Dùng liên hợp.

pt <=> \(\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(1+\sqrt{3}\right)\)

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

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

<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left[\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)-\left(x-1\right)\left(\sqrt{2}+\sqrt{3}\right)\right]\)

\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left[\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)-\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)\right]\)

\(=3x-1\)

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

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

<=> \(3-x^2-2\left(1-x^2\right)=3x-1\)

<=> \(x^2-3x+2=0\) phương trình bậc 2.

Em làm tiếp nhé!

#)Hỏi j đi bn, bn ph hỏi cái j chứ làm lun rùi còn để cộng đồng ngắm ak ???

Bó cả tay lẫn chân !!! Bất lực như gặp cực hình !

Chắc là bạn ấy hỏi bạn ấy làm có đúng ko ha gì đó ?

1. ĐKXĐ:...

\(8-2x-\dfrac{2}{x}-2\sqrt{2-x^2}-2\sqrt{2-\dfrac{1}{x^2}}=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(\dfrac{1}{x^2}-\dfrac{2}{x}+1\right)+\left(2-x^2-2\sqrt{2-x^2}+1\right)+\left(2-\dfrac{1}{x^2}-2\sqrt{2-\dfrac{1}{x^2}}+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(\dfrac{1}{x}-1\right)^2+\left(\sqrt{2-x^2}-1\right)^2+\left(\sqrt{2-\dfrac{1}{x^2}}-1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{x}-1=0\\\sqrt{2-x^2}-1=0\\\sqrt{2-\dfrac{1}{x^2}}-1=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

2.

ĐKXĐ:...

Ta có:

\(VT=x\sqrt{x}+1.\sqrt{12-x}\le\sqrt{\left(x^2+1\right)\left(x+12-x\right)}=2\sqrt{3\left(x^2+1\right)}\)

Dấu "=" xảy ra khi và chỉ khi: \(x\sqrt{12-x}=\sqrt{x}\)

\(\Leftrightarrow x^3-12x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=6-\sqrt{35}\\x=6+\sqrt{35}\end{matrix}\right.\)

3. ĐKXĐ: ...

Với \(x=0\) ko phải nghiệm

Với \(x>0\) pt tương đương:

\(\left(\dfrac{x+8\sqrt{x}+4}{\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}+4}{\sqrt{x}}\right)=36\)

\(\Leftrightarrow\left(\sqrt{x}+\dfrac{4}{\sqrt{x}}+8\right)\left(\sqrt{x}+\dfrac{4}{\sqrt{x}}-1\right)=36\)

Đặt \(\sqrt{x}+\dfrac{4}{\sqrt{x}}-1=t\ge3\)

\(t\left(t+9\right)=36\Leftrightarrow t^2+9t-36=0\)

\(\Rightarrow\left[{}\begin{matrix}t=3\\t=-12\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\dfrac{4}{\sqrt{x}}-1=3\)

\(\Leftrightarrow x-4\sqrt{x}+4=0\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)

\(\Leftrightarrow x=4\)

c) Ta có: \(C=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{x-4}\)

\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)

d)

Sửa đề: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)

Ta có: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)

\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)

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

\(=\dfrac{2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2\sqrt{x}+4+x-2\sqrt{x}}{2\left(x-4\right)}\)

\(=\dfrac{x+4}{2x-8}\)