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\(\frac{-5}{6}\)\(+\)\(\frac{4}{9}\)\(\times\)\(\left(\frac{5}{4}-\frac{2}{3}\right)\)\(\times\)\(\left(-3\right)^2\)\(+\)\(\frac{5}{9}\)\(\times\)\(30\%\)
\(=\)\(\frac{-5}{6}\)\(+\)\(\frac{4}{9}\)\(\times\)\(\frac{7}{12}\)\(\times\)\(9\)\(+\)\(\frac{5}{9}\)\(\times\)\(\frac{3}{10}\)
\(=\)\(\frac{-5}{6}\)\(+\)\(\frac{7}{3}\)\(+\)\(\frac{5}{9}\)\(\times\)\(\frac{3}{10}\)
\(=\)\(\frac{-5}{6}\)\(+\)\(\frac{7}{3}\)\(+\)\(\frac{1}{6}\)
\(=\)\(\frac{-5}{6}\)\(+\)\(\frac{1}{6}\)\(+\)\(\frac{7}{3}\)
\(=\)\(\frac{-2}{3}\)\(+\)\(\frac{7}{3}\)
\(=\)\(\frac{5}{3}\)
Bài làm:
Xét: \(\frac{1}{5^2}>\frac{1}{5.6}\) ; \(\frac{1}{6^2}>\frac{1}{6.7}\) ; ... ; \(\frac{1}{100^2}>\frac{1}{100.101}\)
=> \(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\) (1)
Lại có: \(\frac{1}{5^2}< \frac{1}{4.5}\) ; \(\frac{1}{6^2}< \frac{1}{5.6}\) ; ... ; \(\frac{1}{100^2}< \frac{1}{99.100}\)
=> \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\) (2)
Từ (1) và (2) => \(\frac{1}{6}< A< \frac{1}{4}\)
Đặt A=2+24+26+....+22022
=> 22.A=22.(2+24+26+....+22022)
=>4A=24+26+....+22024
=>4A-A=(24+26+....+22024)-(2+24+26+....+22022)
=>3A =24+26+....+22024-2-24-26-....-22022
=>3A=22024-2
=>A=\(\frac{2^{2024}-2}{3}\)
Vậy =>A=\(\frac{2^{2024}-2}{3}\)
TK MK NHA BN
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Ta có:\(\left(-12\right)+\left(-13\right)+36+\left(-11\right)=\left(-12-13-11\right)+36\)
\(=\left(-36\right)+36\)
\(=0\)
( x - 6 )2 - 9 = 16
( x - 6 )2 = 7
x - 6 = \(\sqrt{7}\)
x = \(\sqrt{7}+6\)
x \(\approx\)8,6
(x - 6) ^ 2 - 9 = 16
(x - 6) ^ 2 = 16 + 9
(x - 6) ^ 2 = 25
x - 6 = 5
x = 5 + 6
x = 11
\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{99\cdot100}\)
\(=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{99\cdot100}\)
\(=2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\right)\)
\(=2\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(=2\cdot\frac{99}{100}\)
\(=\frac{99}{50}\)
2/2+2/6+2/12+...+2/99.100
= 2/3.4+2/3.4+....+2/99.100
=2/2.(2/2.3+2/3.4+....+2/99.100)
=1/2.(1/2.3+2/3.4+....+2/99.100)
=1/2.(1/2-1/3+1/3-1/4+....+1/99-1/100
=1/2.(1/2-1/100)
=1/2.49/100
=49/200