Question

[Nguyễn Minh Huy]

Giải hộ tôi 1 câu bài 1

Answer 1

[Nguyễn Minh Huy]

bài 3 nha tui nhầm

Answer 2

[Yen Nhi]

`Answer:`

6. \(\left(x^2-x+2\right)^4-3x^2.\left(x^2-x+2\right)^2+2x^4=0\)

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

\(\Leftrightarrow\left(x^2-x+2\right)^2.[\left(x^2-x+2\right)-x^2]-2x^2.[\left(x^2-x+2\right)^2-x^2]\)

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

\(\Leftrightarrow\orbr{\begin{cases}\left(x^2-x+2\right)^2-x^2=0\\\left(x^2-x+2\right)^2-2x^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x^2-x+2-x\right).\left(x^2-x+2+x\right)=0\\\left(x^2-x+2-\sqrt{2}x\right).\left(x^2-x+2+\sqrt{2}x\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x^2-2x+2\right).\left(x^2+2\right)=0\text{(Vô nghiệm)}\\\left(x^2-\left(1+\sqrt{2}\right)x+2\right).\left(x^2+\left(\sqrt{2}-1\right)x+2\right)=0\text{(Vô nghiệm)}\end{cases}}\)

7. \(3\left(-x^2+2x+3\right)^4-26x^2.\left(-x^2+2x+3\right)^2-9x^4=0\)

Đặt \(\hept{\begin{cases}a=\left(-x^2+2x+3\right)^2\\b=x^2\end{cases}}\)

\(\Leftrightarrow3a^2-26ab-9b^2=0\)

\(\Leftrightarrow3a^2-27ab+ab-9b^2=0\)

\(\Leftrightarrow3a\left(a-9b\right)+b\left(a-9b\right)=0\)

\(\Leftrightarrow\left(a-9b\right)\left(3a+b\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}a=9b\\b=-3a\end{cases}}\)

Xét `a=9b`

`=>(-x^2+2x+3)^2=9x^2` (Bạn tự giải nốt nhé.)

Xét `b=-3a`

`=>x^2=-3.(-x^2+2x+3)^2` (Bạn tự giải nốt nhé.)