\(\dfrac{25}{30}=\dfrac{15}{X}\)
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1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
\(a,MSC:180\\ \dfrac{17}{12}=\dfrac{17.15}{12.15}=\dfrac{255}{180};\dfrac{31}{18}=\dfrac{31.10}{18.10}=\dfrac{310}{180};\dfrac{8}{15}=\dfrac{8.12}{15.12}=\dfrac{96}{180}\\ b,MSC:75\\ \dfrac{7}{15}=\dfrac{7.5}{15.5}=\dfrac{35}{75};\dfrac{8}{25}=\dfrac{8.3}{25.3}=\dfrac{24}{75};\dfrac{11}{75}=\dfrac{11}{75}\)
\(\Leftrightarrow\left(\dfrac{x-5}{1990}-1\right)+\left(\dfrac{x-15}{1980}-1\right)+\left(\dfrac{x-25}{1970}-1\right)\\ +\left(\dfrac{x-1990}{5}-1\right)+\left(\dfrac{x-1980}{15}-1\right)+\left(\dfrac{x-1970}{25}-1\right)=0\\ \Leftrightarrow\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}+\dfrac{x-1995}{1970}+\dfrac{x-1995}{5}\\ +\dfrac{n-1995}{15}+\dfrac{n-1995}{25}=0\\ \Rightarrow\left(x-1995\right)\left(\dfrac{1}{1990}+\dfrac{1}{1980}+\dfrac{1}{1970}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}\right)=0\)
\(\Rightarrow x-1995=0\\ \Rightarrow x=1995\)
\(\dfrac{30-x}{30}\) = \(\dfrac{5}{15}\)
\(\dfrac{30-x}{30}\) = \(\dfrac{1}{3}\)
30 - \(x\) = \(\dfrac{1}{3}\) \(\times\) 30
30 - \(x\) = 10
\(x\) =30 - 10
\(x\) = 20
\(\dfrac{x+30}{72}\) = \(\dfrac{5}{8}\)
\(x+30\) = \(\dfrac{5}{8}\) \(\times\) 72
\(x+30\) = 45
\(x\) = 45 - 30
\(x\) = 15
\(\dfrac{30-x}{30}=\dfrac{8}{15}\)
\(\Rightarrow\left(30-x\right)15=30\times8\)
\(\Rightarrow\left(30-x\right)15=240\)
\(\Rightarrow30-x=240:15\)
\(\Rightarrow30-x=16\)
\(\Rightarrow x=30-16\)
\(\Rightarrow x=14\)
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\(\dfrac{x+30}{72}=\dfrac{5}{8}\)
\(\Rightarrow\left(x+30\right)8=5\times72\)
\(\Rightarrow\left(x+30\right)8=360\)
\(\Rightarrow x+30=360:8\)
\(\Rightarrow x+30=45\)
\(\Rightarrow x=45-30\)
\(\Rightarrow x=15\)
\(D=\dfrac{2}{3\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{11}{8\cdot19}+\dfrac{13}{19\cdot32}+\dfrac{25}{32\cdot57}+\dfrac{30}{57\cdot87}\)
Áp dụng công thức tổng quát \(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
Ta có:
\(D=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{32}+\dfrac{1}{32}-\dfrac{1}{57}+\dfrac{1}{57}-\dfrac{1}{87}\\ D=\dfrac{1}{3}-\dfrac{1}{87}\\ D=\dfrac{28}{87}\)
Ta có: \(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}\)
\(\Leftrightarrow\)\(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}-3=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}-3\)
\(\Leftrightarrow\)\(\frac{x-5}{1990}-1+\frac{x-15}{1980}-1+\frac{x-25}{1970}-1=\frac{x-1990}{5}-1+\frac{x-1980}{15}-1+\frac{x-1970}{25}-1\)\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)
\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)
\(\Leftrightarrow\)\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)
\(\Leftrightarrow\)\(x-1995=0\)
\(\Leftrightarrow\)\(x=1995\)
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X= \(\dfrac{30.15}{25}=18\)