a) ta có : \(A+B=\sqrt[3]{7-5\sqrt{2}}+\sqrt[3]{20+14\sqrt{2}}\)
\(=\sqrt[3]{\left(1-\sqrt{2}\right)^3}+\sqrt{\left(\sqrt{2}+2\right)^3}=1-\sqrt{2}+\sqrt{2}+2=3\)
b) ở đây : https://hoc24.vn/hoi-dap/question/650070.html
a, Ta có: \(A+B=\sqrt[3]{7-5\sqrt{2}}+\sqrt[3]{20+14\sqrt{2}}\)
\(=\sqrt[3]{\left(1-\sqrt{2}\right)^3+\sqrt[3]{\left(\sqrt{2}+2\right)^3}}\)
\(=1-\sqrt{2}+\sqrt{2}+2=1+2=3\)
Vậy ...
\(b,x=\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)
Đặt \(\sqrt{2+\sqrt{3}=t}\) , ta có:
\(x=\sqrt{6-3.t}-\sqrt{2+t}\)
\(\Rightarrow x^2=2+t+3.\left(2-t\right)-2\sqrt{3}\left(2+t\right)\left(2-t\right)\)
\(=8-2t-2\sqrt{3\left(4-t^2\right)}\)
\(=8-2t-2\sqrt{3\left(4-2-\sqrt{3}\right)}\)
\(=8-2t-\sqrt{6}.\sqrt{4-2\sqrt{3}}\)
\(=8-2\sqrt{2+\sqrt{3}}-\sqrt{6}\left(\sqrt{3}-1\right)\)
\(=8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{2}\left(\sqrt{3}+1\right)-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}\)
\(=8-4\sqrt{2}\)
\(\Rightarrow x^2-8=-4\sqrt{2}\)
\(\Rightarrow\left(x^2-8\right)^2=32\)
\(\Rightarrow x^4-16x^2+64=32\)
\(\Rightarrow x^4-16x^2+64-32=0\)
\(\Rightarrow x^4-16x^2+32=0\) (đpcm)
Chúc bạn hok tốt!!!
