Các câu hỏi tương tự
Tìm a,b>=0 biết:
\(\left(a^2+b+\frac{3}{4}\right)\)\(\left(b^2+a+\frac{3}{4}\right)\)=\(\left(2a+\frac{1}{2}\right)\)\(\left(2b+\frac{1}{2}\right)\)
\(\left(a^2+b+\frac{3}{4}\right)\left(b^2+a+\frac{3}{4}\right)\ge\left(2a+\frac{1}{2}\right)\left(2b+\frac{1}{2}\right)\)
\(\frac{b^2c^3}{a^2+\left(b+c\right)^3}+\frac{c^2a^3}{b^2+\left(c+a\right)^3}+\frac{a^2b^3}{c^2+\left(a+b\right)^3}\ge\frac{9abc}{4\left(3abc+a^2c+b^2a+c^2b\right)}\)voi a,b,c>0
tìm các số thực a và b thỏa mãn
\(\left(a^2+b+\frac{3}{4}\right)\left(b^2+a+\frac{3}{4}\right)=\left(2a+\frac{1}{2}\right)\left(2b+\frac{1}{2}\right)\)
giúp với nha mơn nhiều
a) Tìm m để pt \(\left(x^2-1\right)\left(x+3\right)\left(x+5\right)=m\) có 4 nghiệm thỏa: \(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}=-1\)
b) Tìm các số \(a,b,c\ge0\)sao cho: \(\left(a^2+b+\frac{3}{4}\right)\left(b^2+a+\frac{3}{4}\right)=\left(2a+\frac{1}{2}\right)\left(2b+\frac{1}{2}\right)\)
Cho các số thực a, b, c > 0. Chứng minh rằng :
\(\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}+\frac{b^2}{\left(2b+a\right)\left(2b+c\right)}+\frac{c^2}{\left(2c+a\right)\left(2c+b\right)}\ge\frac{1}{3}\)
Tìm \(a,b\ge0\) thỏa mãn \(\left(a^2+b+\frac{3}{4}\right)\left(b^2+a+\frac{3}{4}\right)=\left(2a+\frac{1}{2}\right)\left(2b+\frac{1}{2}\right)\)
Pfrac{a}{sqrt{left(b+1right)left(b^2-b+1right)}}+frac{b}{sqrt{left(c+1right)left(c^2-c+1right)}}+frac{c}{sqrt{left(a+1right)left(a^2-a+1right)}}gefrac{2a}{b^2+2}+frac{2b}{c^2+2}+frac{2c}{a^2+2}left(a+b+cright)-left(frac{ab^2}{b^2+2}+frac{bc^2}{c^2+2}+frac{ca^2}{a^2+2}right)6-left(frac{2ab^2}{b^2+4+b^2}+frac{2bc^2}{c^2+4+c^2}+frac{2ca^2}{a^2+4+a^2}right)ge6-left(frac{2ab}{b+4}+frac{2bc}{c+4}+frac{2ca}{a+4}right)6-left(2a+2b+2c-frac{8a}{b+4}-frac{8b}{c+4}-frac{8c}{a+4}right)frac{8a}{b+4}+frac{8b}{...
Đọc tiếp
\(P=\frac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}+\frac{b}{\sqrt{\left(c+1\right)\left(c^2-c+1\right)}}+\frac{c}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\)
\(\ge\frac{2a}{b^2+2}+\frac{2b}{c^2+2}+\frac{2c}{a^2+2}=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+2}+\frac{bc^2}{c^2+2}+\frac{ca^2}{a^2+2}\right)\)
\(=6-\left(\frac{2ab^2}{b^2+4+b^2}+\frac{2bc^2}{c^2+4+c^2}+\frac{2ca^2}{a^2+4+a^2}\right)\ge6-\left(\frac{2ab}{b+4}+\frac{2bc}{c+4}+\frac{2ca}{a+4}\right)\)
\(=6-\left(2a+2b+2c-\frac{8a}{b+4}-\frac{8b}{c+4}-\frac{8c}{a+4}\right)\)
\(=\frac{8a}{b+4}+\frac{8b}{c+4}+\frac{8c}{a+4}-6=\frac{8a^2}{ab+4a}+\frac{8b^2}{bc+4b}+\frac{8c^2}{ca+4c}-6\)
\(\ge\frac{8\left(a+b+c\right)^2}{\left(ab+bc+ca\right)+4\left(a+b+c\right)}-6\ge\frac{288}{\frac{\left(a+b+c\right)^2}{3}+24}-6=2\)
cho các số thực a,b không âm:
Chứng minh rằng: \(\left(a^2+b+\frac{3}{4}\right)+\left(b^2+a+\frac{3}{4}\right)\ge\left(2a+\frac{1}{2}\right)\left(2b+\frac{1}{2}\right)\)