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\(B=10^2+8^2+...+2^2-\left(9^2+7^2+5^2+3^2+1^2\right)\)
\(B=\left(10^2-9^2\right)+\left(8^2-7^2\right)+...+\left(2^2-1^2\right)\)
\(B=\left(10+9\right)\left(10-9\right)+\left(8+7\right)\left(8-7\right)+...+\left(2-1\right)\left(2+1\right)\)
\(B=19+15+...+3\)
Đến đây dễ rồi. Câu a) đang suy nghĩ
\(A=1+\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+4\cdot\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^{32}-1\right)\left(5^{32}+1\right)\)
\(4A=4+5^{64}-1\)
\(4A=5^{64}+3\)
\(A=\frac{5^{64}+3}{4}\)
3: =(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1
4:
D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)
=(4^8-1)(4^8+1)*...*(4^64+1)
=...
=4^128-1
5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)
=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1
=5^256-1+5^256-1
=2*5^256-2
\(M=4.6\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)\(=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)=\left(5^{16}-1\right)\left(5^{16}+1\right)=5^{32}-1\)
Vậy M <N
a)\(\left(2+1\right)\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{256}+1\right)-1\)
Tiếp tục như thế, ta được:
\(=\left(2^{256}-1\right)\left(2^{256}+1\right)-1=2^{512}-1-1=2^{512}-2\)
b) \(24\left(5^2+1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^4-1\right)\left(5^4+1\right)...\left(5^{32}+1\right)-5^{64}\)
Tiếp tục như thế, ta được:
\(=\left(5^{32}-1\right)\left(5^{32}+1\right)-5^{64}=5^{64}-1-5^{64}=-1\)
\(\left(2+1\right).\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(\left(2-1\right).\left(2+1\right).\left(2^2+1\right).....\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right).\left(2^2+1\right)....\left(2^{256}+1\right)-1\)
\(=\left(2^{256}-1\right).\left(2^{256}+1\right)+1=2^{512}+1\)
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
\(B=\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(B=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^{32}-1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^{64}-1\right).\frac{1}{24}\)
\(B=\frac{5^{64}-1}{24}\)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(\frac{1}{2}\left(5^{32}+1\right)=\frac{5^{32}+1}{2}\)
a)
Ta có
a chia 5 dư 4
=> a=5k+4 ( k là số tự nhiên )
\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)
Vì 25k^2 chia hết cho 5
40k chia hết cho 5
16 chia 5 dư 1
=> đpcm
2) Ta có
\(12=\frac{5^2-1}{2}\)
Thay vào biểu thức ta có
\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)}{2}\)
\(\Rightarrow P=\frac{\left[\left(5^2\right)^2-1^2\right]\left[\left(5^2\right)^2+1^2\right]\left(5^8+1\right)}{2}\)
\(\Rightarrow P=\frac{\left[\left(5^4\right)^2-1^2\right]\left[\left(5^4\right)^2+1^2\right]}{2}\)
\(\Rightarrow P=\frac{5^{16}-1}{2}\)
3)
\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+b^3+c^2+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
mk ko biết mk mới học lớp nhỏ thôi . Đó là lớp này nè bn...... tự vào trang của mk coi đi nhé
\(B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=5^{32}-1< 5^{32}\)
Vậy \(B< A\)