a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4

=>2*căn(x+5)=4

=>căn (x+5)=2

=>x+5=4

=>x=-1

b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

=>2*căn x-1=16

=>x-1=64

=>x=65

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

TH₁: \(x\ge3\)

\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)

TH₂: \(2\le x< 3\)

\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH₃: \(0\le x< 2\)

\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

TH₄: \(x< 0\)

\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)

a) \(\sqrt{1-6x+9x^2}=9\)

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

\(\Leftrightarrow\left|1-3x\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}1-3x=9\\1-3x=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=1-9\\3x=1+9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=-8\\3x=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=\dfrac{10}{3}\end{matrix}\right.\)

b) \(\sqrt{2x-3}-\sqrt{x+1}=0\) (\(x\ge\dfrac{3}{2}\))

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

\(\Leftrightarrow2x-3=x+1\)

\(\Leftrightarrow2x-x=1+3\)

\(\Leftrightarrow x=4\left(tm\right)\)

c) \(\sqrt{9x^2+12+4}-2=3x\)

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

\(\Leftrightarrow\left|3x+2\right|=3x+2\)

\(\Leftrightarrow3x+2\ge0\)

\(\Leftrightarrow3x\ge-2\)

\(\Leftrightarrow x\ge-\dfrac{2}{3}\)

a: =>|3x-1|=9

=>3x-1=9 hoặc 3x-1=-9

=>x=-8/3 hoặc x=10/3

b: =>căn 2x-3=căn x+1

=>2x-3=x+1

=>x=4

c: =>|3x+2|=3x+2

=>3x+2>=0

=>x>=-2/3

a) \(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x-8}=1+\sqrt{3}\)

\(pt\Leftrightarrow\sqrt{-x^2+2x+1+1}+\sqrt{-x^2-6x-9+1}=1+\sqrt{3}\)

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

Dễ thấy: \(VT\le2< 1+\sqrt{3}=VP\) (vô nghiệm)

b)\(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)

\(pt\Leftrightarrow\sqrt{9x^2-6x+1+1}+\sqrt{45x^2-30x+5+4}=\sqrt{-9x^2+6x-1+9}\)

\(\Leftrightarrow\sqrt{\left(3x-1\right)^2+1}+\sqrt{5\left(3x-1\right)^2+4}=\sqrt{-\left(3x-1\right)^2+9}\)

Dễ thấy: \(VT\ge1+\sqrt{4}=3=VP\)

Đẳng thức xảy ra khi \(x=\dfrac{1}{3}\)

a.

ĐKXĐ: \(x\ge-\dfrac{5}{3}\)

\(9x^2-3x-\left(3x+5\right)-\sqrt{3x+5}=0\)

Đặt \(\sqrt{3x+5}=t\ge0\)

\(\Rightarrow9x^2-3x-t^2-t=0\)

\(\Delta=9+36\left(t^2+t\right)=\left(6t+3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+6t+3}{18}=\dfrac{t+1}{3}\\x=\dfrac{3-6t-3}{18}=-\dfrac{t}{3}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+5}=3x-1\left(x\ge\dfrac{1}{3}\right)\\\sqrt{3x+5}=-3x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+5=9x^2-6x+1\left(x\ge\dfrac{1}{3}\right)\\3x+5=9x^2\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

c.

ĐKXĐ: \(x\ge-5\)

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

Đặt \(\sqrt{x+5}=t\ge0\)

\(\Rightarrow-t^2-t+x^2-3x+2=0\)

\(\Delta=1+4\left(x^2-3x+2\right)=\left(2x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{1+2x-3}{-2}=1-x\\t=\dfrac{1-2x+3}{-2}=x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1-x\left(x\le1\right)\\\sqrt{x+5}=x-2\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2-2x+1\left(x\le1\right)\\x+5=x^2-4x+4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-\dfrac{8}{3}\)

\(\left(3x+2\right)^2-6-\sqrt{3x+8}=0\)

Đặt \(\sqrt{3x+8}=t\ge0\Rightarrow3x+2=t^2-6\)

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

\(\Leftrightarrow t^4-12t^2-t+30=0\)

\(\Leftrightarrow\left(t^2+t-5\right)\left(t^2-t-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+8}=3\\\sqrt{3x+8}=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

a) giải pt ra ta được  : x=-1

b) giải pt ra ta được  : x=2

c)giải pt ra ta được  : x vô ngiệm

d)giải pt ra ta được  : x=vô ngiệm

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

Đăng 1 lúc mà nhiều thế. Lần sau đăng 1 câu thôi b.

b/ \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)

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

Ta có: \(VT\ge1+2+\sqrt{5}=3+\sqrt{5}\)

Dấu = xảy ra khi \(x=2\)

c/ \(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x-8}=\sqrt{3-\left(x-1\right)^2}+\sqrt{1-\left(x+3\right)^2}\)

\(\le1+\sqrt{3}\)

Dấu = không xảy ra nên pt vô nghiệm

Câu d làm tương tự

\(a,\sqrt{x^2-4}-x^2+4=0\) 

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

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

\(\Leftrightarrow x^2-4-x^4+8x^2-16=0\)  

\(\Leftrightarrow-x^4-7x^2-20=0\) 

\(\Leftrightarrow-\left(x^4+7x^2+\frac{49}{4}\right)-\frac{31}{4}=0\) 

\(\Leftrightarrow-\left(x^2+\frac{7}{2}\right)^2=\frac{31}{4}\) 

\(\Leftrightarrow\left(x^2+\frac{7}{2}\right)=-\frac{31}{4}\) 

\(\Rightarrow\)pt vô nghiệm

a/ĐK: \(x\ge2\)

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

Đặt \(x^2-4=t\Rightarrow x^2=t+4\)

Thay vào,phương trình đã cho tương đương với: 

\(t=\sqrt{t}\Leftrightarrow t^2=t\Rightarrow\orbr{\begin{cases}t=1\\t=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-4=1\\x^2-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=5\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}\\x=2\end{cases}}\)  (t/m)