Tuyển Cộng tác viên Hoc24 nhiệm kì 28 tại đây: https://forms.gle/GrfwFgzveoKLVv3p6
Trả lời: pt<=> \(x^2+\frac{x^2}{\left(x+1\right)^2}-2\frac{x^2}{\left(x+1\right)}+2\frac{x^2}{\left(x+1\right)}=3\Leftrightarrow\left(x-\frac{x}{x+1}\right)^2+2\frac{x^2}{x+1}=3\)
\(\Leftrightarrow\left(\frac{x^2}{x+1}\right)^2+2\frac{x^2}{x+1}-3=0\)
Đặt \(a=\frac{x^2}{\left(x+1\right)}\) thì pt<=> \(a^2+2a-3=0\)
nên a=3 hoặc a=-1
GPT còn lại tìm ra nghiệm ngay mà
Gpt
a) \(\left(x-3\right)\left(x+1\right)+4\left(x-3\right)\sqrt{\frac{x+1}{x-3}}=-3\)
b)\(\frac{x\left(x^2+1\right)}{\left(x^2-x+1\right)}=2\)
GPT :\(x^3+\left(\frac{x}{x-1}\right)^3+\frac{3x^2}{x-1}=2\)
Gpt \(x^2+\frac{x^2}{\left(x+1\right)^2}=3\)
gpt : \(x^2-4x+5-\frac{3x}{x^2+x+1}=\left(x-1\right)\left(1-\frac{2\sqrt{1-x}}{\sqrt{x^2+x+1}}\right)\)
gpt\(\left(x+1\right)\left(x^2+1\right)=\left(2y+1\right)^2\)
tìm x,y t/m:\(\orbr{\begin{cases}\frac{x^2}{\left(y+1\right)^2}+\frac{y^2}{\left(x+1\right)^2}=\frac{1}{2}\\3xy=x+y+1\end{cases}}\)
GPT: \(x^3+\frac{x^3}{\left(x-1\right)^3}+\frac{3x^2}{x-1}-2=0\)
gpt:
1, \(\sqrt{x^2+x+2}=\frac{x^2+5x+2}{2\left(x+1\right)}\)
2, \(\left(x+1\right)\sqrt{x^2-2x+3}=x^2+1\).
gpt:
\(3\left(x^2-3x+1\right)+\sqrt{3\left(x^4+x^2+1\right)}=0\)
\(\sqrt[3]{x^3+5x^2}-1=\sqrt{\frac{5x^2-2}{6}}\)
GPT
\(x.\frac{3-x}{x+1}\left(x+\frac{3-x}{x+1}\right)=2\)
