Cho \(a^2+4b+4=0, b^2+4c+4=0, c^2+4a+4=0\). Tính \(P=\left(\frac{a}{b}\right)^{2018}+\left(\frac{b}{c}\right)^{2020}+\left(\frac{c}{a}\right)^{2021}\)
Cho a,b,c >0 và \(\frac{b-20a+16c}{4a}=\frac{c-20b+16a}{4b}=\frac{a-20c+16b}{4c}\)
Tính giá trị \(F=\left(4+\frac{a}{4b}\right).\left(4+\frac{b}{4c}\right).\left(4+\frac{c}{4a}\right)\)
Trừ mỗi vế cho 1, ta có:
\(\frac{b-16a+16c}{4a}=\frac{c-16b+16a}{4b}=\frac{a-16c+16b}{4c}=\frac{a+b+c}{4.\left(a+b+c\right)}=\frac{1}{4}\)(vì a,b,c > 0 nên a+b+c>0)
\(\Leftrightarrow\hept{\begin{cases}b+16c=17a\\c+16a=17b\\a+16b=17c\end{cases}}\Leftrightarrow a=b=c\)
tự thay vào
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 = \(\left(2+\frac{a}{b}\right).\left(3+\frac{b}{c}\right).\left(4+\frac{c}{a}\right)\)
Cho 3 số x,y,z>0 thỏa mãn:
\(\frac{4a^2+\left(b-c\right)^2}{2a^2+b^2+c^2}+\frac{4b^2+\left(c-a\right)^2}{2b^2+c^2+a^2}+\frac{4c^2+\left(a-b\right)^2}{2c^2+a^2+b^2}\ge3\)
\(\frac{1}{a+b+c}=\frac{a+4b-c}{c}=\frac{b+4c-a}{a}=\frac{c+4a-b}{b}\)
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 \(P=\left(2+\frac{a}{b}\right)\left(3+\frac{b}{c}\right)\left(4+\frac{c}{a}\right)\)
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 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. Chứng minh rằng :
\(\frac{a}{\sqrt{4a^2+\left(b+c\right)^2}}+\frac{b}{\sqrt{4b^2+\left(c+a\right)^2}}+\frac{c}{\sqrt{4c^2+\left(a+b\right)^2}}\le\frac{3\sqrt{2}}{4}\)
\(P=\sum\frac{a}{\sqrt{\left(2a\right)^2+\left(b+c\right)^2}}\le\sqrt{2}\sum\frac{a}{2a+b+c}=\sqrt{2}\sum a\left(\frac{1}{a+b+a+c}\right)\le\frac{\sqrt{2}}{4}\sum\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{3\sqrt{2}}{4}\)
Dấu "=" xảy ra khi \(a=b=c\)
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 = \(\left(2+\frac{a}{b}\right)\left(3+\frac{b}{c}\right)\left(4+\frac{c}{a}\right)\)
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 -__-