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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)
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
Q
24 tháng 6 2017
a) \(\sqrt{14-6\sqrt{5}}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
b, c) tương tự câu a.
d) \(\left(3-\sqrt{2}\right)\sqrt{11+6\sqrt{2}}\)
\(=\left(3-\sqrt{2}\right)\sqrt{\left(3+\sqrt{2}\right)^2}\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
\(=9-2\)
\(=7\)
e) \(\sqrt{11-6\sqrt{2}+\sqrt{3-2\sqrt{2}}}\)
\(=\sqrt{11-6\sqrt{2}+\sqrt{\left(1-\sqrt{2}\right)^2}}\)
\(=\sqrt{11-6\sqrt{2}+\sqrt{2}-1}\)
\(=\sqrt{10-5\sqrt{2}}\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 9 2021
ĐKXĐ: \(x\ge0;x\ne4\)
\(A=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
23 tháng 9 2021
\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+7}{4-x}\left(x>0;x\ne4\right)\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ P=\dfrac{\sqrt{x}+6-x-x-3\sqrt{x}-2+2\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ P=\dfrac{-2x+11}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ P=\dfrac{-2x\sqrt{x}+11\sqrt{x}+\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}\left(x-4\right)}\)
\(P=\dfrac{-2x\sqrt{x}+11\sqrt{x}+x\sqrt{x}-4\sqrt{x}+2x-8}{\sqrt{x}\left(x-4\right)}\\ P=\dfrac{-x\sqrt{x}+8\sqrt{x}+2x-8}{\sqrt{x}\left(x-4\right)}\)
24 tháng 9 2021
\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2\sqrt{x}+7}{x-4}\right)\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\)
\(=\dfrac{-x+8\sqrt{x}-15+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{-x+8\sqrt{x}-15+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
24 tháng 9 2021
\(ĐK:x\ge0;x\ne4\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}-5\right)+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{8\sqrt{x}-15-x+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
AH
Akai Haruma
Giáo viên
19 tháng 7 2021
Lần sau bạn chú ý viết đầy đủ đề.
1.
\(\sqrt{9+4\sqrt{5}-\sqrt{9-4\sqrt{5}}}=\sqrt{9+4\sqrt{5}-\sqrt{5-2\sqrt{4.5}+4}}\)
\(=\sqrt{9+4\sqrt{5}-\sqrt{(\sqrt{5}-\sqrt{4})^2}}=\sqrt{9+4\sqrt{5}-(\sqrt{5}-\sqrt{4})}\)
\(=\sqrt{9+4\sqrt{5}-\sqrt{5}+2}=\sqrt{11+3\sqrt{5}}\)
AH
Akai Haruma
Giáo viên
19 tháng 7 2021
2.
\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}=\sqrt{8-2\sqrt{7}-\sqrt{7+2\sqrt{7}+1}}\)
\(=\sqrt{8-2\sqrt{7}-\sqrt{(\sqrt{7}+1)^2}}\)
\(=\sqrt{8-2\sqrt{7}-\sqrt{7}-1}=\sqrt{7-3\sqrt{7}}\)
AH
Akai Haruma
Giáo viên
4 tháng 9 2021
Lời giải:
\(\frac{2x-2\sqrt{x}+2}{x-\sqrt{x}}=\frac{2(x-\sqrt{x})+2}{x-\sqrt{x}}=\frac{2(x-\sqrt{x})+2}{x-\sqrt{x}}=2+\frac{2}{x-\sqrt{x}}\)
4 tháng 9 2021
\(\dfrac{2x-2\sqrt{x}+2}{x\sqrt{x}+1}=\dfrac{2}{\sqrt{x}+1}\)
\(=13+2\sqrt{42}+2\sqrt{42}=13+4\sqrt{42}\)
\(\left(\sqrt{7}+\sqrt{6}\right)^2+\sqrt{168}\)
\(=7+6+2\sqrt{42}+2\sqrt{42}=13+4\sqrt{42}\)