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\(\dfrac{5}{8}x-\dfrac{1}{3}x-\dfrac{1}{6}x=15\)
\(\Rightarrow x\left(\dfrac{5}{8}-\dfrac{1}{3}-\dfrac{1}{6}\right)=15\)
\(\Rightarrow x\left(\dfrac{15}{24}-\dfrac{8}{24}-\dfrac{4}{24}\right)=15\)
\(\Rightarrow x.\dfrac{3}{24}=15\)
\(\Rightarrow x.\dfrac{1}{8}=15\Rightarrow x=15:\dfrac{1}{8}=15.\dfrac{8}{1}=120\)
mình làm được bài tìm x
x.(2/1.3+2/3.5+2/5.7+...+2/97.99)-x=-100/99
x.(1-1/3+1/3-1/4+1/4-1/5+1/5+...+1/97-1/97-1/99)-x=-100/99
x.(1-1/99)-x=-100/99
x.98/99-x=-100/99
x.98/99=-100/99+x
x.x=-100/99-98/99
2x=-198/99
x=-198/99/2
x=-1
Em xem lại đề câu B nhé\(B=\dfrac{3}{2}+\dfrac{3}{6}+\dfrac{3}{12}+\dfrac{3}{20}+...+\dfrac{3}{\left(n-1\right).n}\\ =3.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{\left(n-1\right).n}\right)\\ =3.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)=3.\left(1-\dfrac{1}{n}\right)=3.\dfrac{n-1}{n}=3-\dfrac{3}{n}.\)
\(C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{30.32}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{30}-\dfrac{1}{32}\\ =1-\dfrac{1}{32}=\dfrac{31}{32}.\)
\(D=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n+1}-\dfrac{1}{n+3}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{n+3}\right)=\dfrac{1}{2}.\dfrac{n+2}{n+3}.\)
\(a,\left(\frac{15}{10}x+25\right):\frac{2}{3}=60\)
\(\Rightarrow\frac{15}{10}x+25=60.\frac{2}{3}\)
\(\Rightarrow\frac{15}{10}x+25=40\)
\(\Rightarrow\frac{15}{10}x=40-25\)
\(\Rightarrow\frac{15}{10}x=15\)
\(\Rightarrow x=15:\frac{15}{10}\)
\(\Rightarrow x=15.\frac{10}{15}\)
\(\Rightarrow x=10\)
\(\Leftrightarrow2^{x+1}=8\)
=>x+1=3
hay x=2