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Tìm x
\(a,2x-25\%=\frac{1}{2}\)
\(b,\left(\frac{3x}{7}+1\right).\left(-0,25\right)=\frac{-1}{28}\)
\(\)
Ta có : \(N=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1000.1001}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{1001-1000}{1000.1001}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1000}-\frac{1}{1001}\)
\(=1-\frac{1}{1001}=\frac{1000}{1001}\)
Ta thấy : \(1001< 2020\Rightarrow\frac{1}{1001}>\frac{1}{2020}\)
\(\Rightarrow-\frac{1}{1001}< -\frac{1}{2020}\)
\(\Rightarrow1-\frac{1}{1001}< 1-\frac{1}{2020}\Rightarrow\frac{1000}{1001}< \frac{2019}{2020}\)
Hay : \(N< M\)
1) Ta có: \(\frac{2019}{2020}+\frac{2020}{2021}=\frac{2019}{2020}+\frac{4040}{4042}>\frac{4040}{4042}>\frac{4039}{4041}\)
Mà \(\frac{2019+2020}{2020+2021}=\frac{4039}{4041}\)
\(\Rightarrow\frac{2019}{2020}+\frac{2020}{2021}>\frac{2019+2020}{2020+2021}\)
2) BĐT cần CM tương đương:
\(\frac{a^2+b^2}{ab}\ge2\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\) (Luôn đúng)
Dấu "=" xảy ra khi: a = b
Hoặc có thể sử dụng BĐT Cauchy nếu bạn học cao hơn
Tìm x e Z biết: 2x+1 e Ư (x+5) và x e N
giải giúp mình nhé!
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b)=(2/3 +2/7 - 2/28)/(-3/3 -3/7 + 3/28)
=[2(1/3+1/7-1/28)]/[(-3)(1/3+1/7-1/28)]
=2/-3
=-2/3
Câu 3 : \(2+4+6+.........+2n=156\)
\(\Leftrightarrow2\left(1+2+3+.....+n\right)=156\)
\(\Leftrightarrow1+2+3+.........+n=78\)
\(\Leftrightarrow\frac{n\left(n+1\right)}{2}=78\)\(\Leftrightarrow n\left(n+1\right)=156=12.13\)\(\Leftrightarrow n=12\)
Vậy \(n=12\)
a) \(\frac{21}{52}=\frac{210}{520}=1-\frac{310}{520}\)
\(\frac{213}{523}=1-\frac{310}{523}\)
Vì \(520< 523\)\(\Rightarrow\frac{1}{520}>\frac{1}{523}\)\(\Rightarrow\frac{310}{520}>\frac{310}{523}\)
\(\Rightarrow1-\frac{310}{520}< 1-\frac{310}{523}\)
hay \(\frac{21}{52}< \frac{213}{523}\)
b) \(\frac{1515}{9797}=\frac{15.101}{97.101}=\frac{15}{97}\); \(\frac{171171}{991991}=\frac{171.1001}{991.1001}=\frac{171}{991}\)
Ta có: \(\frac{15}{97}=\frac{150}{970}=1-\frac{820}{970}\); \(\frac{171}{991}=1-\frac{820}{991}\)
Vì \(970< 991\)\(\Rightarrow\frac{1}{970}>\frac{1}{991}\)\(\Rightarrow\frac{820}{970}>\frac{820}{991}\)
\(\Rightarrow1-\frac{820}{970}< 1-\frac{920}{991}\)
hay \(\frac{1515}{9797}< \frac{171171}{991991}\)
c) \(\frac{n+2}{n+3}=1-\frac{1}{n+3}\); \(\frac{n+3}{n+4}=1-\frac{1}{n+4}\)
Vì \(n\inℕ^∗\)\(\Rightarrow n+3< n+4\)\(\Rightarrow\frac{1}{n+3}>\frac{1}{n+4}\)
\(\Rightarrow1-\frac{1}{n+3}< 1-\frac{1}{n+4}\)
hay \(\frac{n+2}{n+3}< \frac{n+3}{n+4}\)
d) \(\frac{n+7}{n+6}=1+\frac{1}{n+6}\); \(\frac{n+1}{n}=1+\frac{1}{n}\)
Vì \(n\inℕ^∗\)\(\Rightarrow n+6>n\)\(\Rightarrow\frac{1}{n+6}< \frac{1}{n}\)
\(\Rightarrow1+\frac{1}{n+6}< 1+\frac{1}{n}\)
hay \(\frac{n+7}{n+6}< \frac{n+1}{n}\)
a)\(A=\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{2}\right)vaB=\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
+)Ta có:\(A=\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{2}\right)\)
\(\Leftrightarrow A=\frac{31}{23}-\left(\frac{7}{32}+\frac{128}{32}\right)\)
\(\Leftrightarrow A=\frac{31}{23}-\frac{135}{32}\)
\(\Leftrightarrow A=\frac{992}{736}-\frac{3105}{736}\)
\(\Leftrightarrow A=\frac{-2113}{736}\left(1\right)\)
+)Ta lại có:\(B=\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(\Leftrightarrow B=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(\Leftrightarrow B=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(\Leftrightarrow B=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)
\(\Leftrightarrow B=\frac{1}{3}+\left(-1\right)+1\)
\(\Leftrightarrow B=\frac{1}{3}\left(2\right)\)
+)Từ (1) và (2)
\(\Leftrightarrow A< 0< B\Leftrightarrow A< B\)
Vậy A<B
b)\(\frac{200420042004}{200520052005}va\frac{2004}{2005}\)
+)Ta có \(\frac{200420042004}{200520052005}=\frac{2004.100010001}{2005.100010001}=\frac{2004}{2005}\)
\(\Leftrightarrow\frac{200420042004}{200520052005}=\frac{2004}{2005}\)
c)\(C=\frac{2020^{2006}+1}{2020^{2007}+1}vaD=\frac{2020^{2005}+1}{2020^{2006}+1}\)
\(C=\frac{2020^{2006}+1}{2020^{2007}+1}< 1\)
\(\Leftrightarrow C< \frac{2020^{2006}+1+2019}{2020^{2007}+1+2019}=\frac{2020^{2006}+2020}{2020^{2007}+2020}=\frac{2020.\left(2020^{2005}+1\right)}{2020.\left(2020^{2006}+1\right)}=\frac{2020^{2005}+1}{2020^{2006}+1}\)
\(\Leftrightarrow C< D\)
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