\(A=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
giúp mình làm cách nào dễ hiểu và giải thích giùm ạ ! sẽ cảm ơn
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NV
30 tháng 6 2016
a/ A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
Vậy A = 1
BA
10 tháng 9 2020
\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\) ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))
\(=\sqrt{\left(2+\sqrt{7}+2-\sqrt{7}\right)\left(2+\sqrt{7}-2+\sqrt{7}\right)}\)
\(=\sqrt{4\cdot\sqrt{7}}\)
\(2,\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}-\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}=\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2=\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2-\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(=\left(3\sqrt{5}-5\sqrt{2}+5\sqrt{2}+3\sqrt{5}\right)\left(3\sqrt{5}-5\sqrt{2}-5\sqrt{2}-3\sqrt{5}\right)\)
\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)
\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}\)
cuối lười tính nên thôi nhá :>
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)
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
1 tháng 8 2017
b/ \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow B^2=8+2\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow B=\sqrt{5}+1\)
1 tháng 8 2017
a/ \(\sqrt{5\sqrt{3}+5\sqrt{48}-10\sqrt{7+4\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+20\sqrt{3}-10\sqrt{\left(2+\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+20\sqrt{3}-20-10\sqrt{3}}\)
\(=\sqrt{15\sqrt{3}-20}\)
23 tháng 10 2016
a/ Đặt \(\hept{\begin{cases}\sqrt{3+\sqrt{5}}=a\\\sqrt{3-\sqrt{5}}=b\end{cases}}\)
Khi đó ta có a2 + b2 = 6; ab = 2; a + b = \(\sqrt{10}\) ; a - b = \(\sqrt{2}\); a2 - b2 = \(2\sqrt{5}\)
Ta có cái ban đầu
\(=\frac{a^2}{\sqrt{10}+a}-\frac{b^2}{\sqrt{10}+b}\)=
\(\frac{\sqrt{10}a^2+a^2b-\sqrt{10}b^2-ab^2}{10+\sqrt{10}a+\sqrt{10}b+ab}\)
\(=\frac{10\sqrt{2}+2\sqrt{2}}{10+10+2}=\frac{6\sqrt{2}}{11}\)
29 tháng 7 2021
Vd1:
d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)
\(\Leftrightarrow x=6\)
MN
14 tháng 7 2021
Bài 1 :
\(a.\sqrt{x^2-1}\)
\(ĐK:\)
\(x^2-1\ge0\)
\(\Leftrightarrow x^2\ge1\)
\(\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)
Bài 2 :
\(2\cdot\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{48}-5\sqrt{50}\)
\(=2\cdot\left|\sqrt{2}-3\right|+4\sqrt{3}-25\sqrt{2}\)
\(=-2\cdot\left(\sqrt{2}-3\right)+4\sqrt{3}-25\sqrt{2}\)
\(=-2\sqrt{2}-6+4\sqrt{3}-25\sqrt{2}\)
\(=-27\sqrt{2}-6+4\sqrt{3}\)
\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+\sqrt{5\left(5-\sqrt{3}\right)}}=\sqrt{5\sqrt{3}+\sqrt{25-5\sqrt{3}}}\)
Trần Đức Thắng lm nốt đi