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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
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 \(\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)
A = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2
1/2^2 < 1/1*2
1/3^2 < 1/2*3
1/4^2 < 1/3*4
...
1/100^2 < 1/99*100
=> A < 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/99*100
=> A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
=> A < 1 - 1/100
=> A < 1
minh deo can ban k dau :((
\(a,\frac{1}{2}x+\frac{3}{5}(x-2)=3\)
\(\Rightarrow\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\Rightarrow\left[\frac{1}{2}+\frac{3}{5}\right]x=3+\frac{6}{5}\)
\(\Rightarrow\left[\frac{5}{10}+\frac{6}{10}\right]x=\frac{21}{5}\)
\(\Rightarrow\frac{11}{10}x=\frac{21}{5}\)
\(\Rightarrow x=\frac{21}{5}:\frac{11}{10}=\frac{21}{5}\cdot\frac{10}{11}=\frac{21}{1}\cdot\frac{2}{11}=\frac{42}{11}\)
Vậy x = 42/11
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 \(\frac{ab}{cd}=\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)
Hai phân số sau có bằng nhau không?
=\(\frac{ab}{cd}\)
\(\Rightarrow\) Hai phan so đều bằng nhau.
a) a b = a . ( − 1 ) b . ( − 1 ) = − a − b
b) ta có:
a b a b ¯ c d c d ¯ = a b a b ¯ : 101 c d c d ¯ : 101 = a b ¯ c d ¯ ; a b a b a b ¯ c d c d ¯ c d = a b a b a b ¯ : 10101 c d c d ¯ c d:10101 = a b ¯ c d ¯
do đó: a b a b ¯ c d c d ¯ = a b a b a b ¯ c d c d ¯ c d
c) a b a b ¯ a b a b a b ¯ = a b a b ¯ : a b ¯ a b a b a b ¯ : a b ¯ = 101 10101
a b a b ¯ c d c d ¯ = 101 a b ¯ 101 c d ¯ = a b ¯ c d ¯ ; a b a b a b ¯ c d c d c d ¯ = 10101 a b ¯ 10101 c d ¯ = a b ¯ c d ¯
⇒ a b a b ¯ c d c d ¯ = a b a b a b ¯ c d c d c d ¯
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....