\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
rút gọn:giải hộ mình nha
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12 tháng 7 2021
Ta có: \(\dfrac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
\(=\dfrac{\sqrt{3}+3+\sqrt{2}-\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}-\sqrt{2}}\)
=3
12 tháng 7 2021
a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
AT
12 tháng 7 2021
\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)
\(=\sqrt{12}+1=2\sqrt{3}+1\)
\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}-1\)
\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
2) biến đổi khúc sau như câu 1:
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
AT
12 tháng 7 2021
1) Ta có: \(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{13+\sqrt{4.12}}}=\sqrt{5-\sqrt{13+2\sqrt{12}}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{12}\right)^2+2.\sqrt{12}+1^2}}=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}=\sqrt{5-\left|\sqrt{4.3}+1\right|}\)
\(=\sqrt{5-\left(2\sqrt{3}+1\right)}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=2\sqrt{3+\sqrt{3}-1}=2\sqrt{2+\sqrt{3}}\)
\(=2\sqrt{\dfrac{4+2\sqrt{3}}{2}}=2\sqrt{\dfrac{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}{2}}=2\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}\)
\(=2.\dfrac{\left|\sqrt{3}+1\right|}{\sqrt{2}}=\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{6}+\sqrt{2}\)
2) Ta có: \(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{3}-1\) (như trên)
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
GB
1
14 tháng 7 2021
Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
=5
HN
1
3 tháng 9 2019
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+2\right)^2}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(\sqrt{3}+2\right)}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
= \(\sqrt{4+\sqrt{5\left(\sqrt{3}+5-\sqrt{3}\right)}}\)
= \(\sqrt{4+\sqrt{25}}\)
= \(\sqrt{4+5}=3\)
NH
1
ML
26 tháng 6 2015
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+\sqrt{4}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3\)
ZG
3
25 tháng 6 2018
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}.\)
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3+4+2\sqrt{12}}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+\sqrt{4}\right)^2}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{25+3-2.\sqrt{25.3}}}\)
\(\sqrt{5\sqrt{3}+5\sqrt{\left(\sqrt{25}-\sqrt{3}\right)^2}}\)
\(\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(\sqrt{25}=5\)
19 tháng 7 2021
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: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)
\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)
19 tháng 7 2021
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
Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{25}=5\)
\(\sqrt{5\sqrt{3+5\sqrt{48}-10\sqrt{7+4\sqrt{3}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48}-10\sqrt{4+2.2\sqrt{3+\left(\sqrt{3}\right)^2}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-10.\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-10.\sqrt{\left(2+\sqrt{3}\right)}}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{48-20+10}\sqrt{3}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{28+10}\sqrt{3}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}\right)^2}+2.5.\sqrt{3}+5^2}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}+5\right)^2}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{\left(\sqrt{3}+5\right)}}}\)
=\(\sqrt{5\sqrt{3+5\sqrt{3+10}}}\)
=\(10\sqrt{3+10}\)
=\(\sqrt{10\left(\sqrt{3+1}\right)}\)