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a)
\(\begin{matrix}N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\\^-M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\\\overline{N\left(x\right)-M\left(x\right)=-3x^4+18x^3-2x^2-4x-1}\end{matrix}\)
b)
\(\begin{matrix}M\left(x\right)=-x^4-9x^3+x^2+9x+\dfrac{4}{3}\\^+N\left(x\right)=-4x^4+9x^3-x^2+5x+\dfrac{1}{3}\\\overline{M\left(x\right)+N\left(x\right)=-5x^4+14x+\dfrac{5}{3}}\end{matrix}\)
Xét số hạng tổng quát \(\frac{n+1}{n}=1+\frac{1}{n}\) . Vì \(0
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
A = 1.2.4 + 2.3.5 + ... + n(n+1)(n+3)
A = 1.2.(3+1) + 2.3.(4+1) + ... + n(n+1)[(n+2)+1]
A = [1.2.3 + 2.3.4 + ... + n(n+1)(n+2)] + [1.2 + 2.3 + ... + n(n+1)]
Đặt B = 1.2.3 + 2.3.4 + ... + n(n+1)(n+2)
4B = 1.2.3.(4-0) + 2.3.4.(5-1) + ... + n(n+1)(n+2)[(n+3)-(n-1)]
4B = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + n(n+1)(n+2)(n+3) - (n-1)n(n+1)(n+2)
4B = n(n+1)(n+2)(n+3)
B = \(\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)
Đặt C = 1.2 + 2.3 + ... + n(n+1)
3C = 1.2.(3-0) + 2.3.(4-1) + ... + n(n+1)[(n+2)-(n-1)]
3C = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + ... + n(n+1)(n+2) - (n-1)n(n+1)
3C = n(n+1)(n+2)
C = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
A = B + C = \(n\left(n+1\right)\left(n+2\right)\left(\frac{n+3}{4}+\frac{1}{3}\right)\)
\(=n\left(n+1\right)\left(n+2\right)\frac{3n+13}{12}\)
tại sao bạn lại rút gọn được A = n(n+1)(n+2)(n+3/4+1/3) vậy