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\(\frac{\left(2n+1\right)^3+n^3}{\left(n+1\right)^3-n^3}=\frac{\left(3n+1\right)\left(3n^2+3n+1\right)}{3n^2+3n+1}=3n+1\)
\(\Rightarrow A=\left(3.1+1\right)+\left(3.2+1\right)+...+\left(3.20+1\right)\)
\(=3\left(1+2+...+20\right)+20\)
\(=\frac{3.20.21}{2}+20=...\)
a)Nhận xét
\(\dfrac{n^3+1}{n^3-1}=\dfrac{\left(n+1\right)\left(n^2-n+1\right)}{\left(n-1\right)\left(n^2+n+1\right)}=\dfrac{\left(n+1\right)\left[\left(n-0,5\right)^2+0;75\right]}{\left(n-1\right)\left[\left(n+0,5\right)^2+0,75\right]}\)
Áp dụng công thức trên:
\(A=\dfrac{2^3+1}{2^3-1}.\dfrac{3^3+1}{3^3-1}....\dfrac{9^3+1}{9^3-1}\)
\(=\dfrac{\left(2+1\right)\left[\left(2-0,5\right)^2+0,75\right]}{\left(2-1\right)\left[\left(2+0,5\right)^2+0,75\right]}.\dfrac{\left(3+1\right)\left[\left(3-0,5\right)^2+0,75\right]}{\left(3-1\right)\left[\left(3+0,5\right)^2+0,75\right]}...\dfrac{\left(9+1\right)\left[\left(9-0,5\right)^2+0,75\right]}{\left(9-1\right)\left[\left(9+0,5\right)^2+0,75\right]}\)
\(=\dfrac{3\left(1,5^2+0,75\right)}{\left(2,5^2+0,75\right)}.\dfrac{4\left(2,5^2+0,75\right)}{2\left(3,5^2+0,75\right)}...\dfrac{10\left(8,5^2+0,75\right)}{8\left(9,5^2+0,75\right)}\)
\(=\dfrac{3.4....10}{1.2.....8}.\dfrac{1,5^2+0,75}{9,5^2+0,75}\)
\(=\dfrac{9.10}{2}.\dfrac{3}{91}\)
\(=\dfrac{3}{2}.\dfrac{90}{91}< \dfrac{3}{2}\)
\(\Rightarrowđpcm\)
b) Làm tương tự
Phần a thành nhân tử sẵn rồi bạn:)
b,\(x^6-9x^3+8=x^6-x^3-8x^3+8\)
\(=x^3\left(x^3-1\right)-8\left(x^2-1\right)\)
\(=x^3\left(x-1\right)\left(x^2+x+1\right)-8\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^4-x^3-8x-1\right)\)
a+b+c+d=0
=>a+b=-(c+d)
=> (a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d)
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d))
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (đpcm)
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