Ui...... người ta nói nó dễ ..................................

\(2\sqrt{x+4}-4\sqrt{2x-6}=x-7\)

\(\Leftrightarrow\sqrt{2^2\left(x+4\right)}-\sqrt{4^2\left(2x-6\right)}=x-7\)

\(\Leftrightarrow\sqrt{4x+16}-\sqrt{32x-96}=x-7\)

\(\Leftrightarrow\left(\sqrt{4x+16}-\sqrt{32x-96}\right)^2=\left(x-7\right)^2\)

\(\Leftrightarrow\sqrt{4x+16}^2-2.\sqrt{4x+16}.\sqrt{32x-96}+\sqrt{32x-96}^2=x^2-14x+49\)

\(\Leftrightarrow\left(4x+16\right)-2.\sqrt{\left(4x+16\right)\left(32x-96\right)}+\left(32x-96\right)=x^2-14x+49\)

\(\Leftrightarrow\left(4x+16\right)-2.\sqrt{128x^2-384x+512x-1536}+\left(32x-96\right)=x^2-14x+49\)

\(\Leftrightarrow\left(-2\sqrt{128x^2-384x+512x-1536}\right)=\left[x^2-14x+49-\left(4x+16\right)-\left(32x-96\right)\right]\)

\(\Leftrightarrow\left(-2\sqrt{128x^2+128x-1536}\right)^2=\left(x^2-50x+129\right)^2\)

\(\Leftrightarrow4.\left(128x^2+128x-1536\right)=\left(x^2-50x\right)^2+2.\left(x^2-50x\right).129+129^2\)

\(\Leftrightarrow512x^2+512x-6144=\left(x^2-50x\right)^2+258.\left(x^2-50x\right)+16641\)

\(\Leftrightarrow512x^2+512x-6144=x^4-100x^3+2500x^2+258x^2-12900x+16641\)

\(\Leftrightarrow-x^4+100x^3-2246x^2+13412x-22785=0\)

\(\Leftrightarrow x_1\approx70,94\) ; \(x_2\approx3,0588\) ; \(x_3=21\) ; \(x_4=5\)

Bài này có 1 nghiệm duy nhất thôi nha : x = 5 

tại máy tính của mình ra sai kết quả 

ĐKXĐ: $x \geq 2$

\(\Leftrightarrow2\left(x-4\right).\sqrt{x-2}-2\left(x-4\right)+\left(x-2\right)\sqrt{x+1}-2\left(x-2\right)+6x-18=0\\ \Leftrightarrow2.\left(x-4\right).\dfrac{x-3}{\sqrt{x-2}+1}+\left(x-2\right).\dfrac{x-3}{\sqrt{x+1}+2}+6.\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(\dfrac{2.\left(x-4\right)}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+6=0\right)\\ \Leftrightarrow x=3\)

Vì \(\dfrac{2.\left(x-4\right)}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+6=\dfrac{2\left(x-4\right)+4.\sqrt{x-2}+4}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+2\\ =\dfrac{2\left(x-2\right)+4.\sqrt{x-2}}{\sqrt{x-2}+1}+\dfrac{x-2}{\sqrt{x+1}+2}+2>0\)

Vậy....

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge2\)

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

\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=2\)