\(\text{Ta có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0.\)

\(\Leftrightarrow bc+ac+ab=0\Rightarrow\hept{\begin{cases}bc=-ac-ab\\ac=-bc-ab\\ab=-bc-ac\end{cases}}\)

\(\Rightarrow BT\text{hức}=\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}\)

\(=\frac{bc}{a^2-ac-ab+bc}+\frac{ac}{b^2-bc-ab+ac}+\frac{ab}{c^2-bc-ac+ab}\)

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

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

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

\(=\frac{b^2c-bc^2-a^2c+ac^2+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{c\left(b^2-a^2\right)-c^2\left(b-a\right)+ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+ab\left(a+b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left(c^2-ac-bc+ab\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

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

Lâu ko lm đổi dấu hơi thừa ra!! ko hiểu chỗ nào thì ib mk giải thích cho

\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right).\)

\(\Rightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+4\left(x^2-5x-2012\right)^2=0\)

\(\Leftrightarrow\left[\left(2x^2+x-2013\right)-2\left(x^2-5x-2012\right)\right]^2=0\)(Hằng đẳng thức)

\(\Leftrightarrow2x^2+x-2013-2x^2+10x+4024=0\)

\(\Leftrightarrow11x=-2011\)

\(\Leftrightarrow x=\frac{-2011}{11}\)

quy đồng nha bạn

\(\left(x^5+x^3+x^2+1\right):\left(x^3+1\right)\)

\(=\left[x^2\left(x^3+1\right)+1\left(x^3+1\right)\right]:\left(x^3+1\right)\)

\(=\left(x^3+1\right)\left(x^2+1\right):\left(x^3+1\right)\)

\(=x^2+1\)

Gọi số cần tìm là \(\overline{abc}\)

Theo bài ra có:

\(\overline{9abc}+\overline{abc9}=10384\Rightarrow9000+\overline{abc}+10.\overline{abc}+9=10384\)

\(\Rightarrow11.\overline{abc}=1375\Rightarrow\overline{abc}=125\)

     \(x^3-7x^2-13x+91=0\)

\(\Rightarrow x^2\left(x-7\right)-13\left(x-7\right)=0\)

\(\Rightarrow\left(x-7\right)\left(x^2-13\right)=0\)

\(\Rightarrow\left(x-7\right)\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)

Tìm được \(x\in\left\{7;\sqrt{13};-\sqrt{13}\right\}\)