(x+2)+(x+4)+(x+6)+...+(x+100)=60001+2+3+4+...+x=15x-(\(\frac{20}{11.13}\)+ \(\frac{20}{13.15}\)+\(\frac{20}{15.17}\)+...+\(\frac{20}{53.55}\))=\(\frac{3}{11}\)\(\frac{7}{4}\)x. (\(\frac{33}{12}\)+ \(\frac{3333}{2020}\)+ \(\frac{333333}{303030}\)+ \(\frac{33333333}{42424242}\)) =...
Đọc tiếp
(x+2)+(x+4)+(x+6)+...+(x+100)=6000
1+2+3+4+...+x=15
x-(\(\frac{20}{11.13}\)+ \(\frac{20}{13.15}\)+\(\frac{20}{15.17}\)+...+\(\frac{20}{53.55}\))=\(\frac{3}{11}\)
\(\frac{7}{4}\)x. (\(\frac{33}{12}\)+ \(\frac{3333}{2020}\)+ \(\frac{333333}{303030}\)+ \(\frac{33333333}{42424242}\)) = 42
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{x\left(x-1\right)}\)= \(\frac{2007}{2009}\)
\(\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{28}+\frac{1}{256}\right)\).x=1
a, (x+2)+(x+4)+(x+6)+...+(x+100)=6000
(x+x+x+...+x)+(2+4+6+...+100)=6000
50.x+2550=6000
50.x=6000-2550
50.x=3450
x=3450:50
x=69
b, 1+2+3+4+...+x=15
10+...+x=15
x=15-10
x=5
Nho **** cho minh nha
Ta có: (x+x+x+...+x) + (2+4+6+...+100) = 6000
Ta thấy vế phải có: (100-2):2+1=50(số hạng)
Tổng của vế phải: [(2+100)*50]:2=2550
\(\Rightarrow\)có 50 số x
\(\Rightarrow\)50*x + 2550 = 6000
\(\Rightarrow\)50*x=6000-2550
\(\Rightarrow\)50*x=3450
\(\Rightarrow\)x=3450:50
\(\Rightarrow\)x=69
Vậy x=69
Mình đúng nè, nhớ k nha