Rút gọn biểu thức:3(22+1)(24+1)(28+1)(216+1)
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rút gọn biểu thức
3)C = (2x-3)^2-(x+4)(2x-1) -(x+3)^2
C = (2x-3)2-(x+4)(2x-1) -(x+3)2
(Chuyển đổi các hằng đẳng thức)
= (4x2-12x+9)-(2x2-x+8x-4)-(x2+6x+9)
= 4x2-12x+9-2x2+x-8x+4-x2-6x-9
(Ta thu gọn các hạng tử đồng dạng với nhau)
= x2-25x-14
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rút gọn biểu thức :
3(2^2+1)(2^4+1)(2^8+1)(2^16+1)
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
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3.(22+1).(24+1).(28+1).(216+1)
=(22-1).(22+1).(24+1).(28+1).(216+1)
=(24-1).(24+1).(28+1).(216+1)
=(28-1).(28+1).(216+1)
=(216-1).(216+1)
=232-1
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Rút gọn biểu thức
3(2^2 +1)(2^4+1)(2^8+1)(2^16+1)
rút gọn biểu thức
a) A=16^8 -1/(2+1)(2^2+1)(2^4+1)(2^8+1(3^16+1)
b) B=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)/9^16-1
giúp mk vs ah mk đang cần gấp ah
a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)
b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)
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Rút gọn biểu thức sau:
A=3.(2^2+1).(2^4+1).(2^8+2).(2^16+1)
Rút gọn biểu thức:
\(B=(2^2+1).(2^4+1).(2^8+1).(2^{16}+1)\)
\(B=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\frac{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\)
\(=\frac{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\)
\(=\frac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\)
\(=\frac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}\)
\(=\frac{2^{32}-1}{3}\)
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rút gọn biểu thức sau:
\((2^2+1).(2^4+1).(2^8+1).(2^{16}+1)\)
\(A=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=> \(3A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
=> \(A=\frac{2^{32}-1}{3}\)
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Rút gọn biểu thức: 3(22 +1)(24+1)(28+1)(216+1)
Đặt A=3(22 +1)(24+1)(28+1)(216+1)
=(4-1)(22+1)(24+1)(28+1)(216+1)
=[(22-1)(22+1)](24+1)(28+1)(216+1)
=(24-1)(24+1)(28+1)(216+1)
=(28-1)(28+1)(216+1)
=(216-1)(216+1)
=232-1
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3(22 +1)(24+1)(28+1)(216+1) = (22 -1)(22 +1)(24+1)(28+1)(216+1) = (24-1)(24+1)(28+1)(216+1) = (28-1)(28+1)(216+1)
= (216-1)(216+1) = 232-1
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Rút gọn biểu thức
a) ( x^2-1)(x+2)-(x-2)(x^2+2x+4)
b) 3(2^2+1)(2^4+1)(2^8+1)(2^16+1)
Đặt A=3(22 +1)(24+1)(28+1)(216+1)
=(4-1)(2^2+1)(2^4+1)(28+1)(2^16+1)
=[(2^2-1)(2^2+1)](2^4+1)(2^8+1)(2^16+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)
Theo mình ý a bn làm đc
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