\(\Leftrightarrow x^2+4=2x+3\)

=>x^2-2x+1=0

=>(x-1)^2=0

=>x=1

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.

\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\Leftrightarrow2x+1=2x^3+x^2+2x+1\)\(\Leftrightarrow2x^3+x^2=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)

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

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

\(x^2+1\ge1\forall x\Rightarrow2x+1\ge0!2x+1!=2x+1\)

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

\(\left(1\right)\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)

\(\left(1\right)\Leftrightarrow2x+1=\left(2x+1\right)\left(x^2+1\right)\Leftrightarrow\left(2x+1\right).\left(1-\left(x^2+1\right)\right)=0\)

\(\hept{\begin{cases}2x+1=0\\-x^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}}\)

Chúc bạn học tốt !!!

Điều kiện x>=-2; y>=0; x>=y-3

Ta xét PT thứ nhất 

Đặt √(x+2) = a; √y = b (a,b>=0)

Thì PT thành a(a² - b² + 1) - b = 0

<=> a³ - ab² + a - b = 0

<=> a(a - b)(a + b) + (a -b) =0

<=> (a - b)(a² + ab + 1)=0

Đễ thấy a² + ab + 1 >0

Nên a =b 

Thế vào ta được y = x + 2

Thay cái này vào PT còn lại là xong

\(\hept{\begin{cases}\sqrt{x+2}\left(x-y+3\right)=\sqrt{y}\left(1\right)\\x^2+\left(x+3\right)\left(2x-y+5\right)=x+16\left(2\right)\end{cases}}\)
DKXD :x>=-2; y>=0
Đặt\(\hept{\begin{cases}\sqrt{x+2=a}\\x-y+3=b\end{cases}\left(a\ge0\right)}\)
Pt 1 có dạng \(ab=\sqrt{a^2-b+1}\Leftrightarrow a^2b^2=a^2-b+1\Leftrightarrow a^2\left(b-1\right)\left(b+1\right)+b-1=0\)
\(\Leftrightarrow\left(b-1\right)\left(a^2b+a^2+1\right)=0\)
+> b-1=0\(\Rightarrow b=1\Leftrightarrow x-y+3=1\)
\(\)Khi đó pt (2) \(\Leftrightarrow x^2+\left(x+3\right)\left(x+2+1\right)=x+16\Leftrightarrow x^2+\left(x+3\right)^2=x+16\)
\(\Leftrightarrow x^2+x^2+6x+9=x+16\Leftrightarrow2x^2+5x-7=0\)
Có : 2+5-7=0
Nên pt trên có 2 no \(x_1=1\left(tm\right);x_2=-\frac{7}{2}\left(ktm\right)\)
\(\Rightarrow1-y+3=1\Leftrightarrow y=3\left(tm\right)\)
+>\(a^2b+a^2+1=0\Leftrightarrow\left(x+2\right)\left(x+3-y\right)+x+3=0\)(3)
Đặt \(x+3=m\). Pt(3) có dạng \(\left(m-1\right)\left(m-y\right)+m=0\Leftrightarrow m^2-m-my+y+m=0\Leftrightarrow m^2=y\left(m-1\right)\)
Nếu \(m-1=0\Leftrightarrow x+3-1=0\Leftrightarrow x=-2\left(tm\right)\Rightarrow y=0\left(tm\right)\)
Nhưng k tm pt 2
\(\Rightarrow m-1\ne0\Rightarrow y=\frac{m^2}{m-1}=\frac{\left(x+3\right)^2}{x+2}\)
Thay vào pt (2) ta được \(x^2+\left(x+3\right)\left(2x+5-\frac{\left(x+3\right)^2}{x+2}\right)=x+16\)
ĐẾn đây tự nhân chéo chuển vế ta được \(2x^3+7x^2-8x-29=0\)

Cảm ơn bạn nhưng mình lỡ k cho bb kia rồi

Xin lỗi nhìu nha 

Do vế trái dương nên pt chỉ có nghiệm khi \(x\ge\dfrac{3}{4}\), kết hợp điều kiện \(2x^4-3x^2+1\ge0\Rightarrow x\ge1\)

Khi đó:

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

\(\Rightarrow4x-3\ge\sqrt{4x^4-4x^2+1}\)

\(\Rightarrow4x-3\ge\left|2x^2-1\right|=2x^2-1\)

\(\Rightarrow2x^2-4x+2\le0\)

\(\Rightarrow2\left(x-1\right)^2\le0\)

\(\Rightarrow x=1\)

Lời giải:

1. ĐKXĐ: $x\geq \frac{-5+\sqrt{21}}{2}$

PT $\Leftrightarrow x^2+5x+1=x+1$

$\Leftrightarrow x^2+4x=0$

$\Leftrightarrow x(x+4)=0$

$\Rightarrow x=0$ hoặc $x=-4$

Kết hợp đkxđ suy ra $x=0$

2. ĐKXĐ: $x\leq 2$

PT $\Leftrightarrow x^2+2x+4=2-x$

$\Leftrightarrow x^2+3x+2=0$

$\Leftrightarrow (x+1)(x+2)=0$

$\Leftrightarrow x+1=0$ hoặc $x+2=0$

$\Leftrightarrow x=-1$ hoặc $x=-2$
3.

ĐKXĐ: $-2\leq x\leq 2$

PT $\Leftrightarrow \sqrt{2x+4}=\sqrt{2-x}$

$\Leftrightarrow 2x+4=2-x$

$\Leftrightarrow 3x=-2$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)