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1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
1: \(=\sqrt{36}=6\)
2: \(=\sqrt{\left(15-9\right)\left(15+9\right)}=\sqrt{24\cdot6}=12\)
3: \(=3\sqrt{5}-1-3\sqrt{5}-1=-2\)
4: \(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
5: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)
1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)
\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)
\(=6-8=-2\)
2) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=3^2-\left(\sqrt{5}\right)^2\)
\(=9-5=4\)
3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)
=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn
Bạn chỉ cần tách chúng thành hằng đẳng thức sau đó áp dụng HĐT: \(\sqrt{A^2}=\left|A\right|\)
1, \(\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}+1\right|=\sqrt{2}+1\)
2, \(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)3, \(\sqrt{5+2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\left|\sqrt{3}+\sqrt{2}\right|=\sqrt{3}+\sqrt{2}\)4, \(\sqrt{7-2\sqrt{10}}=\sqrt{5-2\sqrt{5}.\sqrt{2}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left|\sqrt{5}-\sqrt{2}\right|=\sqrt{5}-\sqrt{2}\)5, \(\sqrt{15-6\sqrt{6}}=\sqrt{9-2.3.\sqrt{6}+6}=\sqrt{\left(3+\sqrt{6}\right)^2}=\left|3+\sqrt{6}\right|=3+\sqrt{6}\)Các câu còn lại tương tự nha!
6, \(\sqrt{8+2\sqrt{15}}=\sqrt{5+2\sqrt{5}.\sqrt{3}+3}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\left|\sqrt{5}+\sqrt{3}\right|=\sqrt{5}+\sqrt{3}\)7, \(\sqrt{10-2\sqrt{21}}=\sqrt{7-2\sqrt{7}.\sqrt{3}+3}=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\left|\sqrt{7}-\sqrt{3}\right|=\sqrt{7}-\sqrt{3}\)8, \(\sqrt{11+2\sqrt{18}}=\sqrt{9+2\sqrt{9}.\sqrt{2}+2}=\sqrt{\left(\sqrt{9}+\sqrt{2}\right)^2}=\left|3+\sqrt{2}\right|=3+\sqrt{2}\)9, \(\sqrt{14-2\sqrt{33}}=\sqrt{11-2\sqrt{11}.\sqrt{3}+3}=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\left|\sqrt{11}-\sqrt{3}\right|=\sqrt{11}-\sqrt{3}\)Thử tự làm những câu còn lại rồi kiểm tra xem đúng hay sai nha!!!
Chúc bạn học tốt!!!
\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)
a) \(A=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+-\sqrt{3}-1=-2\)
b) \(B=\sqrt{11-6\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{3^2-2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)
\(=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\left|3-\sqrt{2}\right|-\left|\sqrt{2}-1\right|\)
\(=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)
c) \(C=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left(\sqrt{5}+\sqrt{3}\right)\left|\sqrt{5}-\sqrt{2}\right|\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{2}\right)=5-\sqrt{10}+\sqrt{15}-\sqrt{6}\)
1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)