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b) \(\dfrac{\left(n+1\right)!-n!}{\left(n+1\right)!+n!}=\dfrac{n!.\left(n+1\right)-n!}{n!\left(n+1\right)+n!}=\dfrac{n!\left(n+1-1\right)}{n!\left(n+1+1\right)}=\dfrac{n}{n+2}\)
a) \(\dfrac{8a^{n+2}+a^{n-1}}{16a^{n+4}+4a^{n+2}+a^n}=\dfrac{8a^{n-1+3}+a^{n-1}}{16a^{n-1+5}+4a^{n-1+3}+a^{n-1+1}}\)
\(=\dfrac{8a^{n-1}.a^3+a^{n-1}}{16a^{n-1}a^5+4a^{n-1}a^3+a^{n-1}a}=\dfrac{a^{n-1}\left(8a^3+1\right)}{a^{n-1}\left(16a^5+4a^3+a\right)}\)
\(=\dfrac{8a^3+1}{16a^5+4a^3+a}\)
a) \(\left(3x^{n+1}-y^{n-1}\right)-3\left(x^{n+1}+5y^{n-1}\right)-4\left(x^{n+1}+2y^{n-1}\right)\)
\(=3x^{n+1}-y^{n-1}-3x^{n+1}-15y^{n-1}+4x^{n+1}+8y^{n-1}\)
\(=-8y^{n-1}+4x^{n+1}\)
b) \(\left(\dfrac{3}{4}x^{n+1}-\dfrac{1}{2}y^n\right)\cdot2xy-\left(\dfrac{2}{3}x^{n+1}-\dfrac{5}{6}y^n\right)\cdot7xy\)
\(=\dfrac{3}{2}x^{n+2}y-xy^{n+1}+\left(-\dfrac{2}{3}x^{n+1}-\dfrac{5}{6}y^n\right)\cdot7xy\)
\(=\dfrac{3}{2}x^{n+2}y-xy^{n+1}-\dfrac{14}{3}x^{n+2}y+\dfrac{35}{6}xy^{n+1}\)
\(=-\dfrac{19}{6}x^{n+2}y+\dfrac{29}{6}xy^{n+1}\)
a)\(\left(3x^{n+1}-y^{n-1}\right)-3\left(x^{n+1}+5y^{n-1}\right)+4\left(x^{n+1}+2y^{n-1}\right)\)
\(=3x^{n+1}-y^{n-1}-3x^{n+1}-15y^{n-1}+4x^{n+1}+8y^{n-1}\)
\(=4x^{n+1}-8y^{n-1}\) \(\left(=4\left(x^{n+1}-2y^{n-1}\right)\right)\)
Với mọi k thuộc N và k > 2 thì ta có :
\(1-\frac{1}{1+2+....+k}=1-\frac{1}{\frac{k\left(k+1\right)}{2}}=1-\frac{2}{k\left(k+1\right)}=\frac{k^2+k-2}{k\left(k+1\right)}=\frac{\left(k+2\right)\left(k-1\right)}{k\left(k+1\right)}\)
Áp dụng vào A ta được :
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+....+n}\right)\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(=\frac{\left[1.2.3....\left(n-1\right)\right]\left[4.5.6.....\left(n+2\right)\right]}{\left(2.3.4......n\right)\left[3.4.5.....\left(n+1\right)\right]}\)
\(=\frac{n+2}{n.3}=\frac{n+2}{3n}\)
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
Đặt \(\left(a-1\right)^2=t\)
Ta có: \(\left(a-1\right)^4-11\left(a-1\right)^2+30\)
\(=t^2-11t+30\)
\(=t\left(t-5\right)-6\left(t-5\right)=\left(t-5\right)\left(t-6\right)\)
\(=\left[\left(a-1\right)^2-5\right]\left[\left(a-1\right)^2-6\right]\)
\(=\left(a^2-2a-4\right)\left(a^2-2a-5\right)\)
Đặt \(a^2-2a=k\)
Ta có: \(3\left(a-1\right)^4-18\left(a^2-2a\right)-3\)
\(=3\left(a^2-2a+1\right)^2-18\left(a^2-2a\right)-3\)
\(=3\left(k+1\right)^2-18k-3\)
\(=3k^2+6k+3-18k-3\)
\(=3k^2-12k=3k\left(k-4\right)\)
\(=3\left(a^2-2a\right)\left(a^2-2a-4\right)\)(Ở đây bạn ghi thêm điều kiện nhé)
Khi đó: \(N=\frac{\left(a^2-2a-4\right)\left(a^2-2a-5\right)}{3\left(a^2-2a\right)\left(a^2-2a-4\right)}=\frac{a^2-2a-5}{3\left(a^2-2a\right)}\)
\(A=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(A=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{2n+1}{n^2\cdot\left(n^2+2n+1\right)}\)
\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)
\(A=1-\dfrac{1}{n^2+2n+1}\)
\(A=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)
a) \(\frac{\left(n+1\right)!}{n!\left(n+2\right)}=\frac{n!\left(n+1\right)}{n!\left(n+2\right)}=\frac{n+1}{n+2}\)
b)\(\frac{n!}{\left(n+1\right)!-n!}=\frac{n!}{n!\left(n+1\right)-n!}=\frac{n!}{n!\left(n+1-1\right)}=\frac{1}{n}\)
c)\(\frac{\left(n+1\right)!-\left(n+2\right)!}{\left(n+1\right)!+\left(n+2\right)!}=\frac{n!\left(n+1\right)-n!\left(n+1\right)\left(n+2\right)}{n!\left(n+1\right)+n!\left(n+1\right)\left(n+2\right)}=\frac{n!\left(n+1\right)\left(1-n-2\right)}{n!\left(n+1\right)\left(1+n+2\right)}=\frac{-n-1}{n+3}\)
( Kí hiệu n!=1.2.3.4...n)