Tính A=(1/1-x)+(1/1+x)+(2/1+x^2)+(4/1+x^4)+(8/1+x^8)+(16/1+x^6). 18/08/2021 Bởi Adalynn Tính A=(1/1-x)+(1/1+x)+(2/1+x^2)+(4/1+x^4)+(8/1+x^8)+(16/1+x^6).
Đáp án: (32/1-x^32) Giải thích các bước giải: A =(1/1-x)+(1/1+x)+(2/1+x^2)+(4/1+x^4)+(8/1+x^8)+(16/1+x^6). = (2/1-x^2) + (2/1-x^2) + (4/1+x^4) + (8/1+x^8) + (16/1+x^16) = (4/1-x^4) + (4/1-x^4) + (8/1+x^8) + (16/1+x^16) = (8/1-x^8) + (8/1+x^8) + (16/1+x^16) = (16/1-x^16) + (16/1+x^16) = (32/1-x^32) (Mik làm có j sai sót mong bn thông cảm nha) Chúc bn học tốt Bình luận
Đáp án: \(A = \dfrac{{32}}{{1 – {x^{32}}}}\) Giải thích các bước giải: \(\begin{array}{l}A = \dfrac{1}{{1 – x}} + \dfrac{1}{{1 + x}} + \dfrac{2}{{1 + {x^2}}} + \dfrac{4}{{1 + {x^4}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\A = \dfrac{{1 + x + 1 – x}}{{\left( {1 – x} \right)\left( {1 + x} \right)}} + \dfrac{2}{{1 + {x^2}}} + \dfrac{4}{{1 + {x^4}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\A = \dfrac{2}{{1 – {x^2}}} + \dfrac{2}{{1 + {x^2}}} + \dfrac{4}{{1 + {x^4}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\A = \dfrac{{2 + 2{x^2} + 2 – 2{x^2}}}{{\left( {1 – {x^2}} \right)\left( {1 + {x^2}} \right)}} + \dfrac{4}{{1 + {x^4}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\A = \dfrac{4}{{1 – {x^4}}} + \dfrac{4}{{1 + {x^4}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\A = \dfrac{{4 + 4{x^4} + 4 – 4{x^4}}}{{\left( {1 – {x^4}} \right)\left( {1 + {x^4}} \right)}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\A = \dfrac{8}{{1 – {x^8}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\A = \dfrac{{8 + 8{x^8} + 8 – 8{x^8}}}{{\left( {1 – {x^8}} \right)\left( {1 + {x^8}} \right)}} + \dfrac{{16}}{{1 + {x^{16}}}}\\A = \dfrac{{16}}{{1 – {x^{16}}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\A = \dfrac{{16 + 16{x^{16}} + 16 – 16{x^{16}}}}{{\left( {1 – {x^{16}}} \right)\left( {1 + {x^{16}}} \right)}}\\A = \dfrac{{32}}{{1 – {x^{32}}}}\end{array}\) Bình luận
Đáp án: (32/1-x^32)
Giải thích các bước giải:
A =(1/1-x)+(1/1+x)+(2/1+x^2)+(4/1+x^4)+(8/1+x^8)+(16/1+x^6).
= (2/1-x^2) + (2/1-x^2) + (4/1+x^4) + (8/1+x^8) + (16/1+x^16)
= (4/1-x^4) + (4/1-x^4) + (8/1+x^8) + (16/1+x^16)
= (8/1-x^8) + (8/1+x^8) + (16/1+x^16)
= (16/1-x^16) + (16/1+x^16)
= (32/1-x^32)
(Mik làm có j sai sót mong bn thông cảm nha)
Chúc bn học tốt
Đáp án:
\(A = \dfrac{{32}}{{1 – {x^{32}}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{1}{{1 – x}} + \dfrac{1}{{1 + x}} + \dfrac{2}{{1 + {x^2}}} + \dfrac{4}{{1 + {x^4}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\
A = \dfrac{{1 + x + 1 – x}}{{\left( {1 – x} \right)\left( {1 + x} \right)}} + \dfrac{2}{{1 + {x^2}}} + \dfrac{4}{{1 + {x^4}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\
A = \dfrac{2}{{1 – {x^2}}} + \dfrac{2}{{1 + {x^2}}} + \dfrac{4}{{1 + {x^4}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\
A = \dfrac{{2 + 2{x^2} + 2 – 2{x^2}}}{{\left( {1 – {x^2}} \right)\left( {1 + {x^2}} \right)}} + \dfrac{4}{{1 + {x^4}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\
A = \dfrac{4}{{1 – {x^4}}} + \dfrac{4}{{1 + {x^4}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\
A = \dfrac{{4 + 4{x^4} + 4 – 4{x^4}}}{{\left( {1 – {x^4}} \right)\left( {1 + {x^4}} \right)}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\
A = \dfrac{8}{{1 – {x^8}}} + \dfrac{8}{{1 + {x^8}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\
A = \dfrac{{8 + 8{x^8} + 8 – 8{x^8}}}{{\left( {1 – {x^8}} \right)\left( {1 + {x^8}} \right)}} + \dfrac{{16}}{{1 + {x^{16}}}}\\
A = \dfrac{{16}}{{1 – {x^{16}}}} + \dfrac{{16}}{{1 + {x^{16}}}}\\
A = \dfrac{{16 + 16{x^{16}} + 16 – 16{x^{16}}}}{{\left( {1 – {x^{16}}} \right)\left( {1 + {x^{16}}} \right)}}\\
A = \dfrac{{32}}{{1 – {x^{32}}}}
\end{array}\)