a, ĐK: \(x\ge2\)

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

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

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

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

Phương trình vô nghiệm.

b, ĐK: \(x\ge-1\)

\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+3=4x^2\end{matrix}\right.\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

c, ĐK: \(x\ge-3\)

\(2\sqrt{x+3}=9x^2-x-4\)

\(\Leftrightarrow x+3+2\sqrt{x+3}+1=9x^2\)

\(\Leftrightarrow\left(\sqrt{x+3}+1\right)^2=9x^2\)

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

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

TH1: \(\left\{{}\begin{matrix}3x-1\ge0\\x+3=9x^2-6x+1\end{matrix}\right.\Leftrightarrow...\)

TH2: \(\left\{{}\begin{matrix}-3x-1\ge0\\x+3=9x^2+6x+1\end{matrix}\right.\Leftrightarrow...\)

Tự giải nha, t kh có máy tính ở đây.

b.

ĐKXĐ: \(x\ge-1\)

\(\sqrt{\left(x+1\right)\left(x+35\right)}-14\sqrt{x+35}+84-6\sqrt{x+1}=0\)

\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+35}-14\right)-6\left(\sqrt{x+35}-14\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-6\right)\left(\sqrt{x+35}-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=6\\\sqrt{x+35}=14\end{matrix}\right.\)

\(\Leftrightarrow...\)

a. ĐKXĐ: \(-1\le x\le1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a+2a^2=-b^2+b+3ab\)

\(\Leftrightarrow\left(2a^2-3ab+b^2\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(2a-b+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a+1=b\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x+5+4\sqrt{x+1}=1-x\left(1\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow4\sqrt{x+1}=-4-5x\) \(\left(x\le-\dfrac{4}{5}\right)\)

\(\Leftrightarrow16\left(x+1\right)=25x^2+40x+16\)

\(\Leftrightarrow25x^2+24x=0\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{24}{25}\end{matrix}\right.\)

c.

ĐKXĐ: \(x\ge-\dfrac{3}{2}\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1-\sqrt{x+5}=0\\\sqrt{2x+3}-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5-\sqrt{x+5}-6=0\\\sqrt{2x+3}=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=-2\left(loại\right)\\\sqrt{x+5}=3\\\sqrt{2x+3}=3\end{matrix}\right.\)

\(\Leftrightarrow...\)

Điều kiện xác định của pt : \(\hept{\begin{cases}\frac{x^3+1}{x+3}\ge0\\x+1\ge0\\x+3\ge0\end{cases}}\) \(\Leftrightarrow x\ge-1\)

Ta có : \(\sqrt{\frac{x^3+1}{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}-\sqrt{x+3}=0\\\sqrt{x^2-x+1}+\sqrt{x+3}=0\end{cases}}\)

Từ điều kiện : Với \(x\ge-1\)thì \(\sqrt{x+3}\ge\sqrt{2}>0\); 

 \(\sqrt{x^2-x+1}=\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge\frac{\sqrt{3}}{2}>0\)

Do đó pt (1) vô nghiệm.

Vậy pt ban đầu vô nghiệm.

Điều kiện xác định của pt : \(\hept{\begin{cases}\frac{x^3+1}{x+3}\ge0\\x+1\ge0\\x+3\ge0\end{cases}}\) \(\Leftrightarrow x\ge-1\)

Ta có : \(\sqrt{\frac{x^3+1}{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}-\sqrt{x+3}=0\\\sqrt{x^2-x+1}+\sqrt{x+3}=0\end{cases}}\)

Do đó pt (1) vô nghiệm.

Vậy pt ban đầu vô nghiệm.

\(VT=\sqrt{\left(\sqrt{x+2}-1\right)^2+3}+\sqrt{\left(\sqrt{x+2}-3\right)^2}>\sqrt{3}>1\) pt vô nghiệm 

Từ dòng thứ hai của ngọc vĩ 

=> \(x^2-2x\sqrt{1-x^2}+1-x^2+x^2=0\)

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

=> \(x-\sqrt{1-x^2}=0\)  và x = 0 

=> \(x=\sqrt{1-x^2}vàx=0\)

=>  \(x=\frac{\sqrt{2}}{2}vàx=0\) ( vô lí )

=> Pt vô nghiệm 

Ngu Người ukm , đừng có ý nghĩ lun cho tui l-ike , áy náy lm ^^

Hoặc có thể làm tiếp của Ngọc Vĩ như sau 

Đặt x^2 = t 

<=> 5t^2 - 2t + 1 = 0 

=> 5 ( t^2 - 2/5t + 1/5 ) = 0 

=> 5(t^2 - 2.t.1/5 + 1/25 +4/25 ) = 0

=> 5 ( t + 1/5 )^2+ 4/5 = 0 

Vô lí vì 5(t+1/5)^2 + 4/5 > 0 

Vậy pt vô nghiệm  

câu 2 có nghiệm x=2 , liên hợp đi 

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\\\sqrt{x^2+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2\\x^2=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}∃x̸\\x=\pm\sqrt{8}\end{matrix}\right.\)

`x^2 + 3x + 1 = (x + 3) \sqrt{x^2 + 1}`

Nghiệm của pt là `x = +- 2 \sqrt{2}`