b/ \(\frac{\left(b-c\right)\left(1+a\right)^2}{x+a^2}+\frac{\left(c-a\right)\left(1+b\right)^2}{x+b^2}+\frac{\left(a-b\right)\left(1+c\right)^2}{x+c^2}=0\)

\(\Leftrightarrow x^2-\left(ab+bc+ca+2a+2b+2c+1\right)x+2abc+ab+bc+ca=0\)

Đặt: \(\hept{\begin{cases}ab+bc+ca+2a+2b+2c+1=m\\2abc+ab+bc+ca=n\end{cases}}\) (đặt cho gọn)

\(\Leftrightarrow x^2-mx+n=0\)

\(\Leftrightarrow\left(x^2-\frac{2m}{2}x+\frac{m^2}{4}\right)-\frac{m^2}{4}+n=0\)

\(\Leftrightarrow\left(x-\frac{m}{2}\right)^2=\frac{m^2}{4}-n\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{m^2}{4}-n}+\frac{m}{2}\\x=-\sqrt{\frac{m^2}{4}-n}+\frac{m}{2}\end{cases}}\)

a/ \(\frac{1}{a+b-x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}\)

\(\Leftrightarrow\left(a+b\right)x^2-\left(a^2+b^2\right)x-ab\left(a+b\right)=0\)

\(\Leftrightarrow\left(\left(a+b\right)x^2-\frac{2x\sqrt{a+b}.\left(a^2+b^2\right)}{2\sqrt{a+b}}+\frac{\left(a^2+b^2\right)^2}{4\left(a+b\right)}\right)-\frac{\left(a^2+b^2\right)^2}{4\left(a+b\right)}-ab\left(a+b\right)=0\)

\(\Leftrightarrow\left(\sqrt{a+b}x-\frac{a^2+b^2}{2\sqrt{a+b}}\right)^2=\frac{\left(a^2+b^2\right)^2}{4\left(a+b\right)}+ab\left(a+b\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{\frac{\left(a^2+b^2\right)^2}{4\left(a+b\right)}+ab\left(a+b\right)}+\frac{a^2+b^2}{2\sqrt{a+b}}}{\sqrt{a+b}}\\x=\frac{-\sqrt{\frac{\left(a^2+b^2\right)^2}{4\left(a+b\right)}+ab\left(a+b\right)}+\frac{a^2+b^2}{2\sqrt{a+b}}}{\sqrt{a+b}}\end{cases}}\)

a) sửa đề: \(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)

=\(\frac{-x^2\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-y^2\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-z^2\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(-\frac{\left\lbrace x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

xét tử số:

Tử=\(x^2y-x^2z+y^2z-y^2x+z^2x-z^2y\)

=\(x^2\left(y-z\right)-x\left(y^2-z^2\right)+yz\left(y-z\right)\)

=\(x^2\left(y-z\right)-x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\)

=\(\left(y-z\right)\left\lbrace x^2-x\left(y+z\right)+yz\right\rbrace\)

=\(\left(y-z\right)\left\lbrace x\left(x-y\right)-z\left(x-y\right)\right\rbrace\)

=\(\left(y-z\right)\left(x-y\right)\left(x-z\right)\)

=\(-\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

thay lại vào biểu thức cũ:

\(\Rightarrow-\frac{\left\lbrace-\left(x-y\right)\left(y-z\right)\left(z-x\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(1\)

b) \(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\)

=\(\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

\(\frac{(b-c)(1+a)^2}{x+a^2}+\frac{(c-a)(1+b)^2}{x+b^2}+\frac{(a-b) (1+c)^2}{x+c^2}=0\)

\(\Leftrightarrow \sum (b-c)(1+a)^2(x+b^2)(x+c^2)=0\)

\(\Leftrightarrow (a-b)(b-c)(c-a)(x^2+(-2a-ca-ba-cb-2c-2b-1)x+ba+2acb+cb+ca)=0\)

\(\Leftrightarrow x^2+(-2a-ca-ba-cb-2c-2b-1)x+ba+2acb+cb+ca=0\)

Xét phương trình  \(x^2+(-2a-ca-ba-cb-2c-2b-1)x+ba+2acb+cb+ca=0\)

Ta thấy \(\Delta=(2a+2b+2c+ab+bc+ca-1)^2+8(a+b+c-abc)\)

Nếu \(\Delta <0\) thì phương trình vô nghiệm

Nếu \(\Delta =0\) thì phương trình có nghiệm kép

Nếu \(\Delta >0\) thì phương trình có hai nghiệm