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a) \(\frac{\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\frac{\sqrt{2}.\left(1-\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}-\sqrt{3}\right).\left(1-\sqrt{2}+\sqrt{3}\right)}.\)
\(=\frac{\sqrt{2}.\left(1-\sqrt{2}+\sqrt{3}\right)}{1-\left(\sqrt{2}-\sqrt{3}\right)^2}=\frac{\sqrt{2}.\left(1-\sqrt{2}+\sqrt{3}\right)}{1-\left(5-2\sqrt{6}\right)}\)
\(=\frac{\sqrt{2}.\left(1-\sqrt{2}+\sqrt{3}\right)}{-4+2\sqrt{6}}=\frac{1-\sqrt{2}+\sqrt{3}}{-2\sqrt{2}+2\sqrt{3}}\)
\(=\frac{\left(1-\sqrt{2}+\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}{-2\left(\sqrt{2}-\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\left(1-\sqrt{2}+\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}{-2.\left(2-3\right)}\)\(=\frac{\left(1-\sqrt{2}+\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}{2}\)
Căn thức ở mẫu đã được trục rồi.
Nếu cần thì phá ngoặc phần tử số ra.
b) Nhân cả tử số và mẫu số cho \(\sqrt{a+3}-\sqrt{a-3}\)thì mẫu số có giá trị là (a + 3) - (a - 3) = 6; tử số có giá trị là \(\left(\sqrt{a+3}-\sqrt{a-3}\right)^2\). Khi đó, căn thức ở mẫu đã được trục đi rồi. Sau đó bạn phá ngoặc phần tử số ra.
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)
\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}=\frac{(\sqrt{a}+\sqrt{b})^2-(\sqrt{a}-\sqrt{b})^2}{2(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}+\frac{2b}{a-b}=\frac{4\sqrt{ab}}{2(a-b)}+\frac{2b}{a-b}\)
\(=\frac{2\sqrt{ab}+2b}{a-b}=\frac{2\sqrt{b}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
EM thử thôi, ko chắc đâu ạ:( Sai thì xin thông cảm cho ạ.
1) \(\sqrt{\frac{2}{3-\sqrt{5}}}=\sqrt{\frac{2\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}=\sqrt{\frac{6+2\sqrt{5}}{4}}=\frac{\sqrt{6+2\sqrt{5}}}{2}\)
2) \(\sqrt{\frac{a-4}{2\left(\sqrt{a}-2\right)}}=\sqrt{\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{2\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}}\)
\(=\sqrt{\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{2\left(a-4\right)}}\)
3) \(\sqrt{\frac{1}{a\left(1-\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-3\right)}}=\sqrt{-\frac{1+\sqrt{3}}{2a}}\)
4) \(\sqrt{\frac{a}{4-2\sqrt{3}}}=\sqrt{\frac{a\left(4+2\sqrt{3}\right)}{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}}=\sqrt{\frac{4a+2a\sqrt{3}}{16-12}}=\sqrt{\frac{4a+2a\sqrt{3}}{4}}=\frac{\sqrt{4a+2a\sqrt{3}}}{2}\)
b)\(\frac{3\sqrt{2}}{\sqrt{3}+1}\)
\(=\frac{3\sqrt{2}\left(\sqrt{3}-1\right)}{(\sqrt{3}+1)\left(\sqrt{3}-1\right)}\)
\(=\frac{3\left(\sqrt{6}-\sqrt{2}\right)}{3-1}\)
\(=\frac{3\left(\sqrt{6}-\sqrt{2}\right)}{2}\)
a)\(\left(\sqrt{3}+1\right)^2+\left(1-\sqrt{3}\right)^2\)
\(=3+2\sqrt{3}+1+1-2\sqrt{3}+3\)
\(=8\)
b)\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
\(=\sqrt{28.7}-2\sqrt{3.7}+\sqrt{7}.\sqrt{7}+\sqrt{84}\)
\(=\sqrt{196}-2\sqrt{21}+7+\sqrt{4.21}\)
\(=\sqrt{14^2}-2\sqrt{21}+7+2\sqrt{21}\)
\(=14-2\sqrt{21}+7+2\sqrt{21}\)
\(=21\)
a) \(\frac{6}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}\)
b) \(\frac{8}{5\sqrt{2}}=\frac{8\cdot5\sqrt{2}}{50}=\frac{40\sqrt{2}}{50}=\frac{4\sqrt{2}}{5}\)
c) \(\frac{1}{\sqrt{a}}=\frac{\sqrt{a}}{a}\)( a > 0 )
d) \(\frac{\sqrt{3-3}}{\sqrt{3-1}}=0\)
e) \(\frac{1-2\sqrt{a}+a}{1-\sqrt{a}}=\frac{\left(1-\sqrt{a}\right)^2}{1-\sqrt{a}}=1-\sqrt{a}\)(\(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\))
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
a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)
\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)
\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)
\(\frac{x+1}{\sqrt{x^2-1}}=\frac{\left(x+1\right)\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2-1}\right)}\) = \(\frac{\left(\sqrt{x^2-1}\right)\left(x+1\right)}{x^2-1}\)