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a: \(=\dfrac{3^8-3^6+3^6\cdot2^3}{5^3}=\dfrac{3^8-3^6\left(1-2^3\right)}{5^3}=\dfrac{11664}{125}\)
b: \(=\dfrac{7^4\cdot4-7^3}{7^3}=7\cdot4-1=27\)
c: \(=28^4-28^4+1=1\)
d: \(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)+1\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)+1\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)+1\)
\(=3^{32}\)
rút gọn biểu thức
a) (x-2y)2+(x+1)2-(2x+2).(x-2y)
=(x2-4xy+4y2)+(x2+2x+1)-(2x2-4xy+2x-4y)
=x2-4xy+4y2+x2+2x+1-2x2+4xy-2x+4y
=4y2+4y+1
Bài 1 :
\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)
\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)
\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)
Đặt \(a=x^2+6x-7\)
\(A=a\left(a-9\right)+8\)
\(A=a^2-9a+8\)
\(A=a^2-8a-a+8\)
\(A=a\left(a-8\right)-\left(a-8\right)\)
\(A=\left(a-8\right)\left(a-1\right)\)
Thay a vào là xong bạn :)
A= \(\frac{3\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{\left(2^2-1\right)}=2^{32-1}\)
mà B= \(2^{32}\)
=> A<B
\(A=-1^2+2^2-3^2+4^2-...-99^2+100^2\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(=\left(2+1\right)\left(2-1\right)+\left(4+3\right)\left(4-3\right)+...+\left(100+99\right)\left(100-99\right)\)
\(=1+2+3+4+...+99+100\)
\(=\frac{\left(1+100\right)\cdot100}{2}=5050\)
\(C=\left(2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}=\left(2^{64}-1\right)-2^{42}=-1\)
Mk chỉ bt làm câu C thôi tại vì mk chỉ học lớp 7
C=(2+1)(24+1)(28+1)(216+1)(232+1)-264
C=(24-1)(24+1)(28+1)(216+1)(232+1)-264
C=(28-1)(28+1)(216+1)(232+1)-264
C=(216-1)(216+1)(232+1)-264
C=(232-1)(232+1)-264
C=264-1-264
C=-1
\(b.x^4+4x^2-5=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
\(c.x^3-19x-30=x^3-25x+6x-30\)
\(=x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)
A = 12 – 22 + 32 – 42 + … – 20042 + 20052
A = 1 + (32 – 22) + (52 – 42)+ …+ ( 20052 – 20042)
A = 1 + (3 + 2)(3 – 2) + (5 + 4 )(5 – 4) + … + (2005 + 2004)(2005 – 2004)
A = 1 + 2 + 3 + 4 + 5 + … + 2004 + 2005
A = ( 1 + 2002 ). 2005 : 2 = 2011015
b/ B = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = (22 - 1) (22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = ( 24 – 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264
B = …
B =(232 - 1)(232 + 1) – 264
B = 264 – 1 – 264
B = - 1
xin lỗi nha chỗ câu a mình lộn
chỗ (1+2002)x2005:2=2011015 là sai nha
(1+2005)x2005:2= 2011015 là đúng nha