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\(a,A=\frac{1-\sqrt{a^3}}{a-1}=-\frac{\sqrt{a^3}-1}{a-1}.\)
\(=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{a+\sqrt{a}+1}{\sqrt{a}+1}\)
\(b,B=3\sqrt{\frac{12\left(a-2\right)^2}{27}}=\sqrt{9}.\sqrt{\frac{12\left(a-2\right)^2}{27}}\)
\(=\sqrt{\frac{9.3.4.\left(x-2\right)^2}{27}}=2\sqrt{\left(x-2\right)^2}=2.|x-2|\)
\(c,C=\left(a-b\right)\sqrt{\frac{ab}{\left(a-b\right)^2}}=\sqrt{\frac{\left(a-b\right)^2ab}{\left(a-b\right)^2}}=\sqrt{ab}\)
\(P=\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(b+c\right)\left(b+a\right)}+\frac{c^2-ab}{\left(c+a\right)\left(c+b\right)}\)
\(P=\frac{\left(a^2-bc\right)\left(b+c\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}+\frac{\left(b^2-ac\right)\left(c+a\right)}{\left(b+c\right)\left(b+a\right)\left(c+a\right)}+\frac{\left(c^2-ab\right)\left(b+a\right)}{\left(c+a\right)\left(c+b\right)\left(b+a\right)}\)
\(P=\frac{a^2b+a^2c-b^2c-bc^2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}+\frac{b^2a+b^2c-a^2c-ac^2}{\left(b+c\right)\left(b+a\right)\left(c+a\right)}+\frac{c^2a+c^2b-a^2b-b^2a}{\left(c+a\right)\left(c+b\right)\left(b+a\right)}\)
\(P=\frac{0}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(P=0\)
Xét: \(f\left(x\right)=\frac{x^2-bc}{\left(x+b\right)\left(x+c\right)}+\frac{b^2-xc}{\left(b+c\right)\left(b+x\right)}+\frac{c^2-xb}{\left(c+x\right)\left(c+b\right)}\)
\(\Rightarrow f\left(a\right)=P\)
Ta có: \(f\left(b\right)=\frac{b^2-bc}{2b\left(b+c\right)}+\frac{b^2-bc}{2b\left(b+c\right)}+\frac{c^2-b^2}{\left(c+b\right)\left(c+b\right)}\)
\(\Rightarrow f\left(b\right)=\frac{2b\left(b-c\right)}{2b\left(b+c\right)}+\frac{\left(c-b\right)\left(c+b\right)}{\left(c+b\right)\left(c+b\right)}=\frac{b-c}{b+c}+\frac{c-b}{c+b}=0\left(1\right)\)
Chứng minh tương tự ta cũng có: \(f\left(c\right)=0\left(2\right)\)
Từ (1) và (2) suy ra \(f\left(x\right)=0\left(\forall x\right)\Rightarrow f\left(a\right)=0\left(\forall x\right)\)
Vậy A =0
Đây nhé
Đặt b + c = x ; c + a = y ; a + b = z
\(\Rightarrow\hept{\begin{cases}x+y=2c+b+a=2c+z\\y+z=2a+b+c=2a+x\\x+z=2b+a+c=2b+y\end{cases}}\)
\(\Rightarrow\frac{x+y-z}{2}=c;\frac{y+z-x}{2}=a;\frac{x+z-y}{2}=b\)
Thay vào PT đã cho ở đề bài , ta có :
\(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\)
\(=\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}-3\right)\)
\(\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)
( cái này cô - si cho x/y + /x ; x/z + z/x ; y/z + z/y)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(\frac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)