Rút gọn :
\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
rút gọn
\(3.\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right).\left(2^{64}+1\right)\)
3 = 2^2 - 1
Áp dụng HĐT a^2 - b^2
kq : 2^128 - 1
rút gọn biểu thức
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
[Toán 8] Rút gọn $ (3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)$ | HOCMAI Forum - Cộng đồng học sinh Việt Nam
\(A=\left(x^2-x+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\left(x^{16}-x^8+1\right)\left(x^{32}-x^{16}+1\right)\)
Rút gọn A dưới dạng phân thức
-x^61+5*x^60+x^59-5*x^58-x^55+5*x^54+x^53-5*x^52-x^49+5*x^48+x^47-5*x^46x^43+5*x^42+x^41-5*x^40-x^37+5*x^36+x^35-5*x^34-x^49+5*x^48+x^47-5*x^46x^43+5*x^42+x^41-5*x^40-x^37+5*x^36+x^35-5*x^34-x^31+5*x^30+x^27-5*x^26-x^25+5*x^24+x^21-5*x^20-x^19+5*x^18+x^15-5*x^14-x^13+5*x^12+x^9-5*x^8-x^7+5*x^6+x^3-5*x^2-x+5
Rút gọn biểu thức:
\(\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\left(x^{32}+1\right)\)
Rút gọn:
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(3(2^2+1)(2^4+1)(2^8+1)(2^16 +1) \)
\( = (2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)\)
\( = (2^4-1)(2^4+1)(2^8+1)(2^16+1) \)
\( = (2^8-1)(2^8+1)(2^16+1) \)
\(= (2^16 -1)(2^16+1) = 2^32 - 1\)
3(2^2 +1) (2^4 +1 ) (2^8 +1) (2^16 +1)
= (4-1)(2^2+1)(2^4 +1)(2^8+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)(2^16+1)
= 2^23 -1
Chúc bạn học tốt
Bài 1 : rút gọn các biểu thức sau
A = \(\left(3x+1\right)^2-2\left(3x+1\right)\left(5x+5\right)+\left(5x+5\right)^2\)
B = \(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-b-a\right)^2\)
C = \(\left(3x+1\right)\left(3x^2+1\right)\left(3x^4+1\right)\left(3x^8+1\right)\left(3x^{16}+1\right)\left(3x^{32}+1\right)\)
câu a là hằng đẳng thức luôn
A=(2x+4)^2
B khai triển tung tóe ra thì phần sau triệt tiêu hết còn 4(a^2+b^2+c^2)
câu c cảm giác sai đề vì mấy câu này phải là (3x)^ ms ra hdt chứ nhỉ
Rút gọn biểu thức:
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
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)(2^16+1)
=2^32-1
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)(2^16+1)
=2^32-1
chúc bn hok tốt @_@
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) \(\left(2+1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)-2^{64}\)
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
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.
Rút gọn biểu thức:
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
a) (6x + 1)2 + (6x - 1)2 - 2(1 + 6x)(6x - 1)
= (6x + 1 - 6x + 1)2 = 4
b) 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