Cho a,b,c >0 và \(\frac{b-20a+16c}{4a}=\frac{c-20b+16a}{4b}=\frac{a-20c+16b}{4c}\)
Tính giá trị \(F=\left(4+\frac{a}{4b}\right).\left(4+\frac{b}{4c}\right).\left(4+\frac{c}{4a}\right)\)
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh:
a) \(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
b)\(\frac{4a^4+5b^4}{4c^4+5d^4}=\frac{a^2b^2}{c^2d^2}\)
c)\(\left(\frac{a-b}{c-d}\right)^{2005}=\frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}\)
d)\(\frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}=\frac{\left(a+b\right)^{2005}}{\left(c+d\right)^{2005}}\)
e)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
f)\(\frac{\left(20a^{2007}-11c^{2007}\right)^{2006}}{\left(20a^{2006}+11c^{2006}\right)^{2007}}=\frac{\left(20b^{2007}-11d^{2007}\right)^{2006}}{\left(20b^{2006}+11d^{2006}\right)^{2007}}\)
ừ, bạn bik làm thì giúp mình nha ^^
Cho \(a^2+4b+4=0, b^2+4c+4=0, c^2+4a+4=0\). Tính \(P=\left(\frac{a}{b}\right)^{2018}+\left(\frac{b}{c}\right)^{2020}+\left(\frac{c}{a}\right)^{2021}\)
Cộng vế với vế giả thiết:
\(a^2+4b+4+b^2+4c+4+c^2+4a+4=0\)
\(\Leftrightarrow\left(a^2+4a+4\right)+\left(b^2+4b+4\right)+\left(c^2+4c+4\right)=0\)
\(\Leftrightarrow\left(a+2\right)^2+\left(b+2\right)^2+\left(c+2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+2=0\\b+2=0\\c+2=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c=-2\)
\(\Rightarrow P=1+1+1=3\)
Cho a,b,c thực dương .CMR
\(\sqrt{\frac{\left(a+b\right)^3}{ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(b+c\right)^3}{bc\left(4b+4c+a\right)}}+\sqrt{\frac{\left(c+a\right)^3}{ca\left(4c+4c+b\right)}}\ge2\sqrt{2}\)
Gọi A là vế trái của BĐT cần chứng minh. Không mất tính tổng quát, ta giả sử a + b + c = 3. Áp dụng BĐT AM - GM ta có:
\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(a+b\right)^3}{8bc\left(4a+4b+c\right)}}+\frac{ab\left(4a+4b+c\right)}{27}\)\(\ge\frac{1}{2}\left(a+b\right)\)
Suy ra
\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}\)\(+\frac{ab\left(4a+4b+c\right)}{54}\ge\frac{1}{4}\left(a+b\right)\)
Tương tự
\(\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\frac{bc\left(4b+4c+a\right)}{54}\ge\frac{1}{4}\left(b+c\right)\)
và \(\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}+\frac{ca\left(4c+4a+b\right)}{54}\ge\frac{1}{4}\left(c+a\right)\)
Cộng ba BĐT trên ta có:
\(\frac{1}{2\sqrt{2}}A\ge B\)
Với \(A=\frac{1}{54}[ab\left(4a+4b+c\right)+bc\left(4b+4c+a\right)\)
\(+ca\left(4c+4a+b\right)]\)
\(=\frac{1}{54}\left[4ab\left(a+b\right)+4bc\left(b+c\right)+4ca\left(c+a\right)+3abc\right]\)
\(=\frac{1}{54}\left[4\left(a+b+c\right)\left(ab+bc+ca\right)-9abc\right]\)
\(\le\frac{1}{54}\left(a+b+c\right)^3=\frac{1}{2}\)
và \(B=\frac{1}{4}.2\left(a+b+c\right)=\frac{3}{2}\)
Suy ra \(\frac{1}{2\sqrt{2}}A\ge\frac{3}{2}-\frac{1}{2}=1\Rightarrow A\ge2\sqrt{2}\)
Vậy
\(\sqrt{\frac{\left(a+b\right)^3}{ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(a+b\right)^3}{bc\left(4a+4b+c\right)}}+\sqrt{\frac{\left(c+a\right)^3}{ca\left(4c+4a+b\right)}}\ge2\sqrt{2}\)(đpcm)
toán lớp 5 phiên bản hack não
cho a;b;c là các số thực dương thỏa mãn \(a^2+b^2+c^2=\frac{1}{3}\)CMR:\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}\ge a+b+c\)
Cho a,b,c thỏa mãn: \(\frac{1}{a+b+c}=\frac{a+4b-c}{c}=\frac{b+4c-a}{a}=\frac{c+4a-b}{b}\)
Tính P = \(\left(2+\frac{a}{b}\right).\left(3+\frac{b}{c}\right).\left(4+\frac{c}{a}\right)\)
\(\frac{1}{a+b+c}=\frac{a+4b-c}{c}=\frac{b+4c-a}{a}=\frac{c+4a-b}{b}\)
Tính P
\(P=\left(2+\frac{a}{b}\right)\left(3+\frac{b}{c}\right)\left(4+\frac{c}{a}\right)\)
Ta có : \(\frac{1}{a+b+c}=\frac{a+4b-c}{c}=\frac{b+4c-a}{a}=\frac{c+4a-b}{b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{1}{a+b+c}=\frac{a+4b-c}{c}=\frac{b+4c-a}{a}=\frac{c+4a-b}{b}\)
\(=\frac{a+4b-c+b+4c-a+c+4a-b}{a+b+c}=\frac{4\left(a+b+c\right) }{a+b+c}=4\)
Có : \(\frac{1}{a+b+c}=4\Leftrightarrow1=4\left(a+b+c\right)\Rightarrow a+b+c=\frac{1}{4}\)
Đến đây tự làm nốt
cho a,b,c thỏa mãn
\(\frac{1}{a+b+c}=\frac{a+4b-c}{c}=\frac{b+4c-a}{a}=\frac{c+4a-b}{b}\)
tính \(P=\left(2+\frac{a}{b}\right)\left(3+\frac{b}{c}\right)\left(4+\frac{c}{a}\right)\)
đặt \(P=\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}\)
Q=8ab(4a+4b+c)+8bc(4b+4c+a)+8ca(4c+4a+b)
=32(a+b+c)(ab+bc+ca)-72abc
áp dụng holder ta có:
\(P^2Q\ge8\left(a+b+c\right)^3\)
theo schur thì \(\left(a+b+c\right)^3\ge4\left(a+b+c\right)\left(ab+bc+ca\right)-9abc\)
\(\Rightarrow8\left(a+b+c\right)^3\ge32\left(a+b+c\right)\left(ab+bc+ca\right)-72abc\)
\(\Rightarrow P^2\ge\frac{8\left(a+b+c\right)^3}{Q}\ge1\left(Q.E.D\right)\)
Cho a, b,c thỏa mãn: \(\frac{1}{a+b+c}=\frac{a+4b-c}{c}=\frac{b+4c-a}{a}\frac{c+4a-b}{b}\)
Tính P = \(\left(2+\frac{a}{b}\right)\left(3+\frac{b}{c}\right)\left(4+\frac{c}{a}\right)\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{1}{a+b+c}=\frac{a+4b-c+b+4c-a+c+4a-b}{a+b+c}\)
\(=\frac{4\left(a+b+c\right)}{a+b+c}=4\)
\(\Rightarrow\left\{{}\begin{matrix}4c=a+4b-c\\4a=b+4a-a\\4b=c+4a-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5c=a+4b\\5a=b+4c\\5b=c+4a\end{matrix}\right.\)
\(\Rightarrow a=b=c\)
\(P=\left(2+\frac{a}{b}\right)\left(3+\frac{b}{c}\right)\left(4+\frac{c}{a}\right)\)
\(=\left(2+1\right)\left(3+1\right)\left(4+1\right)\)
\(=3.4.5=60\)
Vậy .............
Cái đề thiếu dấu " = " kìa -__-