Chứng minh 2013 < \(\dfrac{2013a}{a+b+c}+\dfrac{2013b}{b+c+d}+\dfrac{2013c}{c+d+a}+\dfrac{2013d}{d+a+b}< 4026\)
Những câu hỏi liên quan
Cho a, b, c, d là các số nguyên dương. Chứng tỏ rằng :
\(2013\) < \(\dfrac{2013a}{a+b+c}\) + \(\dfrac{2013b}{b+c+d}\) + \(\dfrac{2013c}{c+d+a}\) + \(\dfrac{2013d}{d+a+b}\) < 4026
Cho\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\left(a,b,c,d>0\right)\)
Tính A=\(\dfrac{2013a-2012b}{c+d}+\dfrac{2013b-2012c}{a+d}+\dfrac{2013c-2012d}{a+b}+\dfrac{2013d-2012a}{b+c}\)
Cho a,b,c,d là các số dương.Chứng tỏ rằng:
2013<\(\dfrac{2013a}{a+b+c}\)+\(\dfrac{2013b}{b+c+d}\)+\(\dfrac{2013c}{c+d+a}\)+\(\dfrac{2013d}{d+a+b}\)<4026
Giúp mik nha
Chứng minh \(2013< \frac{2013a}{a+b+c}+\frac{2013c}{b+c+d}+\frac{2013d}{d+a+b}< \) \(4026\)
cho 3 số thực a,b,c>o thoả mãn a+b+c=2013.cmr:\(\dfrac{a}{a+\sqrt{2013a+bc}}+\dfrac{b}{b+\sqrt{2013b+ac}}+\dfrac{c}{c+\sqrt{2013c+ab}}\le1\)
\(\sum\dfrac{a}{a+\sqrt{\left(a+b\right)\left(c+a\right)}}\le\sum\dfrac{a}{a+\sqrt{ab}+\sqrt{ac}}=\sum\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)
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cho 3 số thực a,b,c thoả mãn a+b+c=2013.
chứng minh \(\dfrac{a}{a+\sqrt{2013a}+bc}+\dfrac{b}{b+\sqrt{2013c+ab}}+\dfrac{c}{c+\sqrt{2013c+ab}}\le1\)
Cho các số a, b, c khác 0 thỏa mãn: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
Tính \(S=\dfrac{2013a^2-2014}{a^2+2bc}+\dfrac{2013b^2-2014}{b^2+2ca}+\dfrac{2013c^2-2014}{c^2+2ab}\)
Ta có kết quả tổng quát hơn như sau:
Cho $a,b,c \neq 0$ thỏa mãn $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0.$
Chứng minh rằng $$S={\frac {k{a}^{2}-k-1}{{a}^{2}+2\,bc}}+{\frac {{b}^{2}k-k-1}{2\,ac+{b}^{2}}}+{\frac {{c}^{2}k-k-1}{2\,ab+{c}^{2}}}=k$$
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Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\) (a, b, c, d > 0). Tính:
A=\(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)
cho\(\frac{a}{2b}\)=\(\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\)(a, b, c, d > 0). Tính:
A=\(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)
Vì a ; b ; c ; d > 0
=> a + b + c + d > 0
=> 2(a + b + c + d) > 0
=> 2a + 2b + 2c + 2d > 0
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
=> \(\frac{a}{2b}=\frac{1}{2}\Rightarrow2a=2b\Rightarrow a=b\)
Tương tự,ta được a = b = c = d
Khi đó A = \(\frac{2013a-2012b}{c+d}+\frac{2013b-2012c}{a+d}+\frac{2013c-2012d}{a+b}+\frac{2013d-2012a}{b+c}\)
= \(\frac{2013a-2012a}{2a}+\frac{2013b-2012b}{2b}+\frac{2013c-2012c}{2c}+\frac{2013d-2012d}{2d}\)(Vì a = b = c = d)
= \(\frac{a}{2a}+\frac{b}{2b}+\frac{c}{2c}+\frac{d}{2d}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
\(a,b,c,d>0\text{ nên : }a+b+c+d>0\Rightarrow\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
do đó: a=b=c=d hay A=1/2+1/2+1/2+1/2=2