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\(\frac{52}{75}=\frac{52.101}{75.101}=\frac{5252}{7575};\frac{52}{75}=\frac{52.10101}{75.10101}=\frac{525252}{757575}\)
\(\frac{13}{15}=\frac{13.101}{15.101}=\frac{1313}{1515};\frac{13}{15}=\frac{13.10101}{15.10101}=\frac{131313}{151515}\)
\(\frac{ab}{cd}=\frac{101ab}{101cd}=\frac{abab}{cdcd};\frac{ab}{cd}=\frac{10101ab}{10101cd}=\frac{ababab}{cdcdcd}\)
ai k minh minh k lai
1)
a) \(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}=\frac{3^{28}.5^{10}.2^{21}}{2^{21}.3^{24}.5^{12}.3^3.2^9}=\frac{3}{5^2}=\frac{3}{25}\)
Bài 2:
\(\frac{abab}{cdcd}=\frac{ab.101}{cd.101}=\frac{ab}{cd};\frac{ababab}{cdcdcd}=\frac{ab.10101}{cd.10101}=\frac{ab}{cd}\)
Vậy \(\frac{ab}{cd}=\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)
\(a,\frac{x+5}{20}=-\frac{5}{4}\)
\(\left(x+5\right):20=\frac{-5}{4}\)
\(x+5=-\frac{5}{4}\times20\)
\(x+5=-25\)
\(x=-25-5\)
\(x=-30\)
các câu khác thì theo đó tự làm nha
bn ơi các câu a, b thì mk làm đc nhưng vướng câu c đó bn ạ
Ta có abab/cdcd=abab:101/cdcd:101=ab/cd
ababab/cdcdcd=ababab:10101/cdcdcd:10101=ab/cd
Vì ab/cd=ab/cd nên abab/cdcd= ababab/cdcdcd
\(\frac{abab}{cdcd}=\frac{ab.101}{cd.101}=\frac{ab}{cd}\)
\(\frac{ababab}{cdcdcd}=\frac{ab.10101}{cd.10101}=\frac{ab}{cd}\)
VẬY \(\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)
Mình ko bít có đúng ko nên sai đừng trách mình nhé !
\(A=\frac{7^{2011}+1}{7^{2013}+1}\)
\(7^2.A=\frac{7^{2013}+49}{7^{2013}+1}=\frac{7^{2013}+1+48}{7^{2013}+1}=\)\(\frac{7^{2013}+1}{7^{2013}+1}+\frac{48}{7^{2013}+1}=1\frac{48}{7^{2013}+1}\)
\(B=\frac{7^{2013}+1}{7^{2015}+1}\)
\(7^2.B=\)\(=\frac{7^{2015}+49}{7^{2015}+1}=\)\(\frac{7^{2015}+1+48}{7^{2015}+1}=\)\(\frac{7^{2015}+1}{7^{2015}+1}+\frac{48}{7^{2015}+1}=1\frac{48}{7^{2015}+1}\)
\(Vì\) \(1\frac{48}{7^{2013}+1}>1\frac{48}{7^{2013}+1}\)\(\Rightarrow7^2.A>7^2.B\)\(\Rightarrow A>B\)
\(Vậy\) \(A>B\)
Bài 2 nè
ta xét B trước:
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..\)\(.....+\frac{1}{2015}-\frac{1}{2016}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+....+\frac{1}{2015}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{2016}\right)\)
\(=\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}\right)-\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
vậy A:B\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)\(:\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
\(=1\)
1,
ta có : \(\frac{\overline{abab}}{\overline{cdcd}}=\frac{\overline{abab}:101}{\overline{cdcd}:101}=\frac{\overline{ab}}{\overline{cd}}\) ; \(\frac{\overline{ababab}}{\overline{cdcdcd}}=\frac{\overline{ababab}:10101}{\overline{cdcdcd}:10101}=\frac{\overline{ab}}{\overline{cd}}\)
Vậy \(\frac{\overline{abab}}{\overline{cdcd}}=\frac{\overline{ababab}}{\overline{cdcdcd}}\)
2,
\(\frac{1}{2}.\frac{1}{b}=\frac{2}{4}\)
\(\Rightarrow\frac{1.1}{2.b}=\frac{2}{4}\)
\(\Rightarrow\frac{1}{2.b}=\frac{1}{2}\)
\(\Rightarrow2.b=2\)
\(\Rightarrow b=2:2=1\)
\(\frac{abab}{cdcd}=\frac{abab:101}{cdcd:101}=\frac{ab}{cd}\)
mà \(\frac{ababab}{cdcdcd}=\frac{ababab:10101}{cdcdcd:10101}=\frac{ab}{cd}\)
=> \(\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)
vậy...
câu 2
\(\frac{1}{2}.\frac{1}{b}=\frac{2}{4}\\ \Rightarrow\frac{1}{b}=\frac{2}{4}:\frac{1}{2}=1\\ \Rightarrow b=1\)
vậy....