cho a,b,c,d,e dương CMR \(\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+e}+\dfrac{d}{e+a}+\dfrac{e}{a+b}\ge\dfrac{5}{2}\)
1.Cho a,b,c,d,e,f \(\ne\) 0 thoả mãn : \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{e}{f}\)
Cmr:\(\left(\dfrac{a+b+c+d+e}{b+c+d+e+f}\right)^5=\dfrac{a}{f}\) với (a+b+c+d+e+f \(\ne\)0)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{e}{f}=\dfrac{a+b+c+d+e}{b+c+d+e+f}=k\)
Ta có:
\(\dfrac{a}{f}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{e}.\dfrac{e}{f}=k^5=\left(\dfrac{a+b+c+d+e}{b+c+d+e+f}\right)^5\)
CMR nếu \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}thì\dfrac{a}{e}=(\dfrac{a-b+c-d}{b-c+d-e})^4\)
Cho số hực dương a,b,c,d, e khác 0 thỏa mãn\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)
Chứng minh rằng\(\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\)=\(\dfrac{a}{e}\)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=k\Rightarrow a=bk;b=ck;c=dk;d=ek\)
\(\Rightarrow a=bk=ck^2=dk^3=ek^4;b=ek^3\)
\(\Rightarrow\dfrac{a}{e}=\dfrac{ek^4}{e}=k^4\left(1\right)\)
Ta có \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}=\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\left(2\right)\)
Lại có \(\dfrac{a^4}{b^4}=\left(\dfrac{a}{b}\right)^4=\left(\dfrac{ek^4}{ek^3}\right)^4=k^4\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\RightarrowĐpcm\)
Cho các số nguyên dương a,b,c,d,e,f biết :
\(\dfrac{a}{b}>\dfrac{c}{d}>\dfrac{e}{f}\) và \(af-be=1.CMR:d\ge b+f\)
cho \(\dfrac{a}{b}\) =\(\dfrac{c}{d}\) cm rằng
a) \(\dfrac{a}{a-b}\) =\(\dfrac{c}{c-d}\) b)\(\dfrac{a}{b}\) =\(\dfrac{a+c}{b+d}\) c) \(\dfrac{a}{3a+d}\) =\(\dfrac{c}{3c+d}\) d)\(\dfrac{a.c}{b.d}\) =\(\dfrac{a^2+c^2}{b^2+c^2}\) e)\(\dfrac{a.b}{c.d}\) =\(\dfrac{a^2-b^2}{c^2-d^2}\) f)\(\dfrac{a.b}{c.d}\) =\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
mn giúp mk vs ạ! thanks
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)
\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)
\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)
Cho các số thực a,b,c,d,e thỏa mãn \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)chứng minh rằng: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)=\dfrac{a^2}{b.c}\)
Sửa: CMR: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\\ \Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3\left(1\right)\\ \dfrac{a}{b}=\dfrac{b}{c}=k\Rightarrow a=bk;b=ck\Rightarrow a=ck^2\\ \Rightarrow\dfrac{a^2}{bc}=\dfrac{c^2k^4}{ck\cdot c}=k^3=\left(\dfrac{a}{b}\right)^3\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)
cho a,b,c,d là các số dương thỏa mãn a+b+c+d=1.CMR:
\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+d}+\dfrac{d^2}{d+a}\ge\dfrac{1}{2}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+d}+\dfrac{d^2}{a+d}\)
\(\ge\dfrac{\left(a+b+c+d\right)^2}{a+b+b+c+c+d+d+a}\)
\(=\dfrac{\left(a+b+c+d\right)^2}{2\left(a+b+c+d\right)}=\dfrac{a+b+c+d}{2}=\dfrac{1}{2}=VP\)
Đẳng thức xảy ra khi \(a=b=c=d=\dfrac{1}{4}\)
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)
CMR :(\(\dfrac{a+b+c+d}{b+c+d+e}\))4=\(\dfrac{a}{e}\)
giúp mình giải bài này nha
Đặt \(k=\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)
Ta có: \(k=\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{a+b+c+d}{b+c+d+e}\) ( t/c dãy tỉ số bằng nhau )
\(\Rightarrow k^4=\left(\dfrac{a+b+c+d}{b+c+d+e}\right)^4\) (1)
\(k^4=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{e}=\dfrac{a}{e}\) (2)
Từ (1), (2) \(\Rightarrow\left(\dfrac{a+b+c+d}{b+c+d+e}\right)^4=\dfrac{a}{e}\left(đpcm\right)\)
Vậy...
Ta có :
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)
=> \(\left(\dfrac{a}{b}\right)^4=\dfrac{a.b.c.d}{b.c.d.e}=\dfrac{a}{e}\) (1)
Mà :
\(\dfrac{a}{b}=\dfrac{a+b+c+d}{b+c+d+e}\left(TTCCDTSBN\right)\)
=> \(\left(\dfrac{a}{b}\right)^4=\left(\dfrac{a+b+c+d}{b+c+d+e}\right)^4\)(2)
=> từ (1) và (2)
=> \(\left(\dfrac{a+b+c+d}{b+c+d+e}\right)^4=\dfrac{a}{e}\left(đp.....\right)\)
cho a,b,c>0 ,\(a^2+b^2+c^2=1\).CMR
\(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\ge\dfrac{1}{2}\)
Dùng bunhia nhé mn.Giúp e với e cần gấp ạ !
Đặt vế trái BĐT là P
Ta có:
\(\left(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\right)\left(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)\right)\ge\left(a^2+b^2+c^2\right)^2\)
\(\Rightarrow P.\left(2ab+2bc+2ca\right)\ge1\)
\(\Rightarrow P\ge\dfrac{1}{2\left(ab+bc+ca\right)}\ge\dfrac{1}{2\left(a^2+b^2+c^2\right)}=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)