( 1+ \(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\)).(\(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\))
Tính:
1) \(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}-1}\)
2) \(\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\)
3) \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
4) \(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{\sqrt{5}-3}\)
5) \(\dfrac{1}{\sqrt{2}-\sqrt{6}}-\dfrac{1}{\sqrt{6}+\sqrt{2}}\)
LM CHI TIẾT GIÚP MK NHÉ
4: Ta có: \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{3-\sqrt{5}}\)
\(=\dfrac{3-\sqrt{5}-3-\sqrt{5}}{4}\)
\(=\dfrac{-\sqrt{5}}{2}\)
1. \(\dfrac{-2}{\sqrt{3}-1}\)
2. \(\dfrac{5}{1-\sqrt{6}}\)
3. \(\dfrac{2+\sqrt{5}}{2-\sqrt{5}}\)
4. \(\dfrac{1}{5+2\sqrt{6}}\)
5. \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\)
6. \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
7. \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{2}}\)
8. \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}\)
9. \(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}\)
Bài 1: Rút gọn biểu thức:
a) \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}+\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}+\sqrt{5}}\right)\)
b) \(\dfrac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\dfrac{1}{3}}\)
c) \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
d) \(\sqrt{\dfrac{3}{4}}+\sqrt{\dfrac{1}{3}}+\sqrt{\dfrac{1}{12}}\)
Bài 2: Giải các phương trình sau:
a) \(x^2+4x+5=2\sqrt{2x+3}\)
b) \(x^2+9x+20=2\sqrt{3x+10}\)
c) \(x^2+7x+14=2\sqrt{x+4}\)
d) \(4\sqrt{x+1}=x^2-5x+14\)
e) \(\sqrt{6-x}=3x-4\)
f) \(\sqrt{5x-9}=9-2x\)
Mọi người làm ơn giúp mình với. Mình đang cần gấp ạ. Cảm ơn mọi người rất nhiều
Bài 1:
a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)
\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)
\(=\dfrac{5}{2}\sqrt{5}\)
c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)
\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)
Bài 2:
e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)
Ta có: \(\sqrt{6-x}=3x-4\)
\(\Leftrightarrow6-x=\left(3x-4\right)^2\)
\(\Leftrightarrow9x^2-24x+16+6-x=0\)
\(\Leftrightarrow9x^2-25x+22=0\)
\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)
Vậy: Phương trình vô nghiệm
\(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{2}{\sqrt{2}+1}-\dfrac{4}{\sqrt{2}}\)
\(\dfrac{4}{\sqrt{5}+1}+\dfrac{5}{\sqrt{5}+2}+\dfrac{5}{\sqrt{5}+3}\)
\(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{2}{\sqrt{2}+1}-\dfrac{4}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\dfrac{2\sqrt{2}}{2+\sqrt{2}}-\dfrac{4\sqrt{2}+4}{2+\sqrt{2}}\)
\(=\sqrt{3}+\sqrt{3}+\dfrac{-2\sqrt{2}-4}{2+\sqrt{2}}\)
\(=2\sqrt{3}+\dfrac{-2\left(2+\sqrt{2}\right)}{2+\sqrt{2}}\)
\(=2\sqrt{3}-2\)
\(------\)
\(\dfrac{4}{\sqrt{5}+1}+\dfrac{5}{\sqrt{5}+2}+\dfrac{5}{\sqrt{5}+3}\)
\(=\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}+\dfrac{5\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\dfrac{5\left(\sqrt{5}-3\right)}{\left(\sqrt{5}+3\right)\left(\sqrt{5}-3\right)}\)
\(=\dfrac{4\sqrt{5}-4}{5-1}+\dfrac{5\sqrt{5}-10}{5-4}+\dfrac{5\sqrt{5}-15}{5-9}\)
\(=5\sqrt{5}-10+\left(\dfrac{4\sqrt{5}-4}{4}+\dfrac{5\sqrt{5}-15}{-4}\right)\)
\(=\dfrac{4\cdot\left(5\sqrt{5}-10\right)}{4}+\left(\dfrac{4\sqrt{5}-4}{4}-\dfrac{5\sqrt{5}-15}{4}\right)\)
\(=\dfrac{20\sqrt{5}-40}{4}+\dfrac{-\sqrt{5}+11}{4}\)
\(=\dfrac{19\sqrt{5}-29}{4}\)
#Ayumu
Rút gọn biểu thức :
\((5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}+\sqrt{5}}):2\sqrt{5}\) và \(\dfrac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\dfrac{1}{3}}\)
`(5sqrt{1/5}+1/2sqrt{20}-5/4sqrt{4/5}+sqrt{5}):2/5
`=(sqrt5+1/2*2sqrt5-sqrt{5/4}+sqrt5):2/5`
`=(sqrt5+sqrt5+sqrt5-sqrt5/2):2/5`
`=(5/2*sqrt5):2/5`
`=25/4sqrt5`
`1/3sqrt{48}+3sqrt{75}-sqrt{27}-10sqrt{1 1/3}`
`=1/3*4sqrt3+3*5sqrt3-3sqrt3-10sqrt{4/3}`
`=4/sqrt3+15sqrt3-3sqrt3-20/sqrt3`
`=12sqrt3-16/sqrt3`
Tính:
1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{2+\sqrt{5}}\)
2) \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
3) \(\dfrac{1}{\sqrt{5}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)
4) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\)
5) \(-\dfrac{1}{\sqrt{2}-\sqrt{3}}\)\(-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
1: ta có: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)
\(=3+2\sqrt{2}+\sqrt{5}-2\)
\(=2\sqrt{2}+\sqrt{5}+1\)
2: Ta có: \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}\)
\(=3+2\sqrt{2}-3+2\sqrt{2}\)
\(=4\sqrt{2}\)
Làm mất căn mẫu và thu gọn
1) \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
2) \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
3) \(\left(\dfrac{3\sqrt{125}}{15}-\dfrac{10-4\sqrt{5}}{\sqrt{5}-2}\right)\dfrac{1}{\sqrt{5}}\)
4) \(\dfrac{1}{1+\sqrt{2}}-\dfrac{1}{1-\sqrt{2}}\)
5) \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{\sqrt{5}-3}\)
6) \(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\)
7) \(\dfrac{4}{1-\sqrt{3}}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
8) \(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}-\dfrac{3}{\sqrt{2}-1}\)
9) \(\dfrac{\sqrt{2}}{\sqrt{\sqrt{2}+1}-1}-\dfrac{\sqrt{2}}{\sqrt{\sqrt{2}+1}+1}\)
10) \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
11) \(\dfrac{5}{1+\sqrt{6}}-\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}\)
12) \(\dfrac{5}{3-\sqrt{7}}-\dfrac{3}{\sqrt{2}+\sqrt{3}}+\dfrac{-1}{\sqrt{2}-1}\)
Giúp em giải với ạ! Help me~!
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không
Rút gọn:
a)\(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}\)
b)\(\dfrac{6}{\sqrt{5}-1}+\dfrac{7}{1-\sqrt{3}}-\dfrac{2}{\sqrt{3}-\sqrt{5}}\)
c)\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right)\div\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
d)\(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
e)\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\times\left(\sqrt{6}+11\right)\)
Lm nhanh giúp mk nhé, mk đang cần gấp!
Bạn chia nhỏ ra để nhận được câu tl sớm nhất nhé!Bạn đặt câu hỏi free mà để dày cộp như này khum ai dám làm =(((
11) \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) + \(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
12) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}\) + \(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\) - \(\dfrac{1}{2-\sqrt{3}}\)
11.
\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\dfrac{\left(5+\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\dfrac{\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\dfrac{25+5+10\sqrt{5}}{20}+\dfrac{25+5-10\sqrt{5}}{20}\)
\(=3\)
12.
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\dfrac{1}{2-\sqrt{3}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{2+\sqrt{3}}{4-3}\)
\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)
\(=\sqrt{2}\)
11: Ta có: \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\dfrac{30+10\sqrt{5}+30-10\sqrt{5}}{20}\)
=3
12: Ta có: \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\dfrac{1}{2-\sqrt{3}}\)
\(=2+\sqrt{3}+\sqrt{2}-2-\sqrt{3}\)
\(=\sqrt{2}\)
\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\) + \(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\) - \(\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)