đkxđ a) \(x\ge\dfrac{3}{2}\)   b)\(\left[{}\begin{matrix}x\ge-13\\x\le-17\end{matrix}\right.\) 

c) x>3    d) \(x\ge1\) 

e) ( ko cần đkxđ)  

f) \(x\ge-2\)

f) \(\dfrac{1}{5}\sqrt{25.\left(x+2\right)}-5\sqrt{x+2}+\sqrt{9.\left(x+2\right)}+9=0\)

\(\Leftrightarrow\dfrac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)

\(\Leftrightarrow\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)

\(\Leftrightarrow9-\sqrt{x+2}=0\) \(\Leftrightarrow9=\sqrt{x+2}\Leftrightarrow81=x+2\Leftrightarrow x=79\)

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

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

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x^2-4}-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4=0\\x^2-4=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm\sqrt{5}\end{matrix}\right.\)

a) \(\Leftrightarrow2x-3=9\Leftrightarrow x=6\)

b)\(\Leftrightarrow x^2-4x+13=9\) \(\Leftrightarrow x^2-4x+4=0\) \(\Leftrightarrow\left(x-2\right)^2=0\) \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)  

c) \(\Leftrightarrow2=4\sqrt{x-3}\) \(\Leftrightarrow4=16x-48\) \(\Leftrightarrow16x=52\Leftrightarrow x=3,25\)

d)\(\Leftrightarrow x^2-3x+2=x-1\) \(\Leftrightarrow x^2-4x+3=0\) \(\Leftrightarrow x^2-4x+4-1=0\)

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

e) \(\Leftrightarrow2x+3=27\Leftrightarrow2x=24\Leftrightarrow x=12\)

giải phương trình sau: 

a) \(4x^2+\left(8x-4\right).\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\) 

b) \(8x^3-36x^2+\left(1-3x\right)\sqrt{3x-2}-3\sqrt{3x-2}+63x-32=0\) 

c) \(2\sqrt[3]{3x-2}-3\sqrt{6-5x}+16=0\) 

d) \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)

giải phương trình sau: \(5x^2-x+5=\sqrt{x^4+x^2+1}\)

đề bài: giải các phương trình sau:

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

b) \(3x^2+2\left(x-1\right)\sqrt{2x^2-3x+1}=5x+2\)

a) Để A có giá trị nguyên thì \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\) có giá trị nguyên 

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

Mà \(\left(\sqrt{x}-1\right)⋮\left(\sqrt{x}-1\right)\) 

\(\Rightarrow\left(\sqrt{x}+2-\sqrt{x}+1\right)⋮\left(\sqrt{x}-1\right)\)\(\Leftrightarrow3⋮\left(\sqrt{x}-1\right)\)

\(\Rightarrow\left(\sqrt{x}-1\right)\inƯ_{\left(3\right)}=\left\{-3;-1;1;3\right\}\)

Ta có bảng sau:

Vậy để A có giá trị nguyên thì \(x\in\left\{0;4;16\right\}\)

b) Để B có giá trị nguyên thì \(\left(2\sqrt{x}-3\right)⋮\left(\sqrt{x}+2\right)\) 

Mà \(\left(\sqrt{x}+2\right)⋮\left(\sqrt{x}+2\right)\) \(\Leftrightarrow2.\left(\sqrt{x}+2\right)⋮\left(\sqrt{x}+2\right)\) \(\Leftrightarrow\left(2\sqrt{x}+4\right)⋮\left(\sqrt{x}+2\right)\) 

\(\Rightarrow\left(2\sqrt{x}+4-2\sqrt{x}+3\right)⋮\left(\sqrt{x}+2\right)\) \(\Leftrightarrow7⋮\left(\sqrt{x}+2\right)\) 

\(\Rightarrow\left(\sqrt{x}+2\right)\inƯ_{\left(7\right)}=\left\{-7;-1;1;7\right\}\)

Ta có bảng sau:

Vậy để B có giá trị nguyên thì \(x=25\)