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Xét A= \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=a.\frac{a}{b+c}+b.\frac{b}{c+a}+c.\frac{c}{a+b}\)
\(=a\left(\frac{a}{b+c}+1-1\right)+b\left(\frac{b}{c+a}+1-1\right)+c\left(\frac{c}{a+b}+1-1\right)\)
\(=a\left(\frac{a+b+c}{b+c}-1\right)+b\left(\frac{a+b+c}{c+a}-1\right)+c\left(\frac{a+b+c}{a+b}-1\right)\)
\(=a.\frac{a+b+c}{b+c}-a+b.\frac{a+b+c}{c+a}-b+c.\frac{a+b+c}{a+b}-c\)
\(=\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(a+b+c\right)\) =0
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{x}{1+y+xz}=\frac{x\left(x^2+y+\frac{z}{x}\right)}{\left(1+y+xz\right)\left(x^2+y+\frac{z}{x}\right)}\le\frac{x^3+xy+z}{\left(x+y+z\right)^2}\)
\(\le\frac{x+y+z}{\left(x+y+z\right)}=\frac{1}{x+y+z}\)
Tương tự ta cũng có: \(\frac{y}{1+z+xy}\le\frac{1}{x+y+z};\frac{z}{1+x+yz}\le\frac{1}{x+y+z}\)
Cộng theo vế ta có: \(\frac{x}{1+y+xz}+\frac{y}{1+z+xy}+\frac{z}{1+x+yz}\le\frac{1+1+1}{x+y+z}=\frac{3}{x+y+z}\)
Bài 3:
a: ĐKXĐ: x<>2
b: \(M=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
c: Khi x=4001/2000 thì \(M=\dfrac{3}{\dfrac{4001}{2000}-2}=3:\dfrac{1}{2000}=6000\)
Bài 2:
a: \(=6x^2+30x+x+5-\left(6x^2-3x-10x+5\right)\)
\(=6x^2+31x+5-6x^2+13x-5=18x⋮6\)
b: \(=x^3+2x^2+3x^2+6x-x-2-x^3+2\)
\(=5x^2+5x=5x\left(x+1\right)⋮2\)