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1.(11/12+11/23+11/23+11/24+...+11/89+11/100)+x=5/3
1.(11/12+11/100)+x=5/3
77/75+x=5/3
x=5/3-77/75
x=16/25
de ot
=> ( 11/12+ 1/12- 1/23+ 1/23- 1/34+...+ 1/89-1/100 ) +x=5/3
=> (11/12+ 1/12-1/100 ) + x=5/3
=> (11/12+ 11/150) + x=5/3
=>99/100 +x = 5/3
=> x = 5/3 - 99/100
=> x= 203 / 300
K dung NHA
\(a)\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{-11.13^7}{11^5.13^8}=\frac{-1}{11^4.13}\) (Bạn xem thử xem có sai đề không nhé)
\(c)\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3+1\right)}{2^9.3^{10}}=\frac{2.4}{3}=\frac{8}{3}\)
\(d)\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.4+9\right)}=\frac{8}{20+9}=\frac{8}{29}\)
\(a)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{\left(-11\right)\cdot13^7}{11^5\cdot13^8}=\frac{-1}{11^4\cdot13}=\frac{-1}{14641\cdot13}=\frac{-1}{190333}\)
\(c)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\cdot2}{2^9\cdot3^{10}}=\frac{2\cdot2}{3}=\frac{4}{3}\)
=> 11/12 + 1/12 - 1/23 + 1/23 - 1/34 + ..... + 1/89 - 1/100 + x = 5/3
=> 11/12 + 1/12 - 1/100 + x = 5/3
=> 99/100 + x = 5/3
=> x = 5/3 - 99/100 = 203/300
Tk mk nha
=\(\frac{-11}{23}\)\(\times\)\(\frac{10}{-13}\)\(+\)\(\frac{11}{-13}\)\(\times\)\(\frac{-3}{23}\)\(+\)\(\frac{12}{23}\)
=\(\frac{110}{299}\)\(+\)\(\frac{33}{299}\)\(+\)\(\frac{12}{23}\)
=\(\frac{143}{299}\)\(+\)\(\frac{12}{23}\)
= 1
Trước hết tính tổng :
\(\frac{11}{12}+\frac{11}{12\times13}+...+\frac{11}{89\times100}=1-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{89}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Do đó \(\frac{99}{100}+x=\frac{5}{3}\)
Vậy \(x-\frac{5}{3}-\frac{99}{100}=\frac{500-297}{300}=\frac{203}{300}\)
Vậy...
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