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a) Sửa đề: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)
Ta có: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)
\(=\sqrt{7+2\cdot\sqrt{7}\cdot1+1}-\sqrt{7}\)
\(=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}\)
\(=\left|\sqrt{7}+1\right|-\sqrt{7}\)
\(=\sqrt{7}+1-\sqrt{7}\)
=1
b) Ta có: \(B=\sqrt{7+4\sqrt{3}}-2\sqrt{3}\)
\(=\sqrt{4+2\cdot2\cdot\sqrt{3}+3}-2\sqrt{3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}\)
\(=\left|2+\sqrt{3}\right|-2\sqrt{3}\)
\(=2+\sqrt{3}-2\sqrt{3}\)
\(=2-\sqrt{3}\)
c) Ta có: \(C=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\)
\(=\sqrt{13-2\cdot\sqrt{13}\cdot1+1}+\sqrt{13+2\cdot\sqrt{13}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\)
\(=\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\)
\(=\sqrt{13}-1+\sqrt{13}+1\)
\(=2\sqrt{13}\)
d) Ta có: \(D=\sqrt{22-2\sqrt{21}}-\sqrt{22+2\sqrt{21}}\)
\(=\sqrt{21-2\cdot\sqrt{21}\cdot1+1}-\sqrt{21+2\cdot\sqrt{21}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{21}-1\right)^2}-\sqrt{\left(\sqrt{21}+1\right)^2}\)
\(=\left|\sqrt{21}-1\right|-\left|\sqrt{21}+1\right|\)
\(=\sqrt{21}-1-\left(\sqrt{21}+1\right)\)
\(=\sqrt{21}-1-\sqrt{21}-1\)
=-2
sữa lại câu cuối cho Nhã Doanh
\(\sqrt{22-2\sqrt{21}-\sqrt{22+2\sqrt{21}}}=\sqrt{22-2\sqrt{21}-\sqrt{\left(\sqrt{21}+1\right)^2}}\)
\(=\sqrt{22-2\sqrt{21}-\sqrt{21}-1}=\sqrt{21-3\sqrt{21}}\)
\(a.\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)
\(b.\sqrt{7+4\sqrt{3}}-2\sqrt{3}=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}=2+\sqrt{3}-2\sqrt{3}=2-\sqrt{3}\)
\(c.\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)\(d.\sqrt{22-2\sqrt{21}-\sqrt{22+2\sqrt{21}}}=\sqrt{\left(\sqrt{21}-1\right)^2-\sqrt{\left(\sqrt{21}+1\right)^2}}=\sqrt{21}-1-\sqrt{\sqrt{21}+1}\)
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)
\(|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=|x-2|+\frac{\sqrt{(x-2)^2}}{x-2}=|x-2|+\frac{|x-2|}{x-2}\)
\(=(x-2)+\frac{(x-2)}{x-2}=x-2+1=x-1\)
a, \(\sqrt{8-2\sqrt{15}}\)
= \(\sqrt{3-2\sqrt{15}+5}\)
= \(\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
= |\(\sqrt{3}-\sqrt{5}\)| = \(\sqrt{5}-\sqrt{3}\) (Do \(\sqrt{5}>\sqrt{3}\))
b, \(\sqrt{9-4\sqrt{5}}\)
= \(\sqrt{5-4\sqrt{5}+4}\)
= \(\sqrt{\left(\sqrt{5}-2\right)^2}\)
= \(\sqrt{5}-2\) (Lười quá bỏ trị tuyệt đối cũng được :v)
Phần c sao sao ý (chắc do mk ngu :v)
d, \(\sqrt{7-2\sqrt{10}}+\sqrt{20}+\frac{1}{2}\sqrt{8}\)
= \(\sqrt{5-2\sqrt{10}+2}+\sqrt{20}+\sqrt{2}\)
= \(\sqrt{5}-\sqrt{2}+\sqrt{20}+\sqrt{2}\)
= \(\sqrt{5}+\sqrt{20}\)
= \(\sqrt{5}\left(1+\sqrt{4}\right)\) = \(3\sqrt{5}\)
Chúc bn học tốt! (Sorry phần c mk thấy sao sao ý nên chịu :v)
a) ĐKXĐ : \(0\le a\ne1\)
\(\frac{\sqrt{a}-a}{\sqrt{a}-1}=\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\)
b) ĐKXĐ : \(b\ne0,a\ne-\sqrt{b}\)
\(\frac{a-\sqrt{b}}{\sqrt{b}}:\frac{\sqrt{b}}{a+\sqrt{b}}=\frac{a-\sqrt{b}}{\sqrt{b}}.\frac{a+\sqrt{b}}{\sqrt{b}}=\frac{a^2-b}{b}=\frac{a^2}{b}-1\)
c) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=\sqrt{5}\left(2-5-4+11\right)\)\(=4\sqrt{5}\)
d) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{2}.\sqrt{7}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
\(=7\left(2-2\sqrt{2}+1\right)+14\sqrt{2}=7\left(2-2\sqrt{2}+1+2\sqrt{2}\right)=7.3=21\)
e) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
b) \(\sqrt{\left(7-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=7-\sqrt{3}+\sqrt{3}+1\)
\(=8\)