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Bài 2 : đề bài này chỉ cần a,b>0 , ko cần phải thuộc N* đâu
a, Áp dụng bất đẳng thức AM-GM cho 2 số lhoong âm a,b ta được :
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\) . Dấu "=" xảy ra khi a=b
b , Áp dụng BĐT AM-GM cho 2 số không âm ta được : \(a+b\ge2\sqrt{ab}\)
\(\dfrac{1}{a}+\dfrac{1}{b}\ge2\sqrt{\dfrac{1}{ab}}=\dfrac{2}{\sqrt{ab}}\)
Nhân vế với vế ta được :
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge2.2.\dfrac{\sqrt{ab}}{\sqrt{ab}}=4\left(đpcm\right)\)
Dấu "="xảy ra tại a=b
Bài 1.
Vì a, b, c, d \(\in\) N*, ta có:
\(\dfrac{a}{a+b+c+d}< \dfrac{a}{a+b+c}< \dfrac{a}{a+b}\)
\(\dfrac{b}{a+b+c+d}< \dfrac{b}{a+b+d}< \dfrac{b}{a+b}\)
\(\dfrac{c}{a+b+c+d}< \dfrac{c}{b+c+d}< \dfrac{c}{c+d}\)
\(\dfrac{d}{a+b+c+d}< \dfrac{d}{a+c+d}< \dfrac{d}{c+d}\)
Do đó \(\dfrac{a}{a+b+c+d}+\dfrac{b}{a+b+c+d}+\dfrac{c}{a+b+c+d}+\dfrac{d}{a+b+c+d}< M< \left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)+\left(\dfrac{c}{c+d}+\dfrac{d}{c+d}\right)\)hay 1<M<2.
Vậy M không có giá trị là số nguyên.
Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
Câu 3:
a: \(A=-\left|x-10\right|+2018< =2018\)
Dấu '=' xảy ra khi x=10
\(B=-\left(x+2\right)^2+1999< =1999\)
Dấu '=' xảy ra khi x=-2
b: \(A=\left(2x-8\right)^2+3>=3\)
Dấu '=' xảy ra khi x=4
\(B=\left|x^2-25\right|-2017>=-2017\)
Dấu '=' xảy ra khi x=5 hoặc x=-5
Bài 1:
a: \(\Leftrightarrow\dfrac{2}{3}\cdot\dfrac{6+9-4}{12}< =\dfrac{x}{18}< =\dfrac{7}{13}\cdot\dfrac{3-1}{6}\)
\(\Leftrightarrow\dfrac{22}{36}< =\dfrac{x}{18}< =\dfrac{14}{78}=\dfrac{7}{39}\)
\(\Leftrightarrow\dfrac{11}{9}< =\dfrac{x}{9}< =\dfrac{7}{13}\)
=>143<=x<=63
hay \(x\in\varnothing\)
b: \(\Leftrightarrow\dfrac{31\cdot9-26\cdot4}{180}\cdot\dfrac{-36}{35}< x< \dfrac{153+64+56}{168}\cdot\dfrac{8}{13}\)
\(\Leftrightarrow-1< x< 1\)
=>x=0
a, b, c là ba số nguyên tố khác nhau.
Ta có [a, b]= a.b, [b, c]= b.c, [c.a]= c.a
Do đó \(\dfrac{1}{\left[a,b\right]}+\dfrac{1}{\left[b,c\right]}+\dfrac{1}{[c,a]}=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\)
Ta có: \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\le\dfrac{1}{2.3}+\dfrac{1}{3.5}+\dfrac{1}{5.2}\)
mả \(\dfrac{1}{2.3}+\dfrac{1}{3.5}+\dfrac{1}{5.2}=\dfrac{5+2+3}{30}=\dfrac{1}{3}\).
Do đó \(\dfrac{1}{\left[a,b\right]}+\dfrac{1}{\left[b,c\right]}+\dfrac{1}{\left[c,a\right]}\le\dfrac{1}{3}\).
Bài 1)
Ta có:
A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)
A < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
A < \(1-\dfrac{1}{8}\) = \(\dfrac{7}{8}\) < 1
Vậy A < 1
Bài 2)
Ta thấy:
\(\dfrac{2011}{2012+2013}< \dfrac{2011}{2012};\dfrac{2012}{2012+2013}< \dfrac{2012}{2013}\)
\(\Rightarrow\) \(\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(\Rightarrow\) \(\dfrac{2011+2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(\Rightarrow\) A < B
Bài 3)
Ta có:
B = \(\left(1-\dfrac{1}{1}\right)\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)......\left(1-\dfrac{1}{20}\right)\)
= \(0.\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)......\left(1-\dfrac{1}{20}\right)\)
= 0
Bài 3)
Ta có:
A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\)
\(\Rightarrow\) 2A = \(2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\right)\)
\(\Rightarrow\) 2A = \(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2011}}\)
\(\Rightarrow\) 2A - A = \(\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2011}}\right)\)-\(\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\right)\)
\(\Rightarrow\) A = 2 - \(\dfrac{1}{2^{2012}}\) = \(\dfrac{2^{2013}-1}{2^{2012}}\)
Bài 5)
\(\pi\) + 5 \(⋮\) \(\pi\) - 2
\(\Leftrightarrow\) \(\pi\) - 2 + 7 \(⋮\) \(\pi\) - 2
\(\Leftrightarrow\) 7 \(⋮\) \(\pi\) - 2 (vì \(\pi\) - 2 \(⋮\) \(\pi\) - 2)
\(\Leftrightarrow\) \(\pi\) - 2 \(\in\) Ư(7)
\(\Leftrightarrow\) \(\pi\) - 2 \(\in\) \(\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow\) \(\pi\) \(\in\) \(\left\{1;3;-5;9\right\}\)
a, \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{n+a-n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)
Vậy \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{a}{n\left(n+a\right)}\)
b,
\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(3B=\dfrac{5.3}{1.4}+\dfrac{5.3}{4.7}+...+\dfrac{5.3}{100.103}\)
\(3B=5\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(3B=5\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(3B=5\left(1-\dfrac{1}{103}\right)=5\cdot\dfrac{102}{103}=\dfrac{510}{103}\)
\(B=\dfrac{510}{103}:3=\dfrac{170}{103}\)
\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(C=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(2C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{51}=\dfrac{16}{51}\)
\(C=\dfrac{16}{51}:2=\dfrac{8}{51}\)
1.
A=\(\dfrac{3\left|x\right|+2}{\left|x\right|-5}=\dfrac{3\left|x\right|-15+17}{\left|x\right|-5}=\dfrac{3\left(\left|x\right|-5\right)+17}{\left|x\right|-5}=\dfrac{3\left(\left|x\right|-5\right)}{\left|x\right|-5}+\dfrac{17}{\left|x-5\right|}=3+\dfrac{17}{\left|x\right|-5}\)
Để A \(\in\)Z thì \(\left|x\right|-5\inƯ\left(17\right)=\left\{-17;-1;1;17\right\}\)
Ta có :
\(\left|x\right|-5=-17\Rightarrow\left|x\right|=-12\left(KTM\right)\)
\(\left|x\right|-5=-1\Rightarrow\left|x\right|=4\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
\(\left|x\right|-5=1\Rightarrow\left|x\right|=6\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
\(\left|x\right|-5=17\Rightarrow\left|x\right|=32\Rightarrow\left[{}\begin{matrix}x=32\\x=-32\end{matrix}\right.\)
Vậy để A \(\in\)Z thì x \(\in\) {-32;-6;-4;4;6;32}
Bài 3: A=2018-|x+2019|. Vì |x+2019|\(\ge\)0 nên -|x+2019|\(\le\)0=>2018-|x+2019|\(\le\) 2. Vậy A có GTLN = 2 khi x+2019=0 hay x=-2019. B=-10-\(\left|2x-\dfrac{1}{1009}\right|\). Vì \(\left|2x-\dfrac{1}{1009}\right|\ge0\Rightarrow-\left|2x-\dfrac{1}{1009}\right|\le0\Rightarrow-10-\left|2x-\dfrac{1}{1009}\right|\le-10\). Vậy B có GTLN = -10 khi 2x-\(\dfrac{1}{1009}=0\) => \(2x=\dfrac{1}{1009}\Rightarrow x=\dfrac{1}{1009}:2=\dfrac{1}{2018}\)
Bài 2: A=\(\left|5x+1\right|-\dfrac{3}{8}\). Vì \(\left|5x+1\right|\ge0\Rightarrow\left|5x+1\right|-\dfrac{3}{8}\ge\dfrac{-3}{8}\). Vậy A có GTNN = \(\dfrac{-3}{8}\) khi 5x+1= 0=> 5x= -1=> x = \(\dfrac{-1}{5}\). B=\(\left|2-\dfrac{1}{6}x\right|+0,25\) , vì \(\left|2-\dfrac{1}{6}x\right|\ge0\Rightarrow\left|2-\dfrac{1}{6}x\right|+0,25\ge0,25\) . Vậy B có GTNN = 0,25 khi \(2-\dfrac{1}{6}x=0\Rightarrow\dfrac{x}{6}=2\Rightarrow x=2.6=12\)
a: \(VT=\dfrac{1}{a+1}+\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{1}{a}\)=VP
b: \(VP=\dfrac{a+1-a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}=VP\)