thu gọn :
a) \(\left(2+1\right)\left(2^2+1\right).....\left(2^{256}+1\right)-1\)
b) \(24\left(5^2+1\right)\left(5^4+1\right)......\left(5^{32}+1\right)-5^{64}\)
Những câu hỏi liên quan
Thu gọn a):\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right).....\left(2^{32}+1\right)-2^{64}\)
b)\(\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+\frac{5^{128}-3^{128}}{2}\)
các pn júp mk nha, mk tik cho, mk cần ngày mai nộp roài
a) (2+1)(2^2+1)(2^4+1)...(2^32+1)-2^64
=(2+1)(2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64
=(2^2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64
=(2^4-1)(2^4+1)....(2^32+1)-2^64
=......
=(2^32-1)(2^32+1)-2^64
=2^64-1-2^64=-1
b)Đặt A=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)+(5^128-3^128)/2
đặt B=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)
\(2B=\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)
\(2B=\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)
\(2B=\left(5^4-3^4\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)
\(2B=.......\)
2B=(5^64-3^64)(5^64+3^64)
2B=5^128-3^128
B=(5^128-3^128)/2 (thế vào đề bài)
=> A=B+(5^128-3^128)/2=(5^128-3^128)/2+(5^128-3^128)/2=\(\frac{2\left(5^{128}-3^{128}\right)}{2}=\left(5^{128}-3^{128}\right)\)
Đúng 0
Bình luận (0)
a) A = ( 2-1)(2+1)(22+1)...(232+1)-264
=(22-1)(22+1)(24+1)... -264
=....
=264-1-264=1
câu b tương tự nhá
Đúng 0
Bình luận (0)
BT7: Tính
\(3,C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{16}+1\right)\)
\(4,D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(5,E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
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
Đúng 2
Bình luận (1)
3, \(C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)
\(C=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)
\(C=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)
\(C=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(C=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(C=5^{32}-1\)
4, \(D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^4-1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^8-1\right)\left(4^8+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^{16}-1\right)\left(4^{16}+1\right)...\left(4^{64}+1\right)\)
\(D=\left(4^{32}-1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)
\(D=\left(4^{64}-1\right)\left(4^{64}+1\right)\)
\(D=4^{128}-1\)
5, \(E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{256}+1\right)\)
\(E=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{128}+1\right)\left(5^{256}+1\right)\)
\(E=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{256}+1\right)\)
....
\(E=\left(5^{128}-1\right)\left(5^{128}+1\right)\left(5^{256}+1\right)\)
\(E=\left(5^{256}-1\right)\left(5^{256}+1\right)\)
\(E=5^{512}-1\)
Đúng 1
Bình luận (0)
Thu gọn:
a) \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\times...\times\left(2^{32}+1\right)-2^{64}\)
b) \(\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)\times...\times\left(5^{64}+3^{64}\right)+\dfrac{5^{128}-3^{128}}{2}\)
CÁC BẠN GIÚP MIK VS NHÉ !!!!! ĐAG CẦN GẤP
a, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}=2^{64}-1-2^{64}=-1\)
b,\(B=\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+\dfrac{5^{128}-3^{128}}{2}\)
\(=\dfrac{\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)}{2}+\dfrac{5^{128}-3^{128}}{2}\)\(=\dfrac{\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}\)
\(=\dfrac{\left(5^{64}-3^{64}\right)\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}=\dfrac{2.5^{128}}{2}=5^{128}\)
Đúng 0
Bình luận (0)
Tính
a) \(A=1+\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
b) \(B=10^2+8^2+.....+2^2-\left(9^2+7^2+5^2+3^2+1^2\right)\)
\(B=10^2+8^2+...+2^2-\left(9^2+7^2+5^2+3^2+1^2\right)\)
\(B=\left(10^2-9^2\right)+\left(8^2-7^2\right)+...+\left(2^2-1^2\right)\)
\(B=\left(10+9\right)\left(10-9\right)+\left(8+7\right)\left(8-7\right)+...+\left(2-1\right)\left(2+1\right)\)
\(B=19+15+...+3\)
Đến đây dễ rồi. Câu a) đang suy nghĩ
Đúng 0
Bình luận (0)
\(A=1+\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+4\cdot\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
\(4A=4+\left(5^{32}-1\right)\left(5^{32}+1\right)\)
\(4A=4+5^{64}-1\)
\(4A=5^{64}+3\)
\(A=\frac{5^{64}+3}{4}\)
Đúng 0
Bình luận (0)
Tính:Asqrt{20}-10sqrt{dfrac{1}{5}}+sqrt{left(sqrt{5}-1right)^2}B2sqrt{32}+5sqrt{8}-4sqrt{32}Csqrt{left(3-sqrt{2}^2right)}-sqrt{left(1-sqrt{2}right)^2}Dsqrt{left(5-1right)^2}+sqrt{left(sqrt{5}-3right)^2}Eleft(3+dfrac{5-sqrt{5}}{sqrt{5}-1}right)left(3-dfrac{5+sqrt{5}}{sqrt{5}-1}right)Fsqrt{6+2sqrt{5}}-sqrt{9-4sqrt{5}}Gdfrac{3sqrt{2}-2sqrt{3}}{sqrt{3}-sqrt{2}}+dfrac{sqrt{15}-sqrt{5}}{sqrt{3}-1}Hdfrac{10}{sqrt{3}-1}-dfrac{55}{2sqrt{3}+1} help
Đọc tiếp
Tính:
\(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(C=\sqrt{\left(3-\sqrt{2}^2\right)}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(D=\sqrt{\left(5-1\right)^2}+\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(E=\left(3+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3-\dfrac{5+\sqrt{5}}{\sqrt{5}-1}\right)\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(G=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\)
\(H=\dfrac{10}{\sqrt{3}-1}-\dfrac{55}{2\sqrt{3}+1}\)
help
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: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
Đúng 3
Bình luận (0)
SO SÁNH A VÀ B BIẾT :\(A=5^{32}\)
VÀ \(B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=5^{32}-1< 5^{32}\)
Vậy \(B< A\)
Đúng 0
Bình luận (0)
tính
a) left[dfrac{0.8divleft(dfrac{4}{5}cdot1025right)}{0.64-1}+dfrac{left(1.08-dfrac{2}{25}right)divdfrac{4}{7}}{left(6dfrac{5}{7}-3dfrac{1}{4}right)cdot2dfrac{2}{17}}+left(1.2cdot0.5right)divdfrac{4}{5}right]
b) left(0.2right)^{-3}left[left(-dfrac{1}{5}right)^{-2}right]^{-1}+left[left(dfrac{1}{2}right)^{-3}right]^{-2}divleft(2^{-3}right)^{-1}-left(0.175right)^{-2}
c) 2+dfrac{1}{1+dfrac{1}{2+dfrac{1}{1+dfrac{1}{2}}}}
d) dfrac{1}{90}-dfrac{1}{72}-dfrac{1}{56}-dfrac{1}{42}-dfrac{1}{3}
e) l...
Đọc tiếp
tính
a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)
b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)
c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)
e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)
h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
Đúng 0
Bình luận (0)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
Đúng 0
Bình luận (0)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{5}{8}\)
\(=\dfrac{3}{8}+\dfrac{5}{8}\)
\(=1\)
Đúng 0
Bình luận (0)
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
Đúng 0
Bình luận (0)
Tính nhanh:
a) \(1^2-2^2+3^2-4^2+5^2-6^2+...+2011^2-2012^2\)
b) \(10\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)-9^{64}\)