}\)