B = \(\left(5^2+1\right)\)\(\left(5^4+1\right)\)\(\left(5^8+1\right)\)\(\left(5^{16}+1\right)\) \(\left(5^{32}+1\right)\)
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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
\(B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=5^{32}-1< 5^{32}\)
Vậy \(B< A\)
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
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
Đặt :
\(A=\)\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2A=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^{16}+1\right)\left(5^{16}-1\right)\)
\(=5^{32}-1\)
\(\Leftrightarrow A=\frac{5^{32}-1}{2}\)
a)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\left(\dfrac{4}{1}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\left(\dfrac{20}{5}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\dfrac{8}{5}.\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=1-\dfrac{2}{5}.\dfrac{25}{-4}\)
\(=1-\dfrac{-5}{2}\)
\(=\dfrac{2}{2}-\dfrac{-5}{2}\)
\(=\dfrac{7}{2}\)
dài quá nên mik sẽ giải lần lượt mỗi câu trả lời là một câu nhá bạn!!
Giải:
a)(4-12/5).25/8-2/5:-4/25
=8/5.25/8-(-5/2)
=5+5/2
=15/2
b)(-5/24+3/4-7/12):(-5/16)
=-1/24:(-5/16)
=2/15
c)6/7+5/4:(-5)-(-1/28).(-2)2
=6/7+(-1/4)-(-1/28).4
=6/7-1/4-(-1/7)
=6/7-1/4+1/7
=(6/7+1/7)-1/4
=1-1/4
=3/4
Chúc bạn học tốt!
\(P=12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{5^{32}-1}{2}\)
\(B=\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(B=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^{32}-1\right)\left(5^{32}+1\right).\frac{1}{24}\)
\(B=\left(5^{64}-1\right).\frac{1}{24}\)
\(B=\frac{5^{64}-1}{24}\)