Giải phương trình:a) \(\frac{1}{a+b-x}\)= \(\frac{1}{a}\)+ \(\frac{1}{b}\)+ \(\frac{1}{x}\) ( x là ẩn số)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 ( a,b,c là hằng số và dôi một khác nhau...
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Giải phương trình:
a) \(\frac{1}{a+b-x}\)= \(\frac{1}{a}\)+ \(\frac{1}{b}\)+ \(\frac{1}{x}\) ( x là ẩn số)
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 ( a,b,c là hằng số và dôi một khác nhau )
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/ \(\frac{1}{a+b-x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}\)
\(\Leftrightarrow\left(a+b\right)x^2-\left(a^2+b^2\right)x-ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(\left(a+b\right)x^2-\frac{2x\sqrt{a+b}.\left(a^2+b^2\right)}{2\sqrt{a+b}}+\frac{\left(a^2+b^2\right)^2}{4\left(a+b\right)}\right)-\frac{\left(a^2+b^2\right)^2}{4\left(a+b\right)}-ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(\sqrt{a+b}x-\frac{a^2+b^2}{2\sqrt{a+b}}\right)^2=\frac{\left(a^2+b^2\right)^2}{4\left(a+b\right)}+ab\left(a+b\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{\frac{\left(a^2+b^2\right)^2}{4\left(a+b\right)}+ab\left(a+b\right)}+\frac{a^2+b^2}{2\sqrt{a+b}}}{\sqrt{a+b}}\\x=\frac{-\sqrt{\frac{\left(a^2+b^2\right)^2}{4\left(a+b\right)}+ab\left(a+b\right)}+\frac{a^2+b^2}{2\sqrt{a+b}}}{\sqrt{a+b}}\end{cases}}\)