\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\) giải giùm .......
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Những câu hỏi liên quan
19 tháng 7 2021
1, \(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}=\dfrac{3+2\sqrt{2}}{9-8}-\dfrac{3-2\sqrt{2}}{9-8}\)
\(=3+2\sqrt{2}-3+2\sqrt{2}=4\sqrt{2}\)
2, \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+\sqrt{12}}\)
\(=\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{6}.\left(-1\right)}-\dfrac{3\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}.\left(-1\right)}\)
\(=\dfrac{2\sqrt{3}+3\sqrt{2}-3\sqrt{2}+3\sqrt{3}}{-\sqrt{6}}=\dfrac{5\sqrt{3}}{-\sqrt{6}}=-5\sqrt{18}=-15\sqrt{2}\)
3, \(\dfrac{2}{\sqrt{5}-2}+\dfrac{-2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)}{1}-\dfrac{2\left(\sqrt{5}-2\right)}{1}\)
\(=2\sqrt{5}+4-2\sqrt{5}+4=8\)
tương tự
19 tháng 7 2021
\(\dfrac{1}{3-2\sqrt{2}}-\dfrac{1}{3+2\sqrt{2}}=3+2\sqrt{2}-3+2\sqrt{2}=4\sqrt{2}\)
13 tháng 2 2021
Sửa đề: \(\dfrac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)
Ta có: \(\dfrac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)
\(=\dfrac{-x+8\sqrt{x}-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}-\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}+\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{-x+8\sqrt{x}-31-\left(x-25\right)+3x-9\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{2x-2\sqrt{x}-28-x+25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x-3\sqrt{x}+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)
15 tháng 8 2021
Ta có: \(\dfrac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}-\dfrac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\dfrac{6+2\sqrt{5}}{2\sqrt{2}+\sqrt{2}\cdot\left(\sqrt{5}+1\right)}-\dfrac{6-2\sqrt{5}}{2\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)}\)
\(=\dfrac{6+2\sqrt{5}}{2\sqrt{2}+\sqrt{10}+\sqrt{2}}-\dfrac{6-2\sqrt{5}}{2\sqrt{2}-\sqrt{10}+\sqrt{2}}\)
\(=\dfrac{6+2\sqrt{5}}{3\sqrt{2}+\sqrt{10}}-\dfrac{6-2\sqrt{5}}{3\sqrt{2}-\sqrt{10}}\)
\(=\dfrac{\left(6+2\sqrt{5}\right)\left(3\sqrt{2}-\sqrt{10}\right)-\left(6-2\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)}{8}\)
\(=\dfrac{18\sqrt{2}-6\sqrt{10}+6\sqrt{10}-10\sqrt{2}-18\sqrt{2}-6\sqrt{10}+6\sqrt{10}+10\sqrt{2}}{8}\)
\(=0\)
15 tháng 7 2021
1) \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}=9+4\sqrt{5}\)
2) \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}=\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{-\left(\sqrt{5}-\sqrt{2}\right)}=-\sqrt{10}\)
3) \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{5}}=\dfrac{\sqrt{10}\left(\sqrt{5}-3\right)}{-\left(\sqrt{5}-3\right)}=-\sqrt{10}\)
4) \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}=\dfrac{\left(6-2\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}=\dfrac{18-6\sqrt{5}-6\sqrt{5}+10}{4}=\dfrac{28-12\sqrt{5}}{4}=7-3\sqrt{5}\)
5)\(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}=\sqrt{5}+2\)
LP
21 tháng 12 2021
1)\(=\left|\sqrt{3}-3\right|+\sqrt{\left(\sqrt{3}-1\right)^2}=3-\sqrt{3}+\left|\sqrt{3}-1\right|=3-\sqrt{3}+\sqrt{3}-1=2\)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$
$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$
$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$
Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$
$\Leftrightarrow \sqrt{x}=\frac{12}{5}$
$\Leftrightarrow x=5,76$ (thỏa mãn)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
b. ĐKXĐ: $x^2\geq 5$
PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$
$\Leftrightarrow \sqrt{x^2-5}=0$
$\Leftrightarrow x=\pm \sqrt{5}$
H9
HT.Phong (9A5)
CTVHS
18 tháng 7 2023
\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11+\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)
\(=\sqrt{7-2\sqrt{6}}-\dfrac{5\left(\sqrt{2}-1\right)}{\sqrt{5}\left(\sqrt{2}-1\right)}+\left|11+2\sqrt{30}\right|\sqrt{11-2\sqrt{30}}\)
\(=\sqrt{1^2-2\sqrt{6}\cdot1+\left(\sqrt{6}\right)^2}-\dfrac{\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{5}\cdot\sqrt{6}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(1-\sqrt{6}\right)^2}-\sqrt{5}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\)
\(=\left|1-\sqrt{6}\right|-\sqrt{5}+\left(11+2\sqrt{30}\right)\left|\sqrt{6}-\sqrt{5}\right|\)
\(=-1+6-\sqrt{5}+\left(\sqrt{6}+\sqrt{5}\right)^2\left(\sqrt{6}-\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\left[\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2\right]\left(\sqrt{6}+\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\left(6-5\right)\left(\sqrt{6}+\sqrt{5}\right)\)
\(=\sqrt{6}-1-\sqrt{5}+\sqrt{6}+\sqrt{5}\)
\(=2\sqrt{6}-1\)
18 tháng 7 2023
\(=\sqrt{6+1-2\sqrt{6}}-\dfrac{\sqrt{5}\left(\sqrt{10}-\sqrt{5}\right)}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+\sqrt{120}\right)^2}\\ =\sqrt{\left(\sqrt{6}-\sqrt{1}\right)^2}-\sqrt{5}+\sqrt{\left(11^2-120\right)\left(11+2\sqrt{30}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{1\left(6+5+2\sqrt{6\cdot5}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}-\sqrt{1}\)
10 tháng 7 2017
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
29 tháng 8 2021
a: Ta có: \(\dfrac{2}{\sqrt{3}+1}+\dfrac{2}{2-\sqrt{3}}\)
\(=\sqrt{3}-1+2+\sqrt{3}\)
\(=2\sqrt{3}+1\)
b: Ta có: \(\dfrac{4}{\sqrt{5}+2}+\dfrac{2}{3+\sqrt{5}}\)
\(=4\sqrt{5}-8+\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\)
\(=-\dfrac{13}{2}+\dfrac{7}{2}\sqrt{5}\)
25 tháng 6 2021
`(sqrtx+2)/(sqrtx-3)-(sqrtx+1)/(sqrtx-2)-(3(sqrtx-1))/(x-5sqrtx+6)`
đk:`x>=0,x ne 4,x ne 9`
`=((sqrtx+2)^2-(sqrtx+1)(sqrtx+3)-3(sqrtx-1))/(x-5sqrtx+6)`
`=(x+4sqrtx+4-x-4sqrtx-3-3sqrtx+3)/(x-5sqrtx+6)`
`=(4-3sqrtx)/(x-5sqrtx+6)`
\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\) \(=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)