a.

ĐKXĐ: \(x>0\)

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

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

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

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

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

b.

ĐKXĐ: \(x\ge-\dfrac{1}{2};x\ne1-\sqrt{2}\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+1}=\sqrt{x+2}\\\sqrt{x+2}=x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x+2\\x^2-x-2=0\left(x\ge0\right)\end{matrix}\right.\)

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

ĐKXĐ : \(\left\{{}\begin{matrix}2x-1>0\\y+2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{2}\\y>-2\end{matrix}\right.\)

PT ( I ) \(\Leftrightarrow\left(\sqrt{\dfrac{2x-1}{y+2}}+\sqrt{\dfrac{y+2}{2x-1}}\right)^2=4\)

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

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

Từ PT ( II ) ta được : \(x=12-y\)

- Thế x vào PT trên ta được : \(\dfrac{2\left(12-y\right)}{y+2}+\dfrac{y+2}{2\left(12-y\right)}=2\)

\(\Leftrightarrow4\left(y-12\right)^2+\left(y+2\right)^2=4\left(12-y\right)\left(y+2\right)\)

\(\Leftrightarrow4\left(y^2-24y+144\right)+y^2+4y+4=4\left(12y+24-y^2-2y\right)\)

\(\Leftrightarrow4y^2-96y+576+y^2+4y+4-40y-96+4y^2=0\)

\(\Leftrightarrow9y^2-132y+484=0\)

\(\Leftrightarrow y=\dfrac{22}{3}\left(TM\right)\)

- Thay lại vào PT ta được : \(x=\dfrac{14}{3}\)

Vậy phương trình có nghiệm là \(S=\left\{\left(\dfrac{22}{3};\dfrac{14}{3}\right);\left(\dfrac{14}{3};\dfrac{22}{3}\right)\right\}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x+2>=0\\2x+1>=0\\x< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{2}\\x< >0\end{matrix}\right.\)

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

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

=>\(\dfrac{1-x^2}{x^2}+\dfrac{x+2-3}{\sqrt{x+2}+\sqrt{3}}=\dfrac{1-x}{x}+\dfrac{2x+1-3}{\sqrt{2x+1}+\sqrt{3}}\)

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

=>x-1=0

=>x=1

b)đk:\(x\ge\dfrac{1}{2}\)

Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)

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

=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\) 

Dấu = xảy ra\(\Leftrightarrow x=1\)

Vậy....

c) đk: \(x\ge0\)

\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)

Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)

\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)

pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)

\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...

 

a)ĐKXĐ: x≥-1/3; x≤6

<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)

(vì x≥-1/3 nên3x+1≥0 )

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

Điều kiện \(\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}-1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x+2\sqrt[]{x}=8\left(\sqrt[]{x}-1\right)\)

\(\Leftrightarrow x-6\sqrt[]{x}+8=0\left(2\right)\)

Đặt \(t^2=x\Leftrightarrow t=\sqrt[]{x}\)

\(\left(2\right)\Leftrightarrow t^2-6t+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x}=2\\\sqrt[]{x}=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=16\end{matrix}\right.\) (thỏa điều kiện)

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

Điều kiện \(\dfrac{2x-3}{x-1}\ge0\Leftrightarrow\left[{}\begin{matrix}x< 1\\x\ge\dfrac{3}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow2x-3=4\left(x-1\right)\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\) (thỏa điều kiện)

b.

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

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

Đặt \(x\sqrt{2x^2+4x}=t\Rightarrow t^2=x^2\left(2x^2+4x\right)=2\left(x^4+2x^2\right)\)

Pt trở thành:

\(\dfrac{t^2}{2}-4+t=0\)

\(\Leftrightarrow t^2+2t-8=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4x}=2\left(x>0\right)\\x\sqrt{2x^2+4x}=-4\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-2=0\left(x>0\right)\\x^4+2x^2-8=0\left(x< 0\right)\end{matrix}\right.\)

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

a.

ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow\dfrac{9}{x^2}+2+\dfrac{2x}{\sqrt{2x^2+9}}=3\)

\(\Leftrightarrow\dfrac{2x^2+9}{x^2}+\dfrac{2x}{\sqrt{2x^2+9}}=3\)

Đặt \(\dfrac{x}{\sqrt{2x^2+9}}=t\Rightarrow\dfrac{2x^2+9}{x^2}=\dfrac{1}{t^2}\)

Pt trở thành:

\(\dfrac{1}{t^2}+2t=3\)

\(\Rightarrow2t^3-3t^2+1=0\)

\(\Leftrightarrow\left(t-1\right)^2\left(2t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{\sqrt{2x^2+9}}=1\left(x>0\right)\\\dfrac{x}{\sqrt{2x^2+9}}=-\dfrac{1}{2}\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2x^2+9\left(vô-nghiệm\right)\\4x^2=2x^2+9\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow x=-\dfrac{3\sqrt{2}}{2}\)

Kiểm tra lại vế trái đề bài câu b