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\(=\frac{3.8.15........\left(n-1\right)\left(n+1\right)}{\left(2.3.4.....n\right)\left(2.3.4......n\right)}=\frac{1.3.2.4.3.5..............\left(n-1\right)\left(n+1\right)}{\left(2.3.4.....n\right)\left(2.3.4......n\right)}=\frac{\left(1.2.3......\left(n-1\right)\right)\left(3.4.5......\left(n+1\right)\right)}{\left(2.3.4....n\right)\left(2.3.4.......n\right)}\)
\(=\frac{1.\left(n+1\right)}{n.2}=\frac{n+1}{2n}\)
Bài này mình làm rồi còn gì?
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
\(A=\left(1+\frac{1}{3}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(A=\frac{3+1}{3}.\frac{8+1}{8}.\frac{15+1}{15}...\frac{n^2+2n+1}{n^2+2n}\)
\(A=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{\left(n+1\right)^2}{n^2+2n}\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(A=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(A=\left(n+1\right).\frac{2}{n+2}=\frac{2.\left(n+1\right)}{n+2}\)
Ta có : \(1+\frac{1}{k^2+2k}=\frac{k^2+2k+1}{k^2+2k}=\frac{\left(k+1\right)^2}{k\left(k+2\right)}\) với k thuộc N*
Áp dụng với k = 1,2,3,....,n được :
\(A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{n^2+2n}\right)\)
\(=\frac{\left(1+1\right)^2}{1.\left(1+2\right)}.\frac{\left(2+1\right)^2}{2.\left(2+2\right)}.\frac{\left(3+1\right)^2}{3.\left(3+2\right)}...\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(=\frac{\left[2.3.4...\left(n+1\right)\right]^2}{1.2.3...n.3.4.5...\left(n+2\right)}=\frac{\left[\left(n+1\right)!\right]^2}{n!.\frac{\left(n+2\right)!}{2}}\)
Ta có:
\(\left(1-\dfrac{1}{2^2}\right).\left(1-\dfrac{1}{3^2}\right).\left(1-\dfrac{1}{4^2}\right).....\left(1-\dfrac{1}{n^2}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.....\dfrac{n^2-1}{n^2}\)
\(=\dfrac{3.8.15....\left(n^2-1\right)}{4.9.16.....n^2}\)
\(=\dfrac{1.3.2.4.3.5....\left(n-1\right)\left(n+1\right)}{2.2.3.3.4.4....n.n}\)
\(=\dfrac{\left[1.2.3....\left(n-1\right)\right].\left[3.4.5....\left(n+1\right)\right]}{\left(2.3.4....n\right).\left(2.3.4....n\right)}\)
\(=\dfrac{1.\left(n+1\right)}{n.2}=\dfrac{n+1}{2n}\)
Ta có công thức:
\(1-\dfrac{1}{k^2}=\dfrac{k^2-1^2}{k^2}=\dfrac{\left(k+1\right)\left(k+2\right)}{k^2}\)
Áp dụng công thức trên ta đc:
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)....\left(1-\dfrac{1}{n^2}\right)\)
\(=\dfrac{2^2-1^2}{2^2}.\dfrac{3^2-1^2}{3^2}.\dfrac{4^2-1^2}{4^2}....\dfrac{n^2-1^2}{n^2}\)
\(=\dfrac{\left(2+1\right)\left(2-1\right)}{2.2}.\dfrac{\left(3+1\right)\left(3-1\right)}{3.3}.\dfrac{\left(4+1\right)\left(4-1\right)}{4.4}....\dfrac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}....\dfrac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\dfrac{[1.2.3....\left(n+1\right)].[3.4.5....\left(n-1\right)]}{\left(2.3.4....n\right)\left(2.3.4....n\right)}\)
\(=\left(n+1\right).\dfrac{1}{2n}=\dfrac{n+1}{2n}\)
Chúc bạn học tốt!
A= \(\left(\frac{3}{4}\right)\left(\frac{8}{9}\right)\left(\frac{15}{16}\right)......\left(\frac{\left(n-1\right)\left(n+1\right)}{n.n}\right)\)
\(=\frac{3.8.15....\left(n-1\right)\left(n+1\right)}{\left(2.3.4......n\right)\left(2.3.4.......n\right)}=\frac{1.3.2.4.3.5.......\left(n-1\right)\left(n+1\right)}{\left(2.3.4.....n\right)\left(2.3.4..................n\right)}=\frac{\left(1.2.3.......\left(n-1\right)\right)\left(3.4.5........\left(n+1\right)\right)}{\left(2.3.4.....n\right)\left(2.3.4...........n\right)}\)
\(=\frac{1.\left(n+1\right)}{n.2}=\frac{n+1}{2n}\)
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