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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)
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
2 tháng 11 2021
\(1,=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\\ 2,=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right):5=\dfrac{2\sqrt{6}}{5}-\dfrac{2\sqrt{5}}{5}\\ 3,=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-\dfrac{9\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\\ 4,Sửa:\dfrac{1}{\sqrt{5}-\sqrt{3}}-\dfrac{1}{\sqrt{5}+\sqrt{3}}\\ =\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
2 tháng 11 2021
1) \(=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\)
2) \(=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right)=\dfrac{2\sqrt{6}}{5}+\dfrac{2\sqrt{5}}{5}-\dfrac{4\sqrt{5}}{5}\)
3) \(=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)
4) \(=\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{5-3}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
6 tháng 6 2021
Bài 2:
a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: \(x\ge2\))
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+\dfrac{6}{\sqrt{81}}\sqrt{x-2}=-4\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\Leftrightarrow-\sqrt{x-2}=-4\) \(\Leftrightarrow x-2=16\)
\(\Leftrightarrow x=18\) (thỏa)
Vậy...
b)\(\sqrt{9x^2+12x+4}=4x\)(Đk:\(9x^2+12x+4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}4x\ge0\\9x^2+12x+4=16x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+12x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+14x-2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-2\right)\left(-7x-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{7}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=2\) (tm đk)
Vậy...
c) \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\) (đk: \(x\ge1\))
\(\Leftrightarrow x-2\sqrt{x-1}=x-1\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{5}{4}\) (tm)
Vậy...
NN
2
Y
7 tháng 7 2023
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left|2+\sqrt{3}\right|}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{5^2-2.5.\sqrt{3}+\sqrt{3^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left|5-\sqrt{3}\right|}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}\)
\(=\sqrt{9}\\ =3\)
DA
7 tháng 7 2023
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10.|2+\sqrt{3}|}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10.\left(2+\sqrt{3}\right)}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5.|5-\sqrt{3}|}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+5.\left(5-\sqrt{3}\right)}}\)
= \(\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
= \(\sqrt{4+\sqrt{25}}\)
= \(\sqrt{4+5}\)
= \(\sqrt{9}\)
= \(3\)
24 tháng 6 2018
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-6\sqrt{20}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\\ =\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}\\ =\sqrt{1}=1\)
\(B=\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12+1+2\sqrt{12}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\\ =\sqrt{6+2\sqrt{3+1-2\sqrt{3}}}\\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{6+2\sqrt{3}-2}\\ =\sqrt{3+1+2\sqrt{3}}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+4\sqrt{3}}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+2\right)^2}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25+3-10\sqrt{3}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\\ =\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3\)
\(D=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\\ \text{Ta có }:\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)^2\\ =3+\sqrt{5}-2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+3-\sqrt{5}\\ =6-2\sqrt{9-5}=6-2\sqrt{4}=6-4=2\\ \Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{2}\\ \Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
H
9 tháng 10 2023
\(A=\sqrt{243}-\sqrt{27}+\sqrt{3}-\sqrt{48}\\ =\sqrt{81\cdot3}-\sqrt{9\cdot3}+\sqrt{3}-\sqrt{16\cdot3}\\ =9\sqrt{3}-3\sqrt{3}+\sqrt{3}-4\sqrt{4}\\ =\left(9-3+1-4\right)\sqrt{3}\\ =3\sqrt{3}\)
\(B=\dfrac{5+\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{3}+\sqrt{5}\right)\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}}+\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\sqrt{5}-\sqrt{3}\\ =\sqrt{5}+1+\sqrt{3}-\sqrt{5}-\sqrt{3}\\ =1\)
10 tháng 1 2021
a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)
b) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{6+2\sqrt{5}-\left(2\sqrt{5}-3\right)}\)
\(=\sqrt{6+3}=3\)
c) Sửa đề: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
Ta có: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)
\(=\sqrt{2+\sqrt{5+2\sqrt{3}-1}}\)
\(=\sqrt{2+\sqrt{3+2\sqrt{3}\cdot1+1}}\)
\(=\sqrt{2+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\sqrt{3+\sqrt{3}}\)
d) Ta có: \(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)^2\cdot\left(\sqrt{5}+1\right)+\left(\sqrt{5}+1\right)^2\cdot\left(\sqrt{5}-1\right)}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(\sqrt{5}-1+\sqrt{5}+1\right)}{2\sqrt{2}}\)
\(=\dfrac{4\cdot2\sqrt{5}}{2\sqrt{2}}\)
\(=\dfrac{8\sqrt{5}}{2\sqrt{2}}=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
H9
HT.Phong (9A5)
CTVHS
25 tháng 7 2023
\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{2^2+2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{5^2-2\cdot5\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}\)
\(=\sqrt{5^2}\)
\(=5\)
25 tháng 7 2023
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}=\sqrt{25}=5\)
NC
4 tháng 10 2020
a) \(\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{29-6\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-2\sqrt{5}+3}\)
\(=\sqrt{3-\sqrt{3}-\sqrt{5}}\)
\(2\sqrt{3+\sqrt{5-\sqrt{3+\sqrt{48}}}}\)
\(\Leftrightarrow5\sqrt{3}+\sqrt{5}\)
\(\Leftrightarrow\sqrt{10\left(8+\sqrt{15}\right)}\)
\(\Leftrightarrow x^4-160x^2+4900\)
\(\Leftrightarrow x\approx10.896320153441.........\)
10,8963220153