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*\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}+1=2\)
\(\Rightarrow A\in Z\)
* \(B=\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-2\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\) \(=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\) \(=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\) \(=\dfrac{3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}}{9-8}\)
\(=2\)
\(\Rightarrow B\in Z\)
a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=-2
b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)
c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
\(=\sqrt{\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}}-2-\sqrt{10}+3-2\sqrt{2}\)
\(=\sqrt{\sqrt{10}}+1-\sqrt{10}-2\sqrt{2}\)
3 bài đầu dễ tự làm nhé.
Bài 4:
\(B=\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\dfrac{\sqrt{\left(1-\sqrt{2}\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(1+\sqrt{2}\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)
\(=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{1+\sqrt{2}}{3+2\sqrt{2}}\)
\(=\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(1+\sqrt{2}\right)\left(3-2\sqrt{2}\right)\)
\(=3\sqrt{2}+4-3-2\sqrt{2}-\left(3-2\sqrt{2}+3\sqrt{2}-4\right)\)
\(=3\sqrt{2}+4-3-2\sqrt{2}-\left(-1+\sqrt{2}\right)\)
\(=3\sqrt{2}+4-3-2\sqrt{2}+1-\sqrt{2}\)
\(=0+2\)
\(=2\)
Vậy B là số tự nhiên.
1.
a) nhân cả tử lẫn mẫu với 1+ \(\sqrt{2}-\sqrt{5}\)
b) tương tự a
2.
a) tách 29 = 20 + 9 là ra hằng đẳng thức, tiếp tục.
Bài 1 bạn nhóm , trục như thường nhé :D
Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)
\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)
\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)
\(D=-\sqrt{6}\left(do:D< 0\right)\)
1 , \(\left(\sqrt{12}-2\sqrt{75}\right).\sqrt{3}=\sqrt{12.3}-\sqrt{300.3}=6-30=-24\)
2 , \(\sqrt{3}.\left(\sqrt{12}.\sqrt{27}-\sqrt{3}\right)=\sqrt{12.27.3}-\sqrt{3.3}=18\sqrt{3}-3\)
3 , \(\left(7\sqrt{48}+3\sqrt{27}-\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-2\sqrt{3}\right):\sqrt{3}=35\)
4 , bạn làm tương tự nhé
5 , bạn làm tương tự nhé
6 , bạn làm tương tự nhé
\(A=\dfrac{5.\left(38^2-17^2\right)}{8\left(47^2-19^2\right)}\\ =\dfrac{5\left(38-17\right)\left(38+17\right)}{8\left(47-19\right)\left(47+19\right)}\\ =\dfrac{5.21.55}{8.28.66}\\ =\dfrac{5.1155}{8.1848}\\ =\dfrac{5.5}{8.8}\\ =\dfrac{25}{64}\)
\(B=\sqrt{\dfrac{0,2\times1,21\times0,3}{7,5\times3,2\times0,64}}\\ =\sqrt{0,0625\times1,890625\times0,04}\\ =\sqrt{\dfrac{121}{25600}}\\ =\dfrac{11}{160}\)
+) ta có : \(A=\sqrt{13+4\sqrt{10}}-\sqrt{13-4\sqrt{10}}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}-2\sqrt{2}+\sqrt{5}=2\sqrt{5}\) (sữa đề)
+) ta có : \(B=\sqrt{\dfrac{3-2\sqrt{2}}{17-12\sqrt{2}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(3-2\sqrt{2}\right)^2}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)^2}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{1}{\sqrt{2}-1}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\sqrt{\dfrac{\left(2-\sqrt{3}\right)\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=\sqrt{2}+1+2-\sqrt{3}=3-\sqrt{3}+\sqrt{2}\) (sữa đề )
+) đk : \(x\ne-3\)
ta có : \(C=\dfrac{\sqrt{x^2+6x+9}}{x+3}=\dfrac{\sqrt{\left(x+3\right)^2}}{x+3}=\dfrac{\left|x+3\right|}{x+3}\)
\(\left[{}\begin{matrix}C=1\left(x>-3\right)\\C=-1\left(x< -3\right)\end{matrix}\right.\)
+) \(m\ge\dfrac{5}{2}\)
ta có : \(D=\sqrt{2m+4+6\sqrt{2m-5}}-\sqrt{2m-5}\)
\(=\sqrt{\left(\sqrt{2m-5}+3\right)^2}-\sqrt{2m-5}=\left|\sqrt{2m-5}+3\right|-\sqrt{2m-5}\)
\(\Leftrightarrow\left[{}\begin{matrix}C=3\left(m\ge7\right)\\C=-3-2\sqrt{2m-5}\left(\dfrac{5}{2}\le m\le7\right)\end{matrix}\right.\)
\(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{17+12\sqrt{2}}}\\ =\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\\ =\dfrac{\sqrt{5}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{5}+1}{3+2\sqrt{2}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{5}+1\right)\left(3-2\sqrt{2}\right)}{3^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{4\sqrt{10}-6}{9-8}=4\sqrt{10}-6\)