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a) \(10^n+1-6\cdot10^n=\left(1-6\right)10^n+1=-5\cdot10^n+1\)
b) \(90\cdot10^n-10^2-2+10^n+1=\left(90-1+1\right)\cdot10^n-2+1=90\cdot10^n-1\)
c) \(2,5\cdot56^n-3=\frac{5}{2}\cdot56^n-3\)
Bài 1:
a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)
\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)
\(=10x^2+10x^2\)
\(=20x^2\)
b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)
\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)
\(=-4x^4+9x^3+4x^2-44x\)
Bài này làm từng câu thôi :
\(A=1+3^1+3^2+.......+3^{2014}+3^{2015}\)
\(\Rightarrow3A=3+3^2+3^3+......+3^{2015}+3^{2016}\)
\(\Rightarrow3A-A=\left(3+3^2+......+3^{2016}\right)-\left(1+3^1+.....+3^{2015}\right)\)
\(\Rightarrow2A=3^{2016}-1\)
\(\Rightarrow A=\frac{3^{2016}-1}{2}\)
A= (937.1 - 4.5) - (-4.5 + 37.1) -100
= (937-20) - 17 -100
= 917- 17-100
=900-100=800
Rút gọn biểu thức A=(1-1/3).(1-1/6).(1-1/10).(1-1/15).....(1-1/253)
Hãy chứng minh A<2/5
Giải
A=(1-1/3).(1-1/6).(1-1/10).(1-1/15).....(1-1/253)
\(=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)...\left(1-\dfrac{1}{253}\right)\\=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{252}{253}\\ =\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot...\cdot\dfrac{504}{506}\\ =\dfrac{1\cdot4}{2\cdot3}\cdot\dfrac{2\cdot5}{3\cdot4}\cdot...\cdot\dfrac{21\cdot24}{22\cdot23}\\ =\dfrac{1\cdot2\cdot3\cdot4^2\cdot5^2\cdot...\cdot21^2\cdot22\cdot23\cdot24}{2\cdot3^2\cdot4^2\cdot...\cdot22^2\cdot23}\\ =\dfrac{1\cdot24}{3\cdot22}=\dfrac{24}{66}< \dfrac{2}{5}\)