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\(a,A=\frac{1-\sqrt{a^3}}{a-1}=-\frac{\sqrt{a^3}-1}{a-1}.\)
\(=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{a+\sqrt{a}+1}{\sqrt{a}+1}\)
\(b,B=3\sqrt{\frac{12\left(a-2\right)^2}{27}}=\sqrt{9}.\sqrt{\frac{12\left(a-2\right)^2}{27}}\)
\(=\sqrt{\frac{9.3.4.\left(x-2\right)^2}{27}}=2\sqrt{\left(x-2\right)^2}=2.|x-2|\)
\(c,C=\left(a-b\right)\sqrt{\frac{ab}{\left(a-b\right)^2}}=\sqrt{\frac{\left(a-b\right)^2ab}{\left(a-b\right)^2}}=\sqrt{ab}\)
\(BT=\frac{a^2\left(b-c\right)+b^2c-b^2a+c^2a-c^2b}{a^4\left(b^2-c^2\right)+b^4c^2-b^4a^2+c^4a^2-c^4b^2}\)
\(=\frac{a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)}{a^4\left(b^2-c^2\right)+b^2c^2\left(b^2-c^2\right)-\left(b^4-c^4\right)a^2}\)
\(=\frac{\left(b-c\right)\left(a^2+bc-a\left(b+c\right)\right)}{\left(b^2-c^2\right)\left(a^4+b^2c^2-a^2\left(b^2+c^2\right)\right)}\)
\(=\frac{\left(a-b\right)\left(a-c\right)}{\left(b+c\right)\left(a^2-b^2\right)\left(a^2-c^2\right)}\)
\(=\frac{1}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
\(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{a^4\left(b^2-c^2\right)+b^4\left(c^2-a^2\right)+c^4\left(a^2-b^2\right)}\)
= \(\frac{a^2\left(b-c\right)+b^2c-c^2b-a\left(b^2-c^2\right)}{a^4\left(b^2-c^2\right)+b^4c^2-c^4b^2-a^2\left(a^4-b^4\right)}\)
= \(\frac{\left(b-c\right)\left(a-b\right)\left(c-a\right)}{\left(b^2-c^2\right)\left(a^2-b^2\right)\left(c^2-a^2\right)}\)
= \(\frac{1}{\left(b+c\right)\left(a+b\right)\left(c+a\right)}\)
Đặt\(a+b=x\)
\(b+c=y\)
\(c+a=z\)
\(\Rightarrow x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(=\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]\)
\(=\left(a+b+c\right)\left[\left(a-c\right)^2+\left(a-b\right)^2+\left(b-c\right)^2\right]\)
\(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3=\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=a^3+b^3+c^3+3\left(abc+c^2a+b^2c+bc^2+a^2b+ca^2+ab^2+abc\right)\)
\(=a^3+b^3+c^3+3\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ca\left(a+b+c\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\Rightarrow\)\(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ca\right)\)
Lại có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=\left(a-b+b-c+c-a\right)^2\)
\(-2\left[\left(a-b\right)\left(b-c\right)+\left(b-c\right)\left(c-a\right)+\left(c-a\right)\left(a-b\right)\right]\)
\(=-2\left(ab-ca-b^2+bc+bc-ab-c^2+ca+ca-bc-a^2+ab\right)\)
\(=2\left(a^2+b^2+c^2-ab-bc-ca\right)=2\left(a+b+c\right)^2-6\left(ab+bc+ca\right)\)
\(\Rightarrow\)\(P=\frac{\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ca\right)}{2\left(a+b+c\right)^2-6\left(ab+bc+ca\right)}\)
\(=\frac{\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]}{2\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]}=\frac{a+b+c}{2}\)