\(\left(x^2-x+1\right)^4-6x^2\left(x^2-x+1\right)^2+5x^4=0\)

\(\Leftrightarrow\left[\left(x^2-x+1\right)^2\right]^2-2\left(x^2-x+1\right)^2.3x^2+\left(3x^2\right)^2-4x^4=0\)

\(\Leftrightarrow\left[\left(x^2-x+1\right)^2-3x^2\right]^2-\left(2x^2\right)^2=0\)

\(\Leftrightarrow\left[\left(x^2-x+1\right)^2-3x^2+2x^2\right]\left[\left(x^2-x+1\right)^2-3x^2-2x^2\right]=0\)

\(\Leftrightarrow\left[\left(x^2-x+1\right)^2-x^2\right]\left[\left(x^2-x+1\right)^2-5x^2\right]=0\)

\(\Leftrightarrow\left(x^2-x+1+x^2\right)\left(x^2-x+1-x^2\right)\left(x^4-2x^3-4x^2+1\right)=0\)

\(\Leftrightarrow\left(2x^2-x+1\right)\left(1-x\right)\left(x+1\right)\left(x^3-2x^2-x+1\right)=0\)

Mấy bạn cho mình gửi tạm nha, xíu mình nhờ CTV xóa :(

a) \(\dfrac{5x-2}{2-2x}+\dfrac{2x-1}{2}=1-\dfrac{x^2+x-3}{1-x}\)

b) \(\dfrac{5-2x}{3}+\dfrac{\left(x-1\right)\left(x+1\right)}{3x-1}=\dfrac{\left(x+2\right)\left(1-3x\right)}{9x-3}\)

c) \(\dfrac{x}{2x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\)

d) \(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}=\dfrac{2}{x^2-4x+3}\)

\(\sqrt{x+6-2\sqrt{x+2}}=\sqrt{x+2-2\sqrt{x+2}+4}\)

\(=\sqrt{\left(\sqrt{x+2}\right)^2-2\sqrt{x+2}+1^2+3}\)

\(=\sqrt{\left(\sqrt{x}+2-1\right)^2+3}\)

Xin lỗi cho mk đăng thôi ạ, mk ko hỏi.

5 h chieu nay giai cho bai nay trong sach giao khoa