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\(\frac{16.17-5}{16.16+11}\)
\(=\frac{16.16+16-5}{16.16+11}\)
\(=\frac{16.16+11}{16.16+11}\)
\(=1\)
\(\frac{16\cdot17-5}{16\cdot16+11}\)
\(=\frac{16\cdot17-5}{16\cdot17-16-11}\)
\(=\frac{16\cdot17-5}{16\cdot17-5}\)
\(=1\)
mk ko copy của Trần Thùy Dung nhé
\(\frac{16.17-5}{16.16+11}\)
\(=\frac{17-5}{16+11}\)
\(=\frac{12}{27}\)
Bài làm
\(\frac{16x17-5}{16x16+11}\)
\(=\frac{272-5}{256+11}\)
\(=\frac{267}{267}\)
\(=1\)
# Học tốt #
\(\frac{16.17-5}{16.16+11}\)
\(=\frac{16.\left(16+1\right)-5}{16^2+11}\)
\(=\frac{16^2+16-5}{16^2+11}\)
\(=\frac{16^2+11}{16^2+11}\)
\(=1\)
(Nhớ k cho mình với nhé!)
\(\frac{16x17-5}{16x16+11}\)
=\(\frac{16x\left(16+1\right)-5}{16x16+11}\)
=\(\frac{16x16+16x1-5}{16x16+11}\)
= \(\frac{16x16+11}{16x16+11}\)
= 1 ( vì TS = MS )
a: A=16*16+16-5=16*16+11
=>A=B
b: A=2000*2000=2000^2
B=1998*2002=(2000-2)(2000+2)=2000^2-4
=>A<B
d: B=(2020-1)(2020+1)=2020^2-1
=>B<A
a. \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-4}{9}+\frac{5}{6}\right):\frac{7}{12}\)
\(=1.\frac{7}{12}:\frac{7}{12}\)
\(=1\)
b.
\(\frac{5}{9}.\frac{8}{11}+\frac{5}{9}.\frac{9}{11}-\frac{5}{9}.\frac{6}{11}\)
\(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
Tk mk nha!
b) \(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
Ta có :
\(A=100\left(1+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{9899}{9900}\right)\)
\(A=100\left(1+\frac{6-1}{6}+\frac{12-1}{12}+\frac{20-1}{20}+...+\frac{9900-1}{9900}\right)\)
\(A=100\left(1+\frac{6}{6}-\frac{1}{6}+\frac{12}{12}-\frac{1}{12}+\frac{20}{20}-\frac{1}{20}+...+\frac{9900}{9900}-\frac{1}{9900}\right)\)
\(A=100\left(1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\right)\)
\(\frac{A}{100}=1+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{9900}\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{A}{100}=\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}-\frac{1}{100}\right)\)
Do từ \(2\) đến \(99\) có \(99-2+1=98\) số nên có \(98\) số \(1\) suy ra :
\(\frac{A}{100}=98-\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\frac{A}{100}=98-\frac{49}{100}\)
\(\frac{A}{100}=\frac{9751}{100}\)
\(A=\frac{9751}{100}.100\)
\(A=9751\)
Vậy \(A=9751\)
Chúc bạn học tốt ~
a, \(\dfrac{3}{5}\). (\(-\dfrac{9}{11}\)) + \(\dfrac{-3}{5}\).\(\dfrac{2}{11}\) + \(\dfrac{3}{5}\)
= \(\dfrac{3}{5}\).( - \(\dfrac{9}{11}\) - \(\dfrac{2}{11}\) + 1)
= \(\dfrac{3}{5}\).(- \(\dfrac{11}{11}\) + 1)
= \(\dfrac{3}{5}\).(1-1)
= \(\dfrac{3}{5}\).0
= 0
\(\frac{5}{8}+\frac{5}{27}-\frac{5}{49}\)
\(=5\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)\)
\(=\frac{1499}{10584}.5\)
\(=\frac{7495}{10584}\)
\(\frac{11}{8}+\frac{11}{27}-\frac{11}{49}\)
\(=11.\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)\)
\(=\frac{1499}{10584}.11\)
\(=1.557917611\)
\(\frac{16.17-5}{16.16+11}\)
\(=\frac{16.\left(16+1\right)-5}{16^2+11}\)
\(=\frac{16^2+16-5}{16^2+11}\)
\(=\frac{16^2+11}{16^2+11}=1\)
(Nhớ k cho mình với nhé!)
\(\frac{16\cdot17-5}{16\cdot16+11}=1\)