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\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{14}}{2}\)
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ND
25 tháng 7 2018
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\dfrac{\sqrt{9\left(5+3\sqrt{2}\right)}+\sqrt{9\left(5-3\sqrt{2}\right)}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
\(=\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)}{3+\sqrt{2}-3+\sqrt{2}}\)
\(=\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)}{5+3\sqrt{2}-5+3\sqrt{2}}-\dfrac{3+\sqrt{2}+2\sqrt{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}+3-\sqrt{2}}{2\sqrt{2}}\)
\(=\dfrac{3\left[5+3\sqrt{2}+5-3\sqrt{2}+2\sqrt{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}\right]}{6\sqrt{2}}-\dfrac{6+2\sqrt{7}}{2\sqrt{2}}\)
\(=\dfrac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\dfrac{6+2\sqrt{7}}{2\sqrt{2}}=\dfrac{30+6\sqrt{7}-18-6\sqrt{7}}{6\sqrt{2}}=\dfrac{12}{6\sqrt{2}}\)
\(=\sqrt{2}\)
25 tháng 7 2018
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\\ =\dfrac{3\sqrt{5+3\sqrt{2}}+3\sqrt{5-3\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\\ =\dfrac{3\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)\left(\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}\right)}-\dfrac{\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}}{\sqrt{6+3\sqrt{2}}-\sqrt{6-4\sqrt{2}}}\\ =\dfrac{3\left(5+3\sqrt{2}+2\sqrt{25-18}+5-3\sqrt{2}\right)}{5+3\sqrt{2}-5+3\sqrt{2}}-\dfrac{\sqrt{4+2+4\sqrt{2}}+\sqrt{4+2-4\sqrt{2}}}{\sqrt{4+2+4\sqrt{2}}-\sqrt{4+2-4\sqrt{2}}}\\ =\dfrac{3\left(10+2\sqrt{7}\right)}{6\sqrt{2}}-\dfrac{\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}}{\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}}\\ =\dfrac{10+2\sqrt{7}}{2\sqrt{2}}-\dfrac{2+\sqrt{2}+2-\sqrt{2}}{2+\sqrt{2}-2+\sqrt{2}}\\ =\dfrac{10+2\sqrt{7}}{2\sqrt{2}}-\dfrac{4}{2\sqrt{2}}=\dfrac{6+2\sqrt{7}}{2\sqrt{2}}\)
2 tháng 7 2018
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}=\dfrac{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{2\sqrt{2}}-\dfrac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}=\dfrac{4+2\sqrt{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}-2\sqrt{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}}{2\sqrt{2}}\) \(=\dfrac{4+2\sqrt{7}-2\sqrt{7}}{2\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
H9
HT.Phong (9A5)
CTVHS
14 tháng 8 2023
\(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\sqrt{3}+2\sqrt{7}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\)
\(=\dfrac{\sqrt{2}}{2}\)
___________
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
__________
\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\dfrac{3\cdot2\sqrt{2}-2\cdot2\sqrt{3}+2\sqrt{5}}{3\cdot3\sqrt{2}-2\cdot3\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{2}{3}\)
14 tháng 8 2023
a: \(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{2}\)
b: \(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)
c: \(=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\dfrac{2}{3}\)
HP
29 tháng 8 2021
9.
\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\)
\(=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=-\sqrt{5}\)
HP
29 tháng 8 2021
10.
\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\)
\(=5\sqrt{3}-\sqrt{5+\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2+\dfrac{2}{3}}+6\sqrt{3}\)
\(=11\sqrt{3}-\sqrt{\dfrac{16}{3}}+\dfrac{9}{2}\sqrt{\dfrac{8}{3}}\)
\(=11\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}\)
\(=\dfrac{29\sqrt{3}}{3}+3\sqrt{6}\)
22 tháng 7 2023
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
22 tháng 7 2023
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
2
18 tháng 9 2021
\(\dfrac{1}{2}\sqrt{48a}-2\sqrt{45}-\dfrac{a}{3}\sqrt{\dfrac{27}{a}}+5\sqrt{1\dfrac{4}{5}}\)
\(=2\sqrt{3a}-6\sqrt{5}-\sqrt{3a}+3\sqrt{5}\)
=\(=\sqrt{3a}-3\sqrt{5}\)
18 tháng 9 2021
\(=\sqrt{\dfrac{1}{4}.48a}-6\sqrt{5}-\sqrt{\dfrac{a^2}{9}.\dfrac{27}{a}}+\sqrt{25.\dfrac{9}{5}}\)
\(=\sqrt{12a}-6\sqrt{5}-\sqrt{3a}+3\sqrt{5}\)
\(=2\sqrt{3a}-6\sqrt{5}-\sqrt{3a}+3\sqrt{5}\)
\(=\sqrt{3a}-3\sqrt{5}\)
5 tháng 9 2023
a: ĐKXĐ: x>=3
Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)
=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)
=>\(\dfrac{3}{2}\sqrt{x-3}=3\)
=>\(\sqrt{x-3}=2\)
=>x-3=4
=>x=7(nhận)
b: ĐKXĐ: x>=0
\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)
=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)
=>\(7\sqrt{x}-5< =0\)
=>\(\sqrt{x}< =\dfrac{5}{7}\)
=>0<=x<=25/49
c: ĐKXĐ: x>=5
\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)
=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)
=>\(\dfrac{3}{2}\sqrt{x-5}=3\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
ta có:\(\sqrt{45+27\sqrt{2}}\) +\(\sqrt{45-27\sqrt{2}}\) =3(\(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\))
phân thức cần biến đổi trở thành:\(\dfrac{\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}\).3
=\(\dfrac{10+2\sqrt{7}}{6\sqrt{2}}=\dfrac{5+\sqrt{7}}{\sqrt{2}}=\dfrac{5\sqrt{2}+\sqrt{14}}{2}\)
lấy phân số vừa tìm được cộng cho \(\dfrac{\sqrt{14}}{2}\) ta được giá trị biểu thức cần tìm là \(\dfrac{5\sqrt{2}}{2}\)
Tại sao kết quả lại bằng, có thể giải thích hộ mình ko\(\dfrac{10+2\sqrt{7}}{6\sqrt{2}}\)