\(\sqrt{17-3\sqrt{32}}+\sqrt{17-3\sqrt{32}}=2\sqrt{17-3\sqrt{32}}\)
\(=\sqrt{4\left(17-3\sqrt{32}\right)}=\sqrt{68-12\sqrt{32}}=\sqrt{36-12\sqrt{32}+32}\)
\(=\sqrt{6^2-2.6.\sqrt{32}+\left(\sqrt{32}\right)^2}=\sqrt{\left(6-\sqrt{32}\right)^2}=\left|6-\sqrt{32}\right|\)
\(=6-\sqrt{32}=6-4\sqrt{2}\)
Tính thu gọn :
a , \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
b , \(\sqrt{27+12\sqrt{5}}-\sqrt{27-12\sqrt{5}}\)
c , \(\sqrt{15-6\sqrt{6}}+\sqrt{15+\sqrt{6\sqrt{6}}}\)
d , \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
e , \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
f , \(\sqrt{5+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
1.Rút gọn các BT sau:
a) 2\(\sqrt{3}\) +\(\sqrt{\left(2-\sqrt{3}\right)^2}\)
b)\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) +\(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
c) \(\sqrt{17-2\sqrt{32}}\) +\(\sqrt{17+2\sqrt{32}}\)
tính A= \(\sqrt[3]{3-\sqrt{17}}+\sqrt[3]{3-\sqrt{17}}\)
Tính GTBT: \(M=\left(x-y\right)^3+3\left(x-y\right)\left(xy+1\right)\) biết
\(x=\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\)
\(y=\sqrt[3]{17+12\sqrt{2}}-\sqrt[3]{17-12\sqrt{2}}\)
Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
Tính giá trị biểu thức:
\(A=\sqrt{8}+\sqrt{18}-\sqrt{32}\)
\(B=2\sqrt{9}+3\sqrt{36}-\sqrt{64}\)
tính
\(2\sqrt{20}-\sqrt{45}+3\sqrt{18}+3\sqrt{32}-\sqrt{50}\)\(2\sqrt{20}-\sqrt{45}+3\sqrt{18}+3\sqrt{32}-\sqrt{50}\)
1) Thực hiện phép tính
\(\sqrt{50}-3\sqrt{8}+\sqrt{32}\)
2) Giải các phương trình sau:
a)\(\sqrt{x^2-4x+4}=1\)
b)\(\sqrt{x^2-3x}-\sqrt{x-3}=0\)
Tính \(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
