a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)

\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)

\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)

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

bạn làm tương tự nha

câu c chắc là như này

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

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

a) Ta có:

5√15+12√20+√5515+1220+5

=√52.15+√(12)2.20+√5=√25.15+√14.20+√5=√255+√204+√5=√5+√5+√5=(1+1+1)√5=3√5=52.15+(12)2.20+5=25.15+14.20+5=255+204+5=5+5+5=(1+1+1)5=35

b)  Ta có: 

√12+√4,5+√12,512+4,5+12,5

=√12+√92+√252=√12+√9.12+√25.12=√12+√32.12+√52.12=√12+3√12+5√12=(1+3+5).√12=9√12=91√2=9.√22=9√22=12+92+252=12+9.12+25.12=12+32.12+52.12=12+312+512=(1+3+5).12=912=912=9.22=922

c) Ta có:

√20−√45+3√18+√72=√4.5−√9.5+3√9.2+√36.2=√22.5−√32.5+3√32.2+√62.2=2√5−3√5+3.3√2+6√2=2√5−3√5+9√2+6√2=(2√5−3√5)+(9√2+6√2)=(2−3)√5+(9+6)√2=−√5+15√2=15√2−√520−45+318+72=4.5−9.5+39.2+36.2=22.5−32.5+332.2+62.2=25−35+3.32+62=25−35+92+62=(25−35)+(92+62)=(2−3)5+(9+6)2=−5+152=152−5

d) Ta có:

0,1√200+2√0,08+0,4.√50=0,1√100.2+2√0,04.2+0,4√25.2=0,1√102.2+2√0,22.2+0,4√52.2=0,1.10√2+2.0,2√2+0,4.5√2=1√2+0,4√2+2√2=(1+0,4+2)√2=3,4√2

Bạn giải bài đâu vậy? Kiếm điểm hỏi đáp hở, Boy anime?

1) \(=\frac{2\sqrt{3}}{\sqrt{20}}+\frac{1}{\sqrt{60}}-\frac{1}{\sqrt{15}}=\frac{6\sqrt{60}+\sqrt{60}-4\sqrt{15}}{60}=\frac{\sqrt{15}\left(12+2-4\right)}{60}=\frac{\sqrt{15}}{6}\)

a) \(=\frac{9}{\sqrt{3}}=\frac{9\sqrt{3}}{3}\)

b) \(=\frac{12\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\frac{36+12\sqrt{3}}{9-3}=6+2\sqrt{3}\)

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

d) \(=\frac{\left(7\sqrt{3}-5\sqrt{11}\right)\left(8\sqrt{3}+7\sqrt{11}\right)}{\left(8\sqrt{3}-7\sqrt{11}\right)\left(8\sqrt{3}+7\sqrt{11}\right)}=\frac{217-9\sqrt{11}}{347}\)

e) \(=\frac{\left(1-a\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=\frac{1+\sqrt{a}-a\sqrt{a}-a^2}{1-a}=a+\sqrt{a}+1\)

f) \(=\frac{1}{3\sqrt{2}-2\sqrt{2}+\sqrt{8}}=\frac{\sqrt{2}-\sqrt{8}}{\left(\sqrt{2}+\sqrt{8}\right)\left(\sqrt{2}-\sqrt{8}\right)}=\frac{\sqrt{2}}{6}\)

g) \(=\frac{1-\sqrt{2}+\sqrt{3}}{1-\left(\sqrt{2}-\sqrt{3}\right)^2}=\frac{1-\sqrt{2}+\sqrt{3}}{2\sqrt{6}-4}=\frac{\left(1-\sqrt{2}+\sqrt{3}\right)\left(2\sqrt{6}+4\right)}{\left(2\sqrt{6}-4\right)\left(2\sqrt{6}+4\right)}\)

\(=\frac{2\sqrt{6}+4-4\sqrt{3}-4\sqrt{2}+6\sqrt{2}+4\sqrt{3}}{24-16}=\frac{\sqrt{2}+\sqrt{6}+2}{4}\)

f) \(=\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)}=\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{2\sqrt{15}-6}\)

\(=\frac{\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{15}+6\right)}{\left(2\sqrt{15}-6\right)\left(2\sqrt{15}+6\right)}=\frac{2\sqrt{30}+6\sqrt{2}-6\sqrt{5}-6\sqrt{3}+10\sqrt{3}+6\sqrt{5}}{60-36}\)

\(=\frac{\sqrt{30}+3\sqrt{2}+2\sqrt{3}}{12}\)

Bài 1:

a)

\(\frac{\sqrt{2.3}+\sqrt{2.7}}{2\sqrt{3}+2\sqrt{7}}=\frac{\sqrt{2}(\sqrt{3}+\sqrt{7})}{2(\sqrt{3}+\sqrt{7})}=\frac{\sqrt{2}}{2}\)

b)

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

Bài 2:

a)

\(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}\)

\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}=\sqrt{4}-\sqrt{1}=1\) (đpcm)

b)

\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\)

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

c) Sửa đề:

\(\left(\frac{\sqrt{a}}{\sqrt{a}+2}-\frac{\sqrt{a}}{\sqrt{a}-2}+\frac{4\sqrt{a}-1}{a-4}\right):\frac{1}{a-4}=\left[\frac{a-2\sqrt{a}-(a+2\sqrt{a})}{(\sqrt{a}+2)(\sqrt{a}-2)}+\frac{4\sqrt{a}-1}{a-4}\right].(a-4)\)

\(=\left(\frac{-4\sqrt{a}}{a-4}+\frac{4\sqrt{a}-1}{a-4}\right).(a-4)=-4\sqrt{a}+4\sqrt{a}-1=-1\)

d)

\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{(\sqrt{a}+\sqrt{b})^2-(\sqrt{a}-\sqrt{b})^2}{2(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}+\frac{2b}{a-b}=\frac{4\sqrt{ab}}{2(a-b)}+\frac{2b}{a-b}\)

\(=\frac{2\sqrt{ab}+2b}{a-b}=\frac{2\sqrt{b}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

1.Trục căn thức ở mẫu

$\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}$

= \(\dfrac{a-2\sqrt{ab}+b}{a-b}\)