\(\frac{x-a}{a+b}+\frac{x-b}{a-b}=\frac{2ab}{b^2-a^2}\) (ĐKXĐ: a\(\pm\)b)

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

\(\Leftrightarrow\frac{-a^2+xa-xb+ab-b^2+xa+xb-ab+2ab}{a^2-b^2}=0\)

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

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

Vậy phương trình có nghiệm \(x=\frac{\left(a-b\right)^2}{2a}\)

Giải

Điều kiện xác định của phương trình : \(a\ne\pm b\)

Biến đổi phương trình:

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

\(\Leftrightarrow ax-bx-a^2+ab+ax+bx-ab-b^2=-2ab\)

\(\Leftrightarrow2ax=a^2+b^2-2ab\)

\(\Leftrightarrow2ax=\left(a-b\right)^2\)

Nếu \(a\ne0\) thì \(x=\frac{\left(a-b\right)^2}{2a}\)

Nếu a = 0 thì \(2ax=\left(a-b\right)^2\) có dạng \(0x=b^2\). Do \(a\ne b\) nên \(b\ne0\), phương trình vô nghiệm

Kết luận

Nếu \(a\ne0\), \(a\ne\pm b\) thì \(S=\left\{\frac{\left(a-b\right)^2}{2a}\right\}\)

Còn lại, \(S=\varnothing\)

\(\frac{x-a}{a+b}+\frac{x-b}{a-b}=\frac{2ab}{b^2-a^2}.\)

\(\Rightarrow\frac{ax-bx-a^2+ab+ax+bx-ba-b^2}{\left(a+b\right).\left(a-b\right)}=\frac{-2ab}{\left(a+b\right).\left(a-b\right)}\)

\(\Rightarrow2ax=a^2-2ab+b^2\)

=> 2ax = (a-b)²

nếu a=0; \(b\ne0\)

=> \(x\in\varnothing\)

nếu a=0, b=0

=> \(x\in R\)

nếu \(a\ne0;b=0\)

=> x = a/2

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

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

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

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

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

\(\frac{ax-b}{a}+(a+b+1)x>\frac{2b}{a}\)

<=> \(x-\frac{b}{a}+\left(a+b+1\right)x>\frac{2b}{a}\)

<=> \(\left(a+b+2\right)x>\frac{3b}{a}\)

Giờ biện luận theo  a và b thôi

Quy đồng lên, lấy MTC là (a-b)(b-c)(a-c)

PT đâu ????