rut gon bieu thuc2√18+3√32-4√8
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\(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)+11\)
\(=\left(3^{64}-1\right)+11=3^{64}+10\)
A = 8.(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1) + 1
A = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1) + 1
A = (34 - 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1) + 1
A = (38 - 1)(38 + 1)(316 + 1)(332 + 1) + 1
A = (316 - 1)(316 + 1)(332 + 1) + 1
A = (332 - 1)(332 + 1) + 1
A = 364 - 1 + 1
A = 364
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=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=...=2^32-1
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\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
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Bài này có rắc rối đâu em?
Thực hiện phép tính trong ngoặc lại là ra dạng (n+1)/n.
1 dãy các số liên tục kéo dài nhân với nhau thì triệt tiêu là xong!
Chúc em học tốt!
\(2\sqrt{18}+3\sqrt{32}-4\sqrt{8}\)
\(=6\sqrt{2}+12\sqrt{2}-8\sqrt{2}\)
\(=10\sqrt{2}\)
\(2\sqrt{18}+3\sqrt{32}-4\sqrt{8}\)
\(=2.3\sqrt{2}+3.4\sqrt{2}-4.2\sqrt{2}\)(\(3\sqrt{32}=3.4\sqrt{2}\)vì \(\sqrt{32}\)= \(2\sqrt{8}\)mà \(\sqrt{8}=2\sqrt{2}\))
\(=6\sqrt{2}+12\sqrt{2}-8\sqrt{2}\)
\(=\sqrt{2}\left(6+12-8\right)\)
\(=10\sqrt{2}\)