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TÍNH HỢP LÝ
\(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}+\frac{6}{7}\)
a) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2-\frac{11}{5}:\frac{-11}{5}=\left(-\frac{1}{10}\right)^2+1=1\frac{1}{100}\)
b) \(\left(-\frac{5}{7}\right)^2+8.\left(0,5\right)^2+\left(-1\right)^{2010}=\frac{25}{49}+2+1=3\frac{25}{49}\)
c) \(\frac{9999^2}{3333^2}+\left(0,5\right)^2.\left(-2\right)^4-\left(-\frac{4}{3}\right)^2=9+1-\frac{16}{9}=8\frac{2}{9}\)
d) \(\left|-\frac{2}{5}+\frac{1}{7}\right|:\frac{-3}{35}+\frac{-3}{7}.\frac{7}{5}=\frac{9}{35}.\frac{35}{-3}-\frac{3}{5}=-3\frac{3}{5}\)
e) \(\frac{1}{2}-\left(-0,4\right)+\frac{1}{3}+\frac{1}{5}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)
\(=\frac{1}{2}+\frac{2}{5}+\frac{1}{3}+\frac{1}{5}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}=1\frac{732}{1435}\)
Đặt
x/5=y/4=k
khi đó:
x=5k
y=4k
Ta lại có:
x.y=4k.5k=20k^2=20
=> K=+-1
Khi k=1
Khi k=-1
Giải ra nhé
Nhân vô rồi chuyển dấu lên và nhóm nhân -1ra ngoài rồi trg ngoặc là dãy có quy luật giải dãy đó r nhân phá ngoặc
Đặt \(A=\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{101}}\)
\(\Rightarrow25A=5+\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{99}}\)
\(\Rightarrow25A-A=\left(5+\frac{1}{5}+\frac{1}{5^3}+...+\frac{1}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\right)\)
hay \(24A=5-\frac{1}{5^{101}}\)
\(\Rightarrow A=\frac{5-\frac{1}{5^{101}}}{24}\)
\(\Rightarrow A:\left(1-\frac{1}{5^{102}}\right)=\frac{5-\frac{1}{5^{101}}}{24}.\frac{1}{1-\frac{1}{5^{102}}}\)
\(=\frac{5\left(1-\frac{1}{5^{102}}\right)}{24}.\frac{1}{1-\frac{1}{5^{102}}}=\frac{5}{24}\)
A=\([\)\(\frac{2}{7}\)\(\times\)(\(\frac{1}{4}-\frac{1}{3}\))\(]\)\(\div\)\([\)(\(\frac{2}{7}\times\)(\(\frac{3}{9}-\frac{2}{5}\))\(]\)
=(\(\frac{2}{7}\times\)\(\frac{-1}{12}\))\(\div(\)\(\frac{2}{7}\times\)\(\frac{-1}{15}\))
=\(\frac{-1}{42}\)\(\div\)\(\frac{-2}{35}\)
=\(\frac{-1}{42}\)\(\times\)\(\frac{35}{-2}\)
=\(\frac{5}{12}\)
M<1 => \(\frac{x-3}{x+2}\)<1
<=> \(\frac{x-3}{x+2}\)- 1 < 0
<=> \(\frac{x-3}{x+2}\)-\(\frac{x+2}{x+2}\)< 0
<=> \(\frac{x-3-x-2}{x+2}\)< 0
<=> -5 < 0
=> Vô nghiệm
a) \(\frac{45^{10}.5^{20}}{75^{15}}\)
=
\(\frac{\left(5.9\right)^{10}.5^{20}}{\left(5.15\right)^{15}}\)
= \(\frac{5^{10}.9^{10}.5^{20}}{5^{15}.15^{15}}\)
= \(\frac{5^{10}.3^{20}.5^{20}}{5^{15}.15^{15}}\)
= \(\frac{5^{10}.15^{20}}{5^{15}.15^{15}}\)
= \(\frac{15^5}{5^5}\)
= \(\frac{3^5.5^5}{5^5}\)
= \(3^5\)
b) \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}\)
= \(\frac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}\)
= \(\frac{2^5}{0,4}\)
= \(2^5\) : 0,4
(=) 32 : \(\frac{2}{5}\)
= 90
c) \(\frac{2^{15}.9^4}{6^6.8^3}\)
= \(\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}\)
= \(\frac{2^{15}.3^8}{2^6.3^6.2^9}\)
= \(3^2\)
\(x:\frac{1}{3}=\frac{12}{99}:\frac{15}{90}\)
\(x:\frac{1}{3}=\frac{12}{99}:\frac{1}{6}\)
\(x:\frac{1}{3}=\frac{8}{11}\)
\(x=\frac{8}{11}X\frac{1}{3}\)
\(x=\frac{8}{33}\)
tk mk nhe
ló bằng 0
ko nha vì ko cs số nào chia cho 0 cả
HT~