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

\(\Leftrightarrow\left|x+5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)

2) \(ĐK:x\ge2\)

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

\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)

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

4) \(ĐK:x\ge0\)

\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)

1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

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

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)

\(\Leftrightarrow\left|x-5\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

Liên hợp đi

\(\sqrt{4x^2-4x+1}=\sqrt{x^2+10x+25}\)

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

\(\Leftrightarrow\left|2x-1\right|=\left|x+5\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-\left(x+5\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-x-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)

a) 

\(\sqrt{x+3}+2\sqrt{4\left(x+3\right)}-\frac{1}{3}\sqrt{9\left(x+3\right)}=8\)  

\(\sqrt{x+3}+2\cdot2\sqrt{x+3}-\frac{1}{3}\cdot3\sqrt{x+3}=8\)    

\(\sqrt{x+3}+4\sqrt{x+3}-\sqrt{x+3}=8\)    

\(4\sqrt{x+3}=8\)          

\(\sqrt{x+3}=2\) 

\(\orbr{\begin{cases}2\ge0\left(llđ\right)\\x+3=2^2\end{cases}}\) 

\(x+3=4\) 

\(x=1\) 

b) 

\(\orbr{\begin{cases}x^2+10x+25\ge0\\4x^2-4x+1=x^2+10x+25\end{cases}}\) 

\(\orbr{\begin{cases}\left(x+5\right)^2\ge0\left(lld\right)\\3x^2-6x-24=0\end{cases}}\) 

\(\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)        

Bài 1:

\(x^4+2x^3+10x-25=0\)

\(\Leftrightarrow x^4+2x^3-5x^2+5x^2+10x-25=0\)

\(\Leftrightarrow x^2\left(x^2+2x-5\right)+5\left(x^2+2x-5\right)=0\)

\(\Leftrightarrow\left(x^2+5\right)\left(x^2+2x-5\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+5=0\\x^2+2x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+5>0\forall x\rightarrow Vn\\\Delta_{x^2+2x-5}=2^2-\left[-4\left(1.5\right)\right]=24\end{array}\right.\)

\(\Leftrightarrow x_{1,2}=\frac{-2\pm\sqrt{24}}{2}\)

Bài 2:

Đặt \(\begin{cases}\sqrt{x-1}=a\left(a\ge1\right)\\\sqrt{y}=b\left(b\ge0\right)\end{cases}\)(*) hệ đầu thành:

\(\begin{cases}3a+2b=13\left(1\right)\\2a-b=4\left(2\right)\end{cases}\).Từ \(\left(2\right)\Rightarrow b=2a-4\) thay vào (1) ta có:

\(\left(1\right)\Rightarrow3a+2\left(2a-4\right)=13\)

\(\Rightarrow3a+4a-8=13\Rightarrow7a=21\Rightarrow a=3\) (thỏa mãn)

\(a=3\Rightarrow b=2a-4=2\cdot3-4=2\) (thỏa mãn)

Thay \(\begin{cases}a=3\\b=2\end{cases}\) vào (*) ta có:

(*)\(\Leftrightarrow\begin{cases}\sqrt{x-1}=3\\\sqrt{y}=2\end{cases}\)\(\Leftrightarrow\begin{cases}x-1=9\\y=4\end{cases}\)\(\Leftrightarrow\begin{cases}x=10\\y=4\end{cases}\)

a) \(ĐK:y-2x+1\ge0;4x+y+5\ge0;x+2y-2\ge0,x\le1\)

Th1: \(\hept{\begin{cases}y-2x+1=0\\3-3x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}0=0\\-1=\sqrt{10}-1\end{cases}}\)(không thỏa mãn)

Th2: \(x,y\ne1\)

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

Dễ thấy \(\frac{1}{\sqrt{y-2x+1}+\sqrt{3-3x}}+y-2x+1>0\)nên x + y - 2 = 0

Thay y = 2 - x vào phương trình \(x^2-y-1=\sqrt{4x+y+5}-\sqrt{x+2y-2}\), ta được: \(x^2+x-3=\sqrt{3x+7}-\sqrt{2-x}\)\(\Leftrightarrow x^2+x-2=\sqrt{3x+7}-1+2-\sqrt{2-x}\)\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=\frac{3\left(x+2\right)}{\sqrt{3x+7}+1}+\frac{x+2}{2+\sqrt{2-x}}\)\(\Leftrightarrow\left(x+2\right)\left(\frac{3}{\sqrt{3x+7}+1}+\frac{1}{2+\sqrt{2-x}}+1-x\right)=0\)

Vì \(x\le1\)nên\(\frac{3}{\sqrt{3x+7}+1}+\frac{1}{2+\sqrt{2-x}}+1-x>0\)suy ra x = -2 nên y = 4

Vậy nghiệm của hệ phương trình là (x;y) = (-2;4)

b) \(\hept{\begin{cases}x^2+y^2=5\\x^3+2y^3=10x-10y\end{cases}}\Leftrightarrow\hept{\begin{cases}2\left(x^2+y^2\right)=10\left(1\right)\\x^3+2y^3=10\left(x-y\right)\left(2\right)\end{cases}}\)

Thay (1) vào (2), ta được: \(x^3+2y^3=2\left(x^2+y^2\right)\left(x-y\right)\Leftrightarrow\left(2y-x\right)\left(x^2+2y^2\right)=0\)

* Th1: \(x^2+2y^2=0\)(*)

Mà \(x^2\ge0\forall x;2y^2\ge0\forall y\Rightarrow x^2+2y^2\ge0\)nên (*) xảy ra khi x = y = 0 nhưng cặp nghiệm này không thỏa mãn hệ

* Th2: 2y - x = 0 suy ra x = 2y thay vào (1), ta được: \(y^2=1\Rightarrow y=\pm1\Rightarrow x=\pm2\) 

Vậy hệ có 2 nghiệm \(\left(x,y\right)\in\left\{\left(2;1\right);\left(-2;-1\right)\right\}\)

Đặt \(t=6x+1\)và \(h=\sqrt{x^2+3}\)

\(\frac{1}{4}\cdot t^2+h^2-\frac{9}{4}=th\)

\(\Leftrightarrow\left(t-2h\right)^2=9\)

\(\Leftrightarrow t-2h=\pm3\)

Với \(t-2h=3\)ta có

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

\(\Leftrightarrow3x-1=\sqrt{x^2+3}\)

\(\Leftrightarrow\hept{\begin{cases}3x-1\ge0\\x^2+3=\left(3x+2\right)^2\end{cases}\Leftrightarrow x=\frac{\sqrt{7}-3}{4}}\)

Vậy pt có nghiệm là \(x=1;x=\frac{\sqrt{7}-3}{4}\)