\(\dfrac{111\times118-33}{112\times115+185}\)
A = \(\dfrac{2018\times2020+2020\times2022}{2020\times4040}\)
B = \(\dfrac{111\times118-33}{112\times115+185}\)
Ai nhanh mình tick cho
Bài 6. Đúng ghi Đ sai ghi S vào ô trống:
a, \(\dfrac{7}{9}\)\(>\dfrac{7}{8}\) b, \(\dfrac{3}{5}\)\(< \dfrac{5}{7}\) c, \(\dfrac{111}{112}\)\(=\dfrac{112}{113}\)\(\) d, \(\dfrac{13}{9}\)\(=\dfrac{117}{81}\)
mình không có tìm thấy ô trống đâu
mong mọi người thông cảm
a) \(\dfrac{7}{9}>\dfrac{7}{8}\rightarrow Sai\)
b) \(\dfrac{3}{5}=\dfrac{21}{35}< \dfrac{5}{7}=\dfrac{25}{35}\rightarrowĐúng\)
c) \(\dfrac{111}{112}=\dfrac{112}{113}\left(\dfrac{111}{112}< \dfrac{112}{113}\right)\rightarrow Sai\)
\(\dfrac{13}{9}=\dfrac{91}{81}< \dfrac{117}{81}\rightarrow Sai\)
Bài 5:
a) \(\dfrac{x+1}{99}+\dfrac{x+4}{96}+\dfrac{x+8}{92}+\dfrac{x+3}{97}+4=0\)
b) \(\dfrac{x-11}{111}+\dfrac{x-12}{112}=\dfrac{x-23}{123}+\dfrac{x-24}{124}\)
\(\Leftrightarrow\left(\dfrac{x-11}{111}+1\right)+\left(\dfrac{x-12}{112}+1\right)=\left(\dfrac{x-23}{123}+1\right)+\left(\dfrac{x-24}{124}+1\right)\)
=>x+100=0
=>x=-100
a: =>\(\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+8}{92}+1\right)+\left(\dfrac{x+3}{97}+1\right)=0\)
=>x+100=0
=>x=-100
b: =>(x-11/111+1)+(x-12/112+1)=(x-23/123+1)+(x-24/124+1)
=>x+100=0
=>x=-100
So Sánh
a= 112/111+113/112+114/113
b= 111/112+112/111
Ta có:
a=3.026787138
b=2.0000(804375
->a>b
Cho A=\(\dfrac{50}{111}\)+\(\dfrac{50}{112}\)+\(\dfrac{50}{113}+\dfrac{50}{114}\)
CMR : 1<A<2
A<50/100+50/100+50/100+50/100=4.50/100=2
=>A<2
A>4.50/150=4/3+1+1/3>1
=>dccm
Cho A =\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{217.218}\) và B=\(\dfrac{1}{110}+\dfrac{1}{111}+\dfrac{1}{112}+...+\dfrac{1}{218}\)
So sánh A và B.
A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{217.218}\)
A = \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{217}-\dfrac{1}{218}\)
A = 1 - \(\dfrac{1}{218}\)
B = \(\dfrac{1}{110}\) + \(\dfrac{1}{111}\) + \(\dfrac{1}{112}\) + ... + \(\dfrac{1}{218}\)
Xét dãy số 110; 111; 112; ...; 218, dãy số này có số số hạng là:
(218 - 110) : 1 + 1 = 109 (số)
Mặt khác \(\dfrac{1}{110}\) > \(\dfrac{1}{111}>\dfrac{1}{112}>...>\dfrac{1}{218}\)
⇒ B = \(\dfrac{1}{110}\) + \(\dfrac{1}{111}\) + \(\dfrac{1}{112}+...+\dfrac{1}{218}\) < \(\dfrac{1}{110}\) + \(\dfrac{1}{110}\)+ ... +\(\dfrac{1}{110}\)
B < \(\dfrac{1}{110}\) x 109
B < 1 - \(\dfrac{1}{110}\)
\(\dfrac{1}{128}\) < \(\dfrac{1}{110}\) ⇒ A = 1 - \(\dfrac{1}{128}\) > 1 - \(\dfrac{1}{110}\) > B
A > B
Tính nhanh:
\(\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right).\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right).\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right).\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)
\(=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right).\left(\dfrac{4-3-1}{12}\right)\)
\(=\left(\dfrac{66}{111}+\dfrac{2}{33}+\dfrac{15}{117}\right).0\)
\(=0\)
\(\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\\ =\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\\= \left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot\left(\dfrac{4-3-1}{12}\right)\\= \left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\cdot0\\ =0\)
tính bằng cách thuận tiện
111/110 * 112/111 * 113/112 * 114/113 * 55/57
giúp mk nha mk đg cần gấp thạnk you
\(=\dfrac{1}{110}\times\dfrac{1}{1}\times\dfrac{1}{1}\times\dfrac{114}{1}\times\dfrac{55}{57}\\ =\dfrac{114\times55}{110\times57}\\ =\dfrac{2\times1}{2\times1}=\dfrac{2}{2}=1\)
\(\dfrac{111}{110}\times\dfrac{112}{111}\times\dfrac{113}{112}\times\dfrac{114}{113}\times\dfrac{55}{57}=\dfrac{111\times112\times113\times114}{110\times111\times112\times113}\times\dfrac{55}{57}\)
\(=\dfrac{114}{110}\times\dfrac{55}{57}=\dfrac{57}{55}\times\dfrac{55}{57}=1\)
Chứng minh rằng:
a ) A = 1 11 + 1 12 + 1 13 + ... + 1 20 > 1 2 b ) B = 1 5 + 1 6 + 1 7 + ... + 1 16 + 1 17 < 2 c ) C = 1 10 + 1 11 + 1 12 + ... + 1 18 + 1 19 < 1
a) A > 1 20 + 1 20 + ... + 1 20 ⏟ 10 s o = 10 20 = 1 2 .
b) B = 1 5 + ... 1 9 + 1 10 + ... + 1 17 < 1 5 + ... + 1 5 ⏟ 5s o + 1 8 + ... + 1 8 ⏟ 8s o = 2
c) C = 1 10 + 1 11 + 1 12 ... + 1 18 + 1 19 < 1 10 + 1 10 + ... 1 10 ⏟ 9 s o = 1