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Mình làm luôn nhé :
\(\sqrt{45-2.3\sqrt{5}+1}-\sqrt{20-2.3.2\sqrt{5}+9}\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5-\sqrt{45+2.2.\sqrt{2}.3\sqrt{5}+8}}\left(\sqrt{3}+\sqrt{5}\right).\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{7+2.\sqrt{7}.\sqrt{3}+3}\) Tới đây dễ rồi , bạn tự nhóm HĐT là ra ::v
Bài 1:
a) Ta có: \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{45-2\cdot\sqrt{45}\cdot1+1}-\sqrt{9-2\cdot\sqrt{9}\cdot\sqrt{20}+20}\)
\(=\sqrt{\left(\sqrt{45}-1\right)^2}-\sqrt{\left(3-\sqrt{20}\right)^2}\)
\(=\left|\sqrt{45}-1\right|-\left|3-\sqrt{20}\right|\)
\(=\sqrt{45}-1-3+\sqrt{20}\)
\(=\sqrt{45}+\sqrt{20}-4\)
\(=\sqrt{5}\left(3+2\right)-4=5\sqrt{5}-4\)
b) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{8}+8}-\sqrt{45+2\cdot\sqrt{45}\cdot\sqrt{8}+8}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{8}\right)^2}-\sqrt{\left(\sqrt{45}+\sqrt{8}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{8}\right|-\left|\sqrt{45}+\sqrt{8}\right|\)
\(=\sqrt{8}-\sqrt{5}-\sqrt{45}-\sqrt{8}\)
\(=-\sqrt{5}-\sqrt{45}=-\sqrt{5}\left(1+\sqrt{9}\right)=-4\sqrt{5}\)
c) Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{7+4\sqrt{3}}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=\left(3-\sqrt{2}\right)\left(\sqrt{3}+2\right)\)
\(=3\sqrt{3}+6-\sqrt{6}-2\sqrt{2}\)
d) Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10+2\sqrt{21}}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{7+2\cdot\sqrt{7}\cdot\sqrt{3}+3}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\left(\sqrt{7}+\sqrt{3}\right)\)
\(=\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2=7-3=4\)
\(A=\left(2-\sqrt{3}\right)\sqrt{4+2.2.\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)
các câu còn lại làm tương tự nhé bạn !
\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}=6\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)
\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}=2\sqrt{5}+4\sqrt{2}\)
a: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
b: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}\)
=6
c: Ta có: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}\)
=10
d: Ta có: \(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+4\sqrt{90}}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}\)
\(=2\sqrt{5}+4\sqrt{2}\)
\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
\(\sqrt{8-2\sqrt{12}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}=\left|\sqrt{6}-\sqrt{2}\right|=\sqrt{6}-\sqrt{2}\)
\(\sqrt{21+6\sqrt{6}}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\left|3\sqrt{2}-\sqrt{3}\right|=3\sqrt{2}-\sqrt{3}\)
\(\sqrt{15-6\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|=3-\sqrt{6}\)
\(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)
\(\sqrt{41+12\sqrt{5}}=\sqrt{\left(6+\sqrt{5}\right)^2}=6+\sqrt{5}\)
* \(\sqrt{2}\)A = \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+\sqrt{14}=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{14}=\sqrt{7}-1-\left(\sqrt{7}+1\right)+\sqrt{14}=\sqrt{14}-2\)
=> A = \(\sqrt{7}-\sqrt{2}\)
* B là 6,5 hay 6*5 vậy bạn
nếu 6,5 thì : B cũng nhân \(\sqrt{2}\) biểu thức trở thành
\(\sqrt{2}B=\sqrt{13+2\sqrt{12}}+\sqrt{13-2\sqrt{12}}+4\sqrt{3}=\sqrt{\left(1+\sqrt{12}\right)^2}+\sqrt{\left(\sqrt{12}-1\right)^2}+4\sqrt{3}=1+\sqrt{12}+\sqrt{12}-1+4\sqrt{3}=4\sqrt{3}+4\sqrt{3}=8\sqrt{3}\)
=> B = \(\dfrac{8\sqrt{3}}{\sqrt{2}}=4\sqrt{6}\)
nếu 6*5 thì : bạn tách hai căn đầu thành một biểu thức rồi bình phương lên rồi giải , sau đó trục căn , biểu thức luôn dương nhé , mấy bài này nếu không thể tách thì làm cách này cũng được
* C thì mik chỉ bít pt được nhiu đây thôi , bạn thông cảm nhé\(\sqrt{29-6\sqrt{20}}=\sqrt{\left(\sqrt{20}-3\right)^2}=\sqrt{20}+3=2\sqrt{5}-3\)
* D = \(\sqrt{13-2\cdot2\sqrt{2}\cdot\sqrt{5}}-\sqrt{53+2\cdot2\sqrt{2}\cdot3\sqrt{5}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{2}+3\sqrt{5}\right)^2}=2\sqrt{2}-\sqrt{5}-2\sqrt{2}-3\sqrt{5}=-4\sqrt{5}\)
Câu C có sai đề ko? Tui sửa đây!
Ta có: \(C=\sqrt{46+6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
=> \(C=\sqrt{45+2.3\sqrt{5}+1}-\sqrt{20-2.3.2\sqrt{5}+9}\)
=> \(C=\sqrt{\left(3\sqrt{5}+1\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
=> \(C=\left|3\sqrt{5}+1\right|-\left|2\sqrt{5}-3\right|\)
=> \(C=3\sqrt{5}+1-2\sqrt{5}+3=4+\sqrt{5}\)