\(\Leftrightarrow\left(x^2-12x-6\right)\left(x^2-4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-12x-6=0\\x^2-4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-6\right)^2=42\\\left(x-2\right)^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6\in\left\{\sqrt{42};-\sqrt{42}\right\}\\x-2\in\left\{\sqrt{2};-\sqrt{2}\right\}\end{matrix}\right.\Leftrightarrow x\in\left\{\sqrt{42}+6;-\sqrt{42}+6;\sqrt{2}+2;2-\sqrt{2}\right\}\)

Đây đích thực có phải là lớp 1 ko bn?

3x⁴-13x³+16x²-13x+3=0

⇔(x-3)(3x³-4x²+4x-1)=0

⇔(x-3)(x-\(\dfrac {1}{3}\))(3x²-3x+3)=0

⇔3(x-3)(x-\(\dfrac{1}{3}\))(x²-x+1)=0

⇔x-3=0 hoặc x-1/3=0

⇔x=3 hoặc x=1/3

hết cứu đi mà làm

ĐKXĐ: \(x\ge\dfrac{3}{2}\)

\(16x^2-48x+35+\left(\sqrt{6x-9}-\sqrt{2x-2}\right)=0\)

\(\Leftrightarrow\left(4x-7\right)\left(4x-5\right)+\dfrac{4x-7}{\sqrt{6x-9}+\sqrt{2x-2}}=0\)

\(\Leftrightarrow\left(4x-7\right)\left(4x-5+\dfrac{1}{\sqrt{6x-9}+\sqrt{2x-2}}\right)=0\)

\(\Leftrightarrow4x-7=0\)

Tham khảo: https://olm.vn/hoi-dap/detail/254086442152.html

ĐKXĐ: 7<=x<=9

Ta có: \(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)

=>\(\sqrt{x-7}-1+\sqrt{9-x}-1=x^2-16x+64\)

=>\(\frac{x-7-1}{\sqrt{x-7}+1}+\frac{9-x-1}{\sqrt{9-x}+1}=\left(x-8\right)^2\)

=>\(\frac{x-8}{\sqrt{x-7}+1}+\frac{8-x}{\sqrt{9-x}+1}=\left(x-8\right)^2\)

=>\(\left(x-8\right)\left(\frac{1}{\sqrt{x-7}+1}-\frac{1}{\sqrt{9-x}+1}-x+8\right)=0\)

=>x-8=0

=>x=8(nhận)