a.

\(\Leftrightarrow4x^2-6x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(4x^2-2x+1\right)\left(4x^2+2x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2-2x+1}=a>0\\\sqrt{4x^2+2x+1}=b>0\end{matrix}\right.\) ta được:

\(2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

\(\Leftrightarrow\left(a-\dfrac{b}{\sqrt{3}}\right)\left(2a+\sqrt{3}b\right)=0\)

\(\Leftrightarrow a=\dfrac{b}{\sqrt{3}}\)

\(\Leftrightarrow3a^2=b^2\)

\(\Leftrightarrow3\left(4x^2-2x+1\right)=4x^2+2x+1\)

\(\Leftrightarrow...\)

b.

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

Lặp lại cách làm câu a

ĐKXD phức tạp nên ko tìm ngay

Đặt \(x^2=t>0\Rightarrow\sqrt{12-\dfrac{3}{t}}+\sqrt{4t-\dfrac{3}{t}}=4t\)

Đặt \(\sqrt{4t-\dfrac{3}{t}}=a\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{t}=4t-a^2\\3=4t^2-ta^2\end{matrix}\right.\)

\(\Rightarrow\sqrt{4\left(4t^2-ta^2\right)-\left(4t-a^2\right)}+a=4t\)

\(\Rightarrow\sqrt{16t^2-4ta^2-4t+a^2}=4t-a\)

\(\Rightarrow16t^2-4ta^2-4t+a^2=\left(4t-a\right)^2\)

\(\Rightarrow16t^2-4ta^2-4t+a^2=16t^2-8ta+a^2\)

\(\Rightarrow4ta^2-8ta+4t=0\)

\(\Rightarrow4t\left(a-1\right)^2=0\Rightarrow\left[{}\begin{matrix}t=0\left(loại\right)\\a=1\end{matrix}\right.\)

\(\Rightarrow\sqrt{4t-\dfrac{3}{t}}=1\Rightarrow4t^2-t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{3}{4}< 0\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Thử lại 2 nghiệm vào pt ban đầu đều thỏa mãn

Tham khảo:

Giải phương trình: \(\sqrt{12-\dfrac{3}{x^2}}+\sqrt{4x^2-\dfrac{3}{x^2}}=4x^2\) - Hoc24

a) điều kiện xác định : \(x\ge1\)

ta có : \(\sqrt{\dfrac{x-1}{4}}-3=\sqrt{\dfrac{4x-4}{9}}\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-3=\dfrac{2}{3}\sqrt{x-1}\)

\(\Leftrightarrow\dfrac{1}{6}\sqrt{x-1}=-3\left(vôlí\right)\) vậy phương trình vô nghiệm

b) điều kiện xác định \(x\ge3\)

ta có : \(\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}=x-3\)

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

\(\Leftrightarrow x-2+x+3=x-3\Leftrightarrow x=-4\left(L\right)\) vậy phương trình vô nghiệm

c) điều kiện xác định : \(\left[{}\begin{matrix}x\ge\dfrac{3}{2}\\x< 1\end{matrix}\right.\)

ta có : \(\sqrt{\dfrac{2x-3}{x-1}}=2\) \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\Leftrightarrow2x-3=4x-4\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tmđk\right)\) vậy \(x=\dfrac{1}{2}\)

a.

ĐKXĐ: \(x\ne\pm y\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\dfrac{1}{x-y}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u+v=2\\2u+3v=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3u+3v=6\\2u+3v=5\\\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=2-u\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y}=1\\\dfrac{1}{x-y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\x-y=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-4x+7=x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-5x+6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)