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\(a,\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\left(\frac{3}{7}\right)^2\right)^{10}.\frac{3}{7}:\left(\frac{9}{49}\right)^6=\left(\frac{9}{49}\right)^{10}.\frac{3}{7}:\left(\frac{9}{49}\right)^6\)
\(=\left(\left(\frac{9}{49}\right)^{10}:\left(\frac{9}{49}\right)^6\right).\frac{3}{7}=\left(\frac{9}{49}\right)^{10-6}.\frac{3}{7}=\left(\frac{9}{49}\right)^4.\frac{3}{7}=\left(\left(\frac{3}{7}\right)^2\right)^4.\frac{3}{7}\)
\(=\left(\frac{3}{2}\right)^8.\frac{3}{7}=\left(\frac{3}{2}\right)^9\)
\(b,3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}=2+\left(\frac{1}{2}\right)^3=2+\frac{1}{6}=2\frac{1}{6}\)
\(\dfrac{1}{2019^2}-\dfrac{1}{2020^2}=\dfrac{2020^2-2019^2}{2019^2\cdot2020^2}\\ =\dfrac{\left(2020-2019\right)\left(2020+2019\right)}{2019^2\cdot2020^2}=\dfrac{4039}{2019^2\cdot2020^2}\)
Em kiểm tra lại đề bài nhé! Tham khảo link:
Câu hỏi của Phan Thúy Vy - Toán lớp 7 - Học toán với OnlineMath
M = 22010-(22009 + 22008+....+21+20
Đặt A =( 22009+22008+...21 +20)
Suy ra 2A = 22010+22009+22008+...22+2
Suy ra 2A-A = ( 22010+22009+22008+...+22+2) - (22009+ 22008+...+21+20)
Suy ra A= 22010-20
Suy ra M = 22010-A=22010 - 22010+20=1
Vậy M=1
Đúng nha
a: 26⋅33=(22⋅3)3=12326⋅33=(22⋅3)3=123
b: 64⋅83=24⋅34⋅29=213⋅3464⋅83=24⋅34⋅29=213⋅34
c: 16⋅81=36216⋅81=362
d: 254⋅28=1004
M+N=(3/2x6-7x+4x^5+2,5x^2)+(-3x^6+1/2^5-13/2x^2+4x)
M+N=3/2x6-7x+4x^5+2,5x^2+-3x^6+1/2^5-13/2x^2+4x
= (3/2x^6-3x^6)+(7x+4x)+(4x^5+1/2^5)+(2,5x^2-13/2x^2)
=-1,5x^6+11x+4,5x^5-4x^2
M-N=(3/2^6-7x+4x^5+2,5x^2)-(-3x^6+1/2^5-13/2x^2+4x)
=3/2^6-7x+4x^5+2,5x^2+3x^6-1/2^5+13/2x^2-4x
= (3/2x^6+3x^6)+(-7x-4x)+(4x^5-1/2^5)+(2,5x^2+13/2x^2)
= 4,5x^6-11x+3,5x^5+9x^2
N-M=(-3x^6+1/2^5-13/2x^2+4x)-(3/2^6-7x+4x^5+2,5x^2)
= -3x^6+1/2^5-13/2x^2+4x-3/2^6-7x-4x^5-2,5x^2
= (-3x^6-3/2x^6)+(1/2x^5-4x^5)+(-13/2x^2-2,5x^2)+(4x-7x)
= -4,5x^6-3,5x^5-9x^2-3x
sorry nãy sai
\(2:\left(-\frac{1}{6}\right)^3=2\cdot\left(-216\right)=-432\)
\(2:\left(-\frac{1}{6}\right)^3=2:\left(-216\right)=-\frac{2}{216}=-\frac{1}{108}\)