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#Giải:
a)\(\sqrt{27}\)+\(\sqrt{75}\)-\(\sqrt{\dfrac{1}{3}}\)=8\(\sqrt{3}\)-\(\sqrt{\dfrac{1}{3}}\)=\(\dfrac{23\sqrt{3}}{3}\).
b)\(\sqrt{4+2\sqrt{3}}\)-\(\sqrt{4-2\sqrt{3}}\)=2.
c)\(\dfrac{3}{\sqrt{7}+\sqrt{2}}\)+\(\dfrac{2}{3+\sqrt{7}}\)+\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)=1,093+\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)=2,507.
a) = \(3\sqrt{3}+5\sqrt{3}-\dfrac{1}{\sqrt{3}}\)
= \(3\sqrt{3}+5\sqrt{3}-\dfrac{3}{\sqrt{3}}\)
= \(\dfrac{23\sqrt{3}}{3}\)
b) = \(\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
= \(1+\sqrt{3}-\left(\sqrt{3}-1\right)\)
= \(1+\sqrt{3}-\sqrt{3}+1\)
= 2
c) = \(\dfrac{3\left(\sqrt{7}-\sqrt{2}\right)}{5}+\dfrac{2\left(3-\sqrt{7}\right)}{2}+\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)\)
= \(3\sqrt{7}-3\sqrt{2}+3-\sqrt{7}+2\sqrt{2}+2-2-\sqrt{2}\)
= \(\dfrac{3\sqrt{7}-3\sqrt{2}}{5}+3-\sqrt{7}+\sqrt{2}\)
= \(\dfrac{3\sqrt{7}-3\sqrt{2}-5\sqrt{7}+5\sqrt{2}}{5}+3\)
= \(\dfrac{-2\sqrt{7}+2\sqrt{2}}{5}+3\)
\(\approx2,5\)
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\(B=\dfrac{\sqrt{6+2\left(\sqrt{6}+\sqrt{3}+\sqrt{2}\right)}-\sqrt{6-2\left(\sqrt{6}-\sqrt{3}+\sqrt{2}\right)}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}-\sqrt{6-2\sqrt{6}+2\sqrt{3}-2\sqrt{2}}}{\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}-\sqrt{6-2\sqrt{6}+2\sqrt{3}-2\sqrt{2}}\right)\sqrt{2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}\right)\cdot2}-\sqrt{\left(6-2\sqrt{6}+2\sqrt{3}-2\sqrt{2}\right)\cdot2}}{2}\)
\(=\dfrac{\sqrt{12+4\sqrt{6}+4\sqrt{3}+4\sqrt{2}}-\sqrt{12-4\sqrt{6}+4\sqrt{3}-4\sqrt{2}}}{2}\)
\(=\dfrac{4}{2}\)
\(=2\)
\(C=\dfrac{\sqrt{9-6\sqrt{2}}-\sqrt{6}}{\sqrt{3}}\)
\(=\dfrac{\left(\sqrt{9-6\sqrt{2}}-\sqrt{6}\right)\sqrt{3}}{3}\)
\(=\dfrac{\sqrt{\left(9-6\sqrt{2}\right)\cdot3}-3\sqrt{2}}{3}\)
\(=\dfrac{\sqrt{27-18\sqrt{2}}-3\sqrt{2}}{3}\)
\(=\dfrac{\sqrt{\left(3-3\sqrt{2}\right)^2}-3\sqrt{2}}{3}\)
\(=\dfrac{3\sqrt{2}-3-3\sqrt{2}}{3}\)
\(=\dfrac{-3}{3}\)
\(=-1\)
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Xem kỹ lại đề nhé! loại này đề lệch một tý thôi -->Không rút được !
p/s: Tránh truongf hợp làm đến cuối mới biết đề sai.
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Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào bài toán ta được
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)
\(=1-\frac{1}{\sqrt{225}}=1-\frac{1}{15}=\frac{14}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\dfrac{9}{49}}=\sqrt{\dfrac{3}{7}}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}=\dfrac{\sqrt{9}+\sqrt{1521}}{\sqrt{49}+\sqrt{8281}}=\dfrac{3+39}{7+91}=\dfrac{42}{98}\)
c)Tương tự câu b, ta đc:
\(\dfrac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{7^2}-\sqrt{91^2}}=\dfrac{3-39}{7-91}=\dfrac{-36}{86}=\dfrac{3}{7}\)
d)Tương tự câu a, ta đc:
\(\dfrac{\sqrt{39^2}}{\sqrt{91^2}}=\dfrac{39}{91}\)
Chúc Bạn Học Tốt!!!
a) \(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\left(\dfrac{3}{7}\right)^2}=\left|\dfrac{3}{7}\right|=\dfrac{3}{7}\)
b) \(\dfrac{\sqrt{3}^2+\sqrt{39}^2}{\sqrt{7}^2+\sqrt{91}^2}=\dfrac{\left|3\right|+\left|39\right|}{\left|7\right|+\left|91\right|}=\dfrac{3+39}{7+91}=\dfrac{42}{98}=\dfrac{3}{7}\)
c) \(\dfrac{\sqrt{3}^2-\sqrt{39}^2}{\sqrt{7}^2-\sqrt{91}^2}=\dfrac{\left|3\right|- \left|39\right|}{\left|7\right|-\left|91\right|}=\dfrac{3-39}{7-91}=\dfrac{-36}{-84}=\dfrac{3}{7}\)
d) \(\sqrt{\dfrac{39^2}{91^2}}=\sqrt{\left(\dfrac{39}{91}\right)^2}=\left|\dfrac{39}{91}\right|=\dfrac{39}{91}=\dfrac{3}{7}\)
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a) \(\sqrt{\left(-5\right)^2}+\sqrt{5^2}-\sqrt{\left(-3\right)^2}-\sqrt{3^2}\)
\(=5+5-3-3\)
\(=4\)
b) \(\left(\sqrt{4^2}+\sqrt{\left(-4\right)^2}\right).\sqrt{4^{-3}}-\sqrt{3^{-4}}\)
\(=\left(4+4\right).\frac{1}{8}-\frac{1}{9}\)
\(=8.\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
th1:(\(2\sqrt{3}-\sqrt{2}\))2
th2: số lẻ lắm 4,20204...
Máy tính ra là "14 - 4 căn 6"