a) ĐKXĐ: \(x\ge0\)

Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)

\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)

\(a,Đk:x\ge0\\ PT\Leftrightarrow4x-8\sqrt{x}-7\sqrt{x}+14=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(4\sqrt{x}-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{49}{4}\end{matrix}\right.\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\sqrt{x+1}-\sqrt{3x}+1-4x^2=0\\ \Leftrightarrow\dfrac{1-2x}{\sqrt{x+1}+\sqrt{3x}}+\left(1-2x\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(\dfrac{1}{\sqrt{x+1}+\sqrt{3x}}+2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\\dfrac{1}{\sqrt{x+1}+\sqrt{3x}}+2x+1=0\left(1\right)\end{matrix}\right.\)

Với \(x\ge0\Leftrightarrow\left(1\right)>0\)

Vậy PT có nghiệm \(x=\dfrac{1}{2}\)

ĐKXĐ: ...

\(\sqrt{x^2-x-30}-3\sqrt{x+5}-2\sqrt{x-6}=-6\)

\(\Leftrightarrow\sqrt{\left(x+5\right)\left(x-6\right)}-3\sqrt{x+5}-2\sqrt{x-6}=-6\)(*)

đặt \(\sqrt{x+5}=a\ge0;\sqrt{x-6}=b\ge0\)

\(\text{pt(*)}\Leftrightarrow ab-3a-2b=-6\\ \Leftrightarrow\Leftrightarrow ab-3a-2b+6=0\\ \Leftrightarrow a\left(b-3\right)-2\left(b-3\right)=0\\ \Leftrightarrow\left(a-2\right)\left(b-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2\\b=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=2\\\sqrt{x-6}=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+5=4\\x-6=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=15\left(tm\right)\end{matrix}\right.\)

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: \(x\ge2\)

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

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

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

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

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

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

\(\Rightarrow x=2\)

3.

ĐKXĐ: \(x\ge-1\)

\(x^2+x-12+12\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\dfrac{12\left(x-3\right)}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4+\dfrac{12}{\sqrt{x+1}+2}\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

\(x^2-2-2\sqrt{4x-7}=0\)

\(\Leftrightarrow\left(4x-7-2\sqrt{4x-7}+1\right)+\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(\sqrt{4x-7}-1\right)^2+\left(x-2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{4x-7}-1=0\\x-2=0\end{matrix}\right.\)

Tự làm tiếp nhé.

. . .

\(4x^2-5x+1+2\sqrt{x-1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)+2\sqrt{x-1}=0\)

\(\Leftrightarrow\sqrt{x-1}\left[\left(4x-1\right)\sqrt{x-1}+2\right]=0\)

\(\Rightarrow x=1\)

. . .

\(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)

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

\(\Leftrightarrow\left|x-2\right|+\left|x-3\right|=1\)

\(VT=\left|x-2\right|+\left|3-x\right|\ge\left|x-2+3-x\right|=1=VP\)

Dấu "=" xảy ra khi \(\left(x-2\right)\left(3-x\right)\ge0\)

Đến đây lập bảng xét dấu

. . .

\(x^2-x+2=2\sqrt{x^2-x+1}\)

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

Tự làm tiếp nhé.

\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)

\(\Leftrightarrow\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14-5\right)=0\)

\(\Leftrightarrow\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{1-6+x}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)\left(x-5\right)=0\)

\(\Rightarrow x=5\)

. . .

\(\sqrt{2x^2-4x+5}-x+4=0\)

\(\Leftrightarrow\sqrt{2x^2-4x+5}=x-4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-4\ge0\\2x^2-4x+5=x^2-8x+16\end{matrix}\right.\)

Tự làm tiếp nhé.

. . .

\(\sqrt{2x+3}+\sqrt{x-1}=\sqrt{x+6}\)

\(\Leftrightarrow\sqrt{2x+3}=\sqrt{x+6}-\sqrt{x-1}\)

\(\Leftrightarrow2x+3=x+6-2\sqrt{\left(x+6\right)\left(x-1\right)}+x-1\)

\(\Leftrightarrow2\sqrt{x^2+5x-6}=2\)

\(\Leftrightarrow x^2+5x-6=1\)

Tự làm tiếp nhé.

. . .

\(x+y+\dfrac{1}{2}=\sqrt{x}+\sqrt{y}\)

\(\Leftrightarrow\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\left(y-\sqrt{y}+\dfrac{1}{4}\right)=0\)

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

Tự làm tiếp nhé.

- Lô Vỹ Vy Vy Nếu câu hỏi liên quan đến hình học, thì mỗi lần đăng một câu thôi, nếu câu hỏi liên quan đến đại số và số học thì có thể đẳng 3 - 4 câu một lần. Lần sau đừng đăng dày đặc như thế này nữa.