Question

1. Giải ft

3(\(\sqrt{6-5x}-\sqrt{x+3}\) = 3x² - x-5.

2. Cho a,b,c là các số thực dương sao cho a + b + c = 1. Chứng minh rằng :

              \(\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\ge\frac{9}{4}.\)

Answer 1

1)ĐK:\(x\in\left[-3;\frac{6}{5}\right]\)

pt\(\Leftrightarrow3\left(x^2-x+2\right)-3\left[\sqrt{6-5x}-\left(x-2\right)\right]+\left[3\sqrt{x+3}-\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(x^2-x+2\right)\left(\frac{3}{\sqrt{6-5x}+x-2}+\frac{1}{3\sqrt{x+3}+x+5}+3\right)=0\)

\(\Leftrightarrow x^2\)-x+2=0(do(...)>0)

\(\Leftrightarrow x=-2\)hoặc \(x=1\)(t/m)

Answer 2

ÁD BĐT Bunhiacopxki:

\(\left(a+b+c\right)\left[\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\right]\ge\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\)

Lại có:\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)

\(=\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\ge\frac{9}{2\left(a+b+c\right)}-3=\frac{3}{2}\)

\(\Rightarrow VT\ge\left(\frac{3}{2}\right)^2\)=\(\frac{9}{4}\)(đpcm)

Dấu''='' xảy ra\(\Leftrightarrow a=b=c=\frac{1}{3}\)