Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{16.17-5}{16.16+11}\)
\(=\frac{16.16+16-5}{16.16+11}\)
\(=\frac{16.16+11}{16.16+11}\)
\(=1\)
\(\frac{16\cdot17-5}{16\cdot16+11}\)
\(=\frac{16\cdot17-5}{16\cdot17-16-11}\)
\(=\frac{16\cdot17-5}{16\cdot17-5}\)
\(=1\)
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) Mẫu số chung = BCNN(11, 7) = 77
Thừa số phụ: 77: 11= 7; 77:7 = 11.
Ta có:
\(\begin{array}{l}\frac{7}{{11}} + \frac{5}{7} = \frac{{7.7}}{{11.7}} + \frac{{5.11}}{{7.11}}\\ = \frac{{49}}{{77}} + \frac{{55}}{{77}} = \frac{{104}}{{77}}\end{array}\).
b) Mẫu số chung = BCNN(20, 15)= 60
Thừa số phụ: 60:20 = 3; 60:15 = 4
Ta có:
\(\begin{array}{l}\frac{7}{{20}} - \frac{2}{{15}} = \frac{{7.3}}{{20.3}} - \frac{{2.4}}{{15.4}}\\ = \frac{{21}}{{60}} - \frac{8}{{60}} = \frac{{13}}{{60}}\end{array}\).
a) Ta có: BCNN(15,10) = 30 nên ta chọn mẫu số chung là 30
\(\frac{11}{15}+\frac{9}{10}=\frac{22}{30}+\frac{27}{30}=\frac{49}{30}\)
b) Ta có: BCNN(6,9,12) = 36 nên ta chọn mẫu số chung là 36
\(\frac{5}{6} + \frac{7}{9} + \frac{{11}}{{12}} = \frac{{30}}{{36}} + \frac{{28}}{{36}} + \frac{{33}}{{36}} = \frac{{91}}{{36}}\)
c) Ta có: BCNN(24,21) = 168 nên ta chọn mẫu số chung là 168
\(\frac{7}{{24}} - \frac{2}{{21}} = \frac{{49}}{{168}} - \frac{{16}}{{168}} = \frac{{33}}{{168}}=\frac{11}{56}\)
d) Ta có: BCNN(36,24) = 72 nên ta chọn mẫu số chung là 72
\(\frac{{11}}{{36}} - \frac{7}{{24}} = \frac{{22}}{{72}} - \frac{{21}}{{72}} = \frac{1}{{72}}\)
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
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) \(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)
\(=\frac{5}{9}.1\)
\(=\frac{5}{9}\)
\(\frac{16.17-5}{16.16+11}\)
\(=\frac{16.\left(16+1\right)-5}{16^2+11}\)
\(=\frac{16^2+16-5}{16^2+11}\)
\(=\frac{16^2+11}{16^2+11}=1\)
(Nhớ k cho mình với nhé!)
\(\frac{16\cdot17-5}{16\cdot16+11}=1\)