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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)\(C=\frac{2020^{2006}+1}{2020^{2007}+1}vaD=\frac{2020^{2005}+1}{2020^{2006}+1}\)
\(C=\frac{2020^{2006}+1}{2020^{2007}+1}< 1\)
\(\Leftrightarrow C< \frac{2020^{2006}+1+2019}{2020^{2007}+1+2019}=\frac{2020^{2006}+2020}{2020^{2007}+2020}=\frac{2020.\left(2020^{2005}+1\right)}{2020.\left(2020^{2006}+1\right)}=\frac{2020^{2005}+1}{2020^{2006}+1}\)
\(\Leftrightarrow C< D\)
Chúc bạn học tốt
A = 2019 x 2021
A = 2019 x (2020 + 1)
A = 2019 x 2020 + 2019
B = 2020 x (2019 + 1)
B = 2020 x 2019 + 2020
=> B > A
A= 2019 X ( 2020+ 1)
A= 2019x 2020+ 2019
B= 2020 X ( 2019+1)
B= 2020x 2019+ 2020
2019x 2020= 2020x 2019
mà 2019< 2020
nên A< B
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 : \(N< M\)
\(B=\frac{2018+2019}{2019+2020}\)
\(\Rightarrow B=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)
\(\Rightarrow B< \frac{2018}{2019}+\frac{2019}{2020}=A\)
Vậy B < A
\(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Rightarrow B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow B< \frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}=A\)
Vậy B < A
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) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
Sửa \(\frac{a+2003}{a-2003}=\frac{b+2004}{b-2004}\)
Giả sử ngược lại thì ta có \(\frac{a}{2003}=\frac{b}{2004}\)và ta cần chứng minh \(\frac{a+2003}{a-2003}=\frac{b+2004}{b-2004}\)
Đặt \(\frac{a}{2003}=\frac{b}{2004}=k\Rightarrow\hept{\begin{cases}a=2003k\\b=2004k\end{cases}}\)
Khi đó \(\frac{a+2003}{a-2003}=\frac{2003k+2003}{2003k-2003}=\frac{2003\left(k+1\right)}{2003\left(k-1\right)}=\frac{k+1}{k-1}\)(1)
\(\frac{b+2004}{b-2004}=\frac{2004k+2004}{2004k-2004}=\frac{2004\left(k+1\right)}{2004\left(k-1\right)}=\frac{k+1}{k-1}\)(2)
Từ (1) và (2) => \(\frac{a+2003}{a-2003}=\frac{b+2004}{b-2004}\)
=> đpcm
Không hiểu chỗ nào thì ib nhé :)
\(\frac{a+2003}{a-2003}=\frac{b+2004}{b-2004}\Leftrightarrow\frac{\frac{a}{2003}+1}{\frac{a}{2003}-1}=\frac{\frac{b}{2004}+1}{\frac{b}{2004}-1}\)
Đặt \(\frac{a}{2003}=x,\frac{b}{2004}=y\Rightarrow\frac{x+1}{x-1}=\frac{y+1}{y-1}\Leftrightarrow\left(x+1\right)\left(y-1\right)=\left(x-1\right)\left(y+1\right)\)
\(\Leftrightarrow xy-x+y-1=xy+x-y-1\Leftrightarrow2x=2y\Leftrightarrow x=y\)-----> Xooooong :)))
Bài 1:
a) \(\left|2y+1\right|=7\)
\(\Rightarrow2y+1=7\) hoặc \(2y+1=-7\)
\(\Rightarrow2y=6\) hoặc \(2y=-8\)
\(\Rightarrow y=3\) hoặc \(y=-4\)
Vậy................
b) \(\left|y-8\right|-15=22\)
\(\left|y-8\right|=37\)
\(\Rightarrow y-8=37\) hoặc \(y-8=-37\)
\(\Rightarrow y=45\) hoặc \(y=-29\)
Vậy \(y\in\left\{45;-29\right\}\)
Ta có :
\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(A=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(A=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(A=5\left(1-\frac{1}{31}\right)\)
\(A=5.\frac{30}{31}\)
\(A=\frac{150}{31}>1\)
\(\Rightarrow\)\(A>1\)
Vậy \(A>1\)
Chúc bạn học tốt ~
Ko cần dài dòng vậy đâu,A=\(\frac{5^2}{1.6}+\left(\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\right)\)
Ta thấy \(\frac{5^2}{1.6}>1\)và tổng trong ngoặc >0 nên =>A>1
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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