Cho hai so huu ti \(\frac{a}{b}\)va\(\frac{c}{d}\)(b>0, d>0)
Chung minh:\(\frac{a}{b}\)< \(\frac{c}{d}\)
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Cho cac so huu ti x=\(\frac{a}{b}\), y =\(\frac{c}{d}\)b>0,d>0 va cac so tu nhien m,n voi m ,n khac 0 CMR
\(\frac{a}{b}< \frac{c}{d}\) thi \(\frac{a}{b}< \frac{ma+nc}{mb+nd}< \frac{c}{d}\)
Cho hai so huu ti a/b va c/d (b>0, d>0). Chung to rang :
a, Neu a/b < c/d thi ad<cd
b, Neu ad<bc thi a/b < c/d
cho so huu ti a/b va c/d voi b>0 chung to rang neu a/b > c/d thì a/b<a+c/b+d <c/d
C1 : Theo ví dụ trên ta có : \(\frac{a}{b}< \frac{c}{d}\)=> ad < bc
Suy ra :
<=> ad + ab < bc + ba <=> a[b + d] < b[a + c] <=> \(\frac{a}{b}< \frac{a+c}{b+d}\)
Mặt khác ad < bc => ad + cd < bc + cd
<=> d[a + c] < [b + d]c <=> \(\frac{a+c}{b+d}< \frac{c}{d}\)
Từ đó suy ra \(\frac{a}{b}< \frac{a+c}{b+c}< \frac{c}{d}\)
C2 : Xét hiệu : \(\frac{a+c}{b+d}-\frac{a}{b}=\frac{ab+bc-ab-ad}{b(b+d)}=\frac{bc-ad}{b(b+d)}>0\)
\(\frac{c}{d}-\frac{a+c}{b+d}=\frac{bc+cd-ad-cd}{d(b+d)}=\frac{bc-ad}{d(b+d)}>0\)
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a)Chung to rang neu a/b <c/d (b<0,d<0) thi a/b < a+c/d+b < c/d
b)Hay viet 3 so huu ti xen giua -1/3 va -1/4
\(\frac{a}{b}< \frac{c}{d}\) => ad < bc
=> ad + ab < bc + ab
=> a(b + d) < b(a + c)
=> \(\frac{a}{b}< \frac{a+c}{b+d}\)
=> ad < bc
=> ad + cd< bc + cd
=> d(a + c) < c(b + d)
=> \(\frac{a+c}{b+d}< \frac{c}{d}\)
=> đccm
b) \(\frac{-1}{3}=\frac{-16}{48}< \frac{-15}{48}\); \(\frac{-14}{48};\frac{-13}{48}\)\(< \frac{-12}{48}=\frac{-1}{4}\)
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cho so huu ti a/b voi a,b thuoc Z, b>0. Chung minh rang: neu co a<b va >0 thi a/b<a+c/b+c
Ta có a<b
=>ac<bc (c>0)
=> ac+ ab < bc+ ab
=> a(b+c) < b(a+c)
=> a/b< a+c/b+c(đpc/m)
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gia su a,b la 2 so huu ti duong va khong phai la binh phuong cua mot so huu ti
chung minh rang :neu x,y la hai so huu ti sao cho \(m=x\sqrt{a}+y\sqrt{b}\)la so huu ti thi m=0
cho a, b, c, d khac 0 va thoa man
ac=b^2; bd=c^2
chung minh \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
cho 3 so a,b,c khac 0 va (a+b+c)^2=a^2+b^2+c^2 . chung minh \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3abc\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\)
\(\Rightarrow2\left(ab+bc+ac\right)=0\)
\(\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{\left(a+b+c\right)}{abc}=0\)
\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)
\(\Rightarrow\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{-1}{c}\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(\frac{-1}{c}\right)^3\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}.\left(-\frac{1}{c}\right)=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{ab}=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\left(đpcm\right)\)
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\(\left(a+b+c\right)^2=a^2+b^2+c^2\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{ab+bc+ac}{abc}=0\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\left(\frac{1}{a}\right)^3+\left(\frac{1}{b}\right)^3+\left(\frac{1}{c}\right)^3=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
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cho \(\frac{a}{b}\)=\(\frac{c}{d}\) chung minh \(\frac{3a^6+c^6}{3b^6+d^6}\)=\(\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\) va b+d khac 0