\(a,\sqrt{x-2\sqrt{x}-1}-\sqrt{x-1}=1.\)

\(\Rightarrow\sqrt{\left(\sqrt{x}-1\right)^2}-\sqrt{x-1}=1\)

\(\Rightarrow x-1-\sqrt{x-1}=1\)

\(\Rightarrow\sqrt{x-1}=x-1+1\)

\(\Rightarrow x-1=x^2\Rightarrow x^2-x+1=0\) ( vô nghiệm vì nó luôn lớn hơn 0 )

\(đkxđ\Leftrightarrow2x-1\ge0\Rightarrow x\ge\frac{1}{2}\)

\(c,\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}.\)

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

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

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

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

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

\(\Rightarrow\sqrt{2x-1}=1\Rightarrow\sqrt{2x-1}^2=1\)

\(\Rightarrow2x-1=1\Rightarrow2x=2\Leftrightarrow x=1\)\(\left(tm\right)\)

d tương tự nha , nhân thêm 2 vế với \(\sqrt{6}\)là ra

a 

ĐK: \(x^2-2x+1>0\)

PT \(\Leftrightarrow\sqrt{\left(x-1\right)^2}+x-6x+9=0\)

\(\Leftrightarrow\left|x-1\right|-5x+9=0\\ \Leftrightarrow\left|x-1\right|=-9+5x\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-9+5x\\1-x=-9+5x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=\dfrac{10}{6}\left(nhận\right)\end{matrix}\right.\)

b

ĐK: \(\left\{{}\begin{matrix}2x^2-3>0\\4x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>\dfrac{\sqrt{6}}{2}\\x< -\dfrac{\sqrt{6}}{2}\end{matrix}\right.\\x>\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow x>\dfrac{\sqrt{6}}{2}\)

PT \(\Leftrightarrow2x^2-3=4x-3\)

\(\Leftrightarrow2x^2-4x=0\\ \Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

c

ĐK: \(\left\{{}\begin{matrix}1-x^2\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow x=1\)

PT \(\Leftrightarrow1-x^2=x-1\)

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

a) \(\sqrt{\sqrt{2\sqrt{6}+6+2\sqrt{2}+2\sqrt{3}-\sqrt{5+2\sqrt{6}}}}\)

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

b) \(A=\sqrt{x^2-6x+9}-\dfrac{x^2-9}{\sqrt{9-6x+x^2}}\)

\(=\left|x-3\right|-\dfrac{\left(x-3\right)\left(x+3\right)}{\left|x-3\right|}\)

Th1: x-3 < 0

\(A=\left(3-x\right)-\dfrac{\left(x-3\right)\left(x+3\right)}{3-x}=3-x+x-3=0\)

Th2: x-3 > 0

\(A=x-3-\dfrac{\left(x-3\right)\left(x+3\right)}{x-3}=x-3-\left(x+3\right)=-6\)

c)

Đk: x >/ 1 \(B=\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)

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

\(=\dfrac{\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|}{\left|x-2\right|}\cdot\dfrac{x-2}{\sqrt{x-1}}\)

Th1: \(x-2\ge0\Leftrightarrow x\ge2\)

\(B=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}+1}{x-2}\cdot\dfrac{x-2}{\sqrt{x-1}}=\dfrac{2}{\sqrt{x-1}}\)

Th2: \(x-2\le0\Leftrightarrow x\le2\)

kết hợp với đk, ta được: 1 \< x \< 2

\(=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}-1}{2-x}\cdot\dfrac{x-2}{\sqrt{x-1}}=0\)

d) \(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)

chẳng biết có sai sót gì 0 nữa, xin lỗi tớ 0 xem lại đâu vì chán quá!

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

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

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

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

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

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

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

c: Ta có: \(\sqrt{x^2}=x\)

\(\Leftrightarrow\left|x\right|=x\)

hay \(x\ge0\)

d: Ta có: \(\sqrt{x^2-2x+1}=x-1\)

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

hay \(x\ge1\)

a, ĐK: ​\(\left(x+1\right)\left(x^2+2x-1\right)\ge0\)​

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

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

TH1: \(x\ge-1\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-7x-10=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

TH2: \(x< -1\)

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

\(\Leftrightarrow...\)

Bài này dài nên ... cho nhanh nha, đoạn sau dễ rồi

a, ĐKXĐ: \(x^2-4x+4\ge0\Rightarrow\left(x-2\right)^2\ge0\left(luônđúng\right)\)

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

b,\(ĐKXĐ:1-4x+4x^2\ge0\Rightarrow\left(1-2x\right)^2\ge0\left(luônđúng\right)\)

 \(\sqrt{1-4x+4x^2}=5\\ \Rightarrow\left|1-2x\right|=5\\ \Rightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

d, ĐKXĐ: \(\left\{{}\begin{matrix}9x^2\ge0\\2x+1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-\dfrac{1}{2}\end{matrix}\right.\Rightarrow x\ge0\)

\(\sqrt{9x^2}=2x+1\\ \Rightarrow\left|3x\right|=2x+1\\ \Rightarrow\left[{}\begin{matrix}3x=2x+1\\3x=-2x+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

c, ĐKXĐ: \(1-2x+x^2\ge0\Rightarrow\left(1-x\right)^2\ge0\left(luônđúng\right)\)

 \(\sqrt{1-2x+x^2}-6=0\\ \Rightarrow\left|1-x\right|=6\\ \Rightarrow\left[{}\begin{matrix}1-x=-6\\1-x=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)

e, \(\left\{{}\begin{matrix}9-6x+x^2\ge0\\x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(3-x\right)^2\ge0\left(luônđúng\right)\\x\ge0\end{matrix}\right.\)\(\Rightarrow x\ge0\)

\(\sqrt{9-6x+x^2}=x\\ \Rightarrow\left|3-x\right|=x\\ \Rightarrow\left[{}\begin{matrix}3-x=-x\\3-x=x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3=0\left(vôlí\right)\\x=1,5\end{matrix}\right.\)

a) \(\sqrt{x^2-4x+4}=1\)

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

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

b) \(\sqrt{1-4x+4x^2}=5\)

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

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

c) \(\sqrt{x\left(1-2x+x^2\right)}-6=0\)

\(\Leftrightarrow\left(\sqrt{x\left(1-x\right)^2}\right)^2=36\Leftrightarrow x\left(1-x\right)^2=36\)

\(\Leftrightarrow x-2x^2+x^3-36=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2+2x+9\right)=0\)

\(\Leftrightarrow x=4\)(do \(x^2+2x+9=\left(x+1\right)^2+8>0\))

d) \(\sqrt{9x^2}=2x+1\)

\(\Leftrightarrow3\left|x\right|=2x+1\)

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

e) \(\sqrt{9-6x+x^2}=x\left(1\right)\left(đk:x\ge0\right)\)

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

TH1: \(0\le x\le3\)

\(\left(1\right)\Leftrightarrow3-x=x\Leftrightarrow x=\dfrac{3}{2}\)

TH2: \(x>3\)

\(\left(1\right)\Leftrightarrow x-3=x\Leftrightarrow-3=0\left(vn\right)\)