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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 :
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+....+n}\right)\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(=\frac{\left[1.2.3....\left(n-1\right)\right]\left[4.5.6.....\left(n+2\right)\right]}{\left(2.3.4......n\right)\left[3.4.5.....\left(n+1\right)\right]}\)
\(=\frac{n+2}{n.3}=\frac{n+2}{3n}\)
Bài 1.
a. \(3^4.5^4-\left(15^2+1\right)\left(15^2-1\right)=15^4-\left(15^4-1\right)=1\)
b. \(x=11\Rightarrow x+1=12\)
Từ đây, ta có: \(x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+111=-x+111=-11+111=100\)
Bài 2.
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Bài 2
a. (x-2y)2 =2x-4y
b. (2x^2 +3)2 =4x^2+6
c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)
d. (2x-1)3 = 6x-3
Xin lỗi mik chỉ lm ổn bài 2 thôi!
a: \(A=\left(1-\frac49\right)\left(1-\frac{4}{25}\right)\left(1-\frac{4}{49}\right)\cdot\ldots\cdot\left(1-\frac{4}{\left(2n+1\right)^2}\right)\)
\(=\left(1-\frac23\right)\cdot\left(1-\frac25\right)\cdot\ldots\cdot\left(1-\frac{2}{2n+1}\right)\left(1+\frac23\right)\left(1+\frac25\right)\cdot\ldots\cdot\left(1+\frac{2}{2n+1}\right)\)
\(=\frac13\cdot\frac35\cdot\ldots\cdot\frac{2n-1}{2n+1}\cdot\frac53\cdot\frac75\cdot\ldots\cdot\frac{2n+3}{2n+1}\)
\(=\frac{1}{2n+1}\cdot\frac{2n+3}{3}=\frac{2n+3}{3\left(2n+1\right)}\)
b: Ta có công thức tổng quát:
\(1+\frac{1}{n^2-1}\)
\(=\frac{n^2-1+1}{n^2-1}=\frac{n^2}{n^2-1}=\frac{n\cdot n}{\left(n-1\right)\left(n+1\right)}\)
\(B=\left(1+\frac13\right)\left(1+\frac18\right)\cdot\ldots\left(1+\frac{1}{n^2-1}\right)\)
\(=\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\cdot\ldots\cdot\left(1+\frac{1}{n^2-1}\right)\)
\(=\frac{2\cdot2}{\left(2-1\right)\left(2+1\right)}\cdot\frac{3\cdot3}{\left(3-1\right)\left(3+1\right)}\cdot\ldots\cdot\frac{n\cdot n}{\left(n-1\right)\left(n+1\right)}\)
\(=\frac{2\cdot3\cdot\ldots\cdot n}{1\cdot2\cdot\ldots\cdot\left(n-1\right)}\cdot\frac{2\cdot3\cdot\ldots\cdot n}{3\cdot4\cdot\ldots\cdot\left(n+1\right)}=\frac{n}{1}\cdot\frac{2}{n+1}=\frac{2n}{n+1}\)
c: Ta có công thức tổng quát:
\(1-\frac{1}{1+2+\cdots+n}\)
\(=1-\frac{1}{\frac{n\left(n+1\right)}{2}}\)
\(=1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
\(C=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1-\frac{1}{1+2+\cdots+n}\right)\)
\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
\(=\frac{4\cdot5\cdot\ldots\cdot\left(n+2\right)}{3\cdot4\cdot\ldots\cdot\left(n+1\right)}\cdot\frac{1\cdot2\cdot\ldots\cdot\left(n-1\right)}{2\cdot3\cdot\ldots\cdot n}=\frac{n+2}{3}\cdot\frac{1}{n}=\frac{n+2}{3n}\)