Tick cho mình trước khi đọc nha thể nào cũng đúng
Ta có \(x^2+6x^2+6+\left(\frac{x+3}{x+4}\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)^2+\left(\frac{x+3}{x+4}\right)^2-3=0\)
\(\Leftrightarrow\left(x+3\right)^2-2\left(x+3\right)\frac{\left(x+3\right)}{\left(x+4\right)}+\left(\frac{x+3}{x+4}\right)^2+2\frac{\left(x+3\right)^2}{\left(x+4\right)}-3=0\)
\(\Leftrightarrow\left(x+3-\frac{x+3}{x+4}\right)^2+2\frac{\left(x+3\right)^2}{\left(x+4\right)}-3=0\)
\(\Leftrightarrow\left(\frac{\left(x+3\right)\left(x+4\right)-\left(x+3\right)}{x+4}\right)^2+2\frac{\left(x+3\right)^2}{\left(x+4\right)}-3=0\)
\(\Leftrightarrow\left(\frac{x^2+7x+12-\left(x+3\right)}{x+4}\right)^2+2\frac{\left(x+3\right)^2}{\left(x+4\right)}-3=0\)
\(\Leftrightarrow\left(\frac{x^2+6x+9}{x+4}\right)^2+2\frac{\left(x+3\right)^2}{\left(x+4\right)}-3=0\)
\(\Leftrightarrow\left(\frac{\left(x+3\right)^2}{x+4}\right)^2+2\frac{\left(x+3\right)^2}{\left(x+4\right)}-3=0\)
Đặt \(\frac{\left(x+3\right)^2}{x+4}=a\) pt <=> \(a^2+2a-3=0\Leftrightarrow\left(a+3\right)\left(a-1\right)=0\)
nên a=-3 hoặc a=1
Với a=-3 thì \(\frac{\left(x+3\right)^2}{x+4}=-3\Leftrightarrow x^2+6x+9=-3\left(x+4\right)\Leftrightarrow x^2+9x+21=0\)
nên pt này vô nghiệm
Với a=1 thì \(\frac{\left(x+3\right)^2}{x+4}=1\Leftrightarrow x^2+6x+9=\left(x+4\right)\Leftrightarrow x^2+5x+5=0\)
Giải ra được 2 nghiệm
Vậy....
