a) ĐKXĐ: \(x\ge-4\)

a) Ta có: \(\sqrt{6-4x+x^2}=x+4\Rightarrow\left(x+4\right)^2=x^2-4x+6\)

\(\Rightarrow x^2+8x+16=x^2-4x+6\Rightarrow4x+10=0\Rightarrow x=-\frac{5}{2}\left(loại\right)\)

Vậy pt vô nghiệm

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

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

Lời giải:

Bạn cứ nhớ công thức $\sqrt{x^2}=|x|$, rồi dùng điều kiện đề bài để phá dấu trị tuyệt đối là được

a)

\(\sqrt{16a^2}-5a=\sqrt{(4a)^2}-5a=|4a|-5a=4a-5a=-a\)

b)

\(3x+2-\sqrt{9x^2+6x+1}=3x+2-\sqrt{(3x)^2+2.3x.1+1^2}\)

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

c)

\(\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{7+1+2.\sqrt{7}.\sqrt{1}}-\sqrt{7}\)

\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{7}=|\sqrt{7}+1|-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)

d)

\(\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}=\sqrt{13+1-2\sqrt{13}}+\sqrt{13+1+2\sqrt{13}}\)

\(=\sqrt{(\sqrt{13}-1)^2}+\sqrt{(\sqrt{13}+1)^2}=|\sqrt{13}-1|+|\sqrt{13}+1|\)

\(=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)

e)

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

g)

\(|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=|x-2|+\frac{\sqrt{(x-2)^2}}{x-2}=|x-2|+\frac{|x-2|}{x-2}\)

\(=(x-2)+\frac{(x-2)}{x-2}=x-2+1=x-1\)

dell biết

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

\(=2x-1-2x+3=2\)(vì x\(\ge\)1/2 nên 2x-1\(\ge\)0)

b.\(B=\sqrt{\frac{\left(3\sqrt{5}+1\right)\left(2\sqrt{5}+3\right)}{\left(2\sqrt{5}-3\right)\left(2\sqrt{5}+3\right)}}\left(\sqrt{10}-\sqrt{2}\right)\)

\(=\sqrt{\frac{33+11\sqrt{5}}{11}}\left(\sqrt{10}-\sqrt{2}\right)=\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)

\(=\sqrt{6+2\sqrt{5}}\left(\sqrt{5}-1\right)=\sqrt{\left(\sqrt{5}+1\right)^2}\left(\sqrt{5}-1\right)\)

\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4\)