\(\frac{15}{x}\)=\(\frac{25}{35}\)
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\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)
\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)
\(=\frac{1.9}{2.8}=\frac{9}{16}\)
\(\frac{1}{5}+\frac{4}{10}+...+\frac{81}{45}=\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+...+\frac{9}{5}=\frac{1+2+3+...+9}{5}=\frac{45}{5}=9\)
1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
=1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5
=45/5 = 9
a)\(\frac{-5}{6}\).\(\frac{120}{25}\)<x<\(\frac{-7}{15}\).\(\frac{9}{14}\)
-4 <x<\(\frac{-3}{10}\)
\(\frac{-40}{10}\)< x <\(\frac{-3}{10}\)=>x E {-39:-38:-37:.....:-4}
b)\(\left(\frac{-5}{3}\right)^3\)<x<\(\frac{-24}{35}.\frac{-5}{6}\)
\(\frac{-875}{189}< x< \frac{108}{189}\)
=> x E {\(\frac{-874}{189},\frac{-873}{189},......,\frac{107}{189}\)}
Ta có:
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.\frac{35}{36}.\frac{48}{49}=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+\frac{4.6}{5.5}+\frac{5.7}{6.6}+\frac{6.8}{7.7}=\frac{1.2.3.4.5.6}{2.3.4.5.6.7}.\frac{3.4.5.6.7.8}{2.3.4.5.6.7}=\frac{1}{7}.\frac{8}{2}=\frac{4}{7}\)
mình biết đáp án là : \(\frac{9}{16}\)thôi,còn cách giải thì mình không chắc chắn nên không viết ra
\(\frac{3.2.4.3.5.4.6.5.7.6.8.7.9}{4.3.3.4.4.5.5.6.6.7.7.8.8}\)= \(\frac{9}{16}\)
\(\frac{15}{x}=\frac{25}{35}\)
\(\Rightarrow x\times25=15\times35\)
\(\Rightarrow x\times25=525\)
\(\Rightarrow x=525:25\)
\(\Rightarrow x=21\)
\(\frac{15}{x}=\frac{25}{35}\)
\(15:x=\frac{25}{35}\)
\(x=15:\frac{25}{35}\)
\(x=21\)
Vậy \(x=21\)