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\(a,\sqrt{\sqrt{17+12\sqrt{2}}}\)
\(=\sqrt{\sqrt{8+12\sqrt{2}+9}}\)
\(=\sqrt{\sqrt{\left[2\sqrt{2}+3\right]^2}}\)
\(=\sqrt{2\sqrt{2}+3}\)
\(=\sqrt{1+2\sqrt{2}+2}\)
\(=\sqrt{\left[1+\sqrt{2}\right]^2}\)
\(=1+\sqrt{2}\)
\(b,\sqrt{4+2\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)
\(=\sqrt{3+2\sqrt{3}+1}-\sqrt{12-12\sqrt{3}+9}\)
\(=\sqrt{\left[1+\sqrt{3}\right]^2}-\sqrt{\left[2\sqrt{3}-3\right]^2}\)
\(=\left(1+\sqrt{3}\right)-\left(2\sqrt{3}-3\right)\)
\(=1+\sqrt{3}-2\sqrt{3}+3\)
\(=4-\sqrt{3}\)
chúc bn học tốt
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
A = 2 + 2
A = 4
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(B=\sqrt{2}+3+3-\sqrt{2}\)
B = 3 + 3
B = 6
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(C=3+2\sqrt{2}+3-2\sqrt{2}\)
C = 3 + 3
C = 6
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(D=\sqrt{5}+2-\sqrt{5}+2\)
D = 2 + 2
D = 4
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(E=\sqrt{5}+1-\sqrt{5}+1\)
E = 1 + 1
E = 2
\(\sqrt{6-4\sqrt{2}}\)\(+\sqrt{22-12\sqrt{2}}\)
\(=\sqrt{4-4\sqrt{2}+2}\)\(+\sqrt{18-12\sqrt{2}+4}\)
\(=\sqrt{\left(2-\sqrt{2}\right)^2}\)\(+\sqrt{\left(2-3\sqrt{2}\right)^2}\)
\(=2-\sqrt{2}+3\sqrt{2}-2\)
\(=\left(2-2\right)+\left(-\sqrt{2}+3\sqrt{2}\right)\)
\(=0+2\sqrt{2}\)\(=2\sqrt{2}\)
\(\sqrt{17-12\sqrt{2}}\)\(+\sqrt{9+4\sqrt{2}}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}\)\(+\sqrt{\left(2\sqrt{2}+1\right)^2}\)
\(=\left|3-2\sqrt{2}\right|\)\(+\left|2\sqrt{2}+1\right|\)
\(=3-2\sqrt{2}\)\(+2\sqrt{2}+1\)
\(=\left(3+1\right)+\left(-2\sqrt{2}+2\sqrt{2}\right)\)
\(=4+0=4\)
a) \(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2.4\sqrt{2}+2}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}=\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}\)\(=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}=\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}=\sqrt{6-2\left(1+\sqrt{3}\right)}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=1+\sqrt{3}\)
b) Tương tự a) đ/s =5
câu đầu bạn xem lại đề đi nha
các phần còn lại
b)B=\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)=\(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)=\(\sqrt{7}-1-\left(\sqrt{7}+1\right)=-2\)
c)tính từng căn nha
\(\sqrt{13-4\sqrt{3}}=\sqrt{12-2\sqrt{12}+1}=\sqrt{\left(\sqrt{12}-1\right)^2}=\sqrt{12}-1=2\sqrt{3}-1\)
\(\sqrt{22-12\sqrt{2}}=\sqrt{18-4\sqrt{18}+4}=\sqrt{\left(\sqrt{18}-2\right)^2}=\sqrt{18}-2=3\sqrt{2}-3\)
\(\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}=3\sqrt{2}-2\sqrt{3}\)
thay vào tính C đc C=2
d)có \(\sqrt{9+4\sqrt{2}}=\sqrt{8+2\sqrt{8}+1}=\sqrt{\left(\sqrt{8}+1\right)^2}=\sqrt{8}+1\)\(\Rightarrow6\sqrt{2+\sqrt{9+4\sqrt{2}}}=6\sqrt{2+\sqrt{8}+1}=6\sqrt{2+2\sqrt{2}+1}\)
=\(6\sqrt{\left(\sqrt{2}+1\right)^2}=6\left(\sqrt{2}+1\right)=6\sqrt{2}+6\)\(\Rightarrow D=\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2}-6}=\sqrt{11-6\sqrt{2}}=\sqrt{9-6\sqrt{2}+2}\)
=\(\sqrt{\left(3-\sqrt{2}\right)^2}=3-\sqrt{2}\)
a) \(\sqrt{\sqrt{17+12\sqrt{2}}}\)
\(=\sqrt{\sqrt{\left(2\sqrt{2}\right)^2+2\cdot2\sqrt{2}\cdot3+9}}\)
\(=\sqrt{\sqrt{\left(2\sqrt{2}+3\right)^2}}\)
\(=\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{2+2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\sqrt{2}+1\)
b) \(\sqrt{4+2\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)
\(=\sqrt{3+2\sqrt{3}+1}-\sqrt{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2}\cdot3+9}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=\sqrt{3}+1-3+2\sqrt{2}\)
\(=\sqrt{3}+2\sqrt{2}-2\)