Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(T\left(-2\right)=a_0-2a_1+2^2a_2-...-2^{29}a_{29}+2^{30}a_{30}=a_0+H=\left(1+4\right)^{15}\)
\(\Leftrightarrow1+H=5^{15}\)
\(\Leftrightarrow H=5^{15}-1\)
x = \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
x = \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.3+3^2}\) - \(\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.3+3^2}\)
x = \(\sqrt{\left(2\sqrt{5}+3\right)^2}\) - \(\sqrt{\left(2\sqrt{5}-3\right)^2}\)
x = \(|\) \(2\sqrt{5}+3\) \(|\) - \(|\) \(2\sqrt{5}-3\) \(|\)
x = \(\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)\)
x = \(2\sqrt{5}+3-2\sqrt{5}+3\) = 6
\(\sqrt{29+12\sqrt{5}}+\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{5}+3\right)^2}+\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=\left|2\sqrt{5}+3\right|+\left|2\sqrt{5}-3\right|\)
\(=2\sqrt{5}+3+2\sqrt{5}-3=4\sqrt{5}\)
\(\sqrt{29+12\sqrt{5}}+\sqrt{29-12\sqrt{5}}\)
\(\sqrt{29+2.2\sqrt{5}.3}+\sqrt{29-2.2\sqrt{5}.3}\)
\(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.3+3^2}+\sqrt{\left(2\sqrt{5}\right)-2.2\sqrt{5}.3+3^2}\)
\(\sqrt{\left(2\sqrt{5}+3\right)^2}+\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(\left|2\sqrt{5}+3\right|+\left|2\sqrt{5}-3\right|\)
\(2\sqrt{5}+3+2\sqrt{5}-3\)
\(4\sqrt{5}\)
\(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\)
\(=\sqrt[3]{27+27\sqrt{2}+18+2\sqrt{2}}+\sqrt[3]{27-27\sqrt{2}+18-2\sqrt{2}}\)
\(=\sqrt[3]{\left(3+\sqrt{2}\right)^3}+\sqrt[3]{\left(3-\sqrt{2}\right)^3}\)
\(=3+\sqrt{2}+3-\sqrt{2}=6\)
\(F=\sqrt[3]{27-27\sqrt{2}+18-2\sqrt{2}}\)\(+\sqrt[3]{27+27\sqrt{2}+18+2\sqrt{2}}\)
\(F=\sqrt[3]{\left(3-\sqrt{2}\right)^3}+\sqrt[3]{\left(3+\sqrt{2}\right)^3}\)
\(F=3+\sqrt{2}+3-\sqrt{2}=6\)
(-1075) - (29 - 1075) = - 1075 - 29 + 1075 = - 29
(-1075)-(29-1075)
=-1075-29+1075
=(-1075+1075)-29=0-29=-29