Thu gọn
A=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
Những câu hỏi liên quan
Thu gọn
A=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
Rút gọn biểu thức:
A= (2+1)(22+1)(24+1)(28+1)(216+1)(232+1)
Giải:
\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow A=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow A=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow A=2^{64}-1\)
Vậy ...
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Rút gọn biểu thức
3(2^2 +1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
Ta có : $3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$
$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)$
$=(2^{16}-1)(2^{16}+1)(2^{32}+1)$
$=(2^{32}-1)(2^{32}+1)$
$=2^{64}-1$
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3.(22+1)(24+1)(28+1)(216+1)(232+1)
= (22-1)(22+1)(24+1)(28+1)(216+1)(232+1)
= (24-1)(24+1)(28+1)(216+1)(232+1)
= (28-1)(28+1)(216+1)(232+1)
= (216-1)(216+1)(232+1)
= (232-1)(232+1)
= 264 - 1
Gợi ý: Áp dụng hằng đẳng thức số 3 trong 7 hằng đẳng thức đáng nhớ
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Thu gọn biểu thức sau :
a) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot\left(\frac{1}{16}+1\right)\cdot\cdot\cdot\left(1+\frac{1}{2^{2n}}\right)\)
b) \(\left(2+1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)-2^{64}\)
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
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a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)
Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)
\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)
b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.
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Rút gọn biểu thức:
a) (x-2)(x^2-2x+4)(x+2)(x^2+2x+4)
b)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
(x^2+x+1)(x^4+x^2+1)(x^8+x^4+1)(x^16+x^8+1)(x^32+x^16+1) rút gọn zùm mình với
https://www.youtube.com/watch?v=cFZDEMTQQCs
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rút gọn biểu thức : A = 3( 22 + 1)( 24 + 1)( 28 + 1)( 216 + 1)( 232 + 1)
\(A=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(=2^{64}-1\)
A = 3( 22 + 1 )( 24 + 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 22 - 1 )( 22 + 1 )( 24 + 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 24 - 1 )( 24 + 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 28 - 1 )( 28 + 1 )( 216 + 1 )( 232 + 1 )
= ( 216 - 1 )( 216 + 1 )( 232 + 1 )
= ( 232 - 1 )( 232 + 1 )
= 264 - 1
Xem thêm câu trả lời
Rút gọn biểu thức
a/(a-b+c)^2-(b-c)^2+2ab-2ac
b/(3x+1)^2-2(3x+1)(3x+5)+(3x+5)^2
c/ (2+1)(2^2+1)(2^4+1)(2^8+1)...(2^64+1)
d/ (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\) =\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\) =\(a^2\) b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(\left(3x+3\right)^2-4\right)+\left(3x+5\right)^2\) =\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\) =25 c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\) =\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)....\left(2^{64}+1\right)\) =\(\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)\) =... =\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\) \)
d)Tương tự
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Rút gọn biểu thức
a/(a-b+c)^2-(b-c)^2+2ab-2ac
b/(3x+1)^2-2(3x+1)(3x+5)+(3x+5)^2
c/ (2+1)(2^2+1)(2^4+1)(2^8+1)...(2^64+1)
d/ (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
=\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\)
=\(a^2\)
b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
=\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\)
=\(\left(3x+1\right)^2-2\left(\left(3x+3\right)^2-4\right)+\left(3x+5\right)^2\)
=\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\)
=25
c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\)
=\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)....\left(2^{64}+1\right)\)
=\(\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)\)
=...
=\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\)
d)Tương tự
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