Câu hỏi của I lay my love on you - Toán lớp 8 - Học toán với OnlineMath    dv

\(n^5-n=n\left(n^4-1\right)=n\left(n^2-1\right)\left(n^2+1\right)=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)^{\left(1\right)}\)

              \(=n\left(n-1\right)\left(n+1\right)\left(n^2-4+5\right)\)

              \(=n\left(n-1\right)\left(n+1\right)\left[\left(n-2\right)\left(n+2\right)+5\right]\)

               \(=\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)+5n\left(n-1\right)\left(n+1\right)\)

Vì n(n-2)(n+2)(n - 1)(n + 1) chia hết cho 5 

    5n(n - 1)(n + 1) chia hết cho 5

=>  n(n-2)(n+2)(n - 1)(n + 1) + 5n(n - 1)(n + 1) chia hết cho 5 

=> \(n^5-n⋮5\)(2)

 Vì n , (n-1) , (n+1) là 3 số tự nhiên liên tiếp nên luôn tồn tại 1 số chia hết cho 2 và 3 trong 3 số này

Mà ( 2 ; 3 ) = 1

=> n(n+1)(n-1) chia hết cho 2.3=6

=> n(n+1)(n-1)(n²+1 ) chia hết cho 6

Hay n^5 - n chia hết cho 6 (3)

Từ (2) , (3) và ( 5 ; 6 ) = 1

=> n^5 -n chia hết cho 5.6 = 30

Vậy n^5 - n chia hết cho 30

\(a,a^5+b^5=\left(a+b\right)^5\)

\(b,a^4-b^6=\left(a-b\right)^{4-6}\)

\(d,a^6-b^6=\left(a-b\right)^6\)

mk chỉ lm đc 2 phần cuối thôi nha !

\(b,a^4-b^4=\left(a^2+b^2\right)\left(a^2-b^2\right)=\left(a^2+b^2\right)\left(a-b\right)\left(a+b\right)\)

\(c,a^6-b^6=\left(a^3-b^3\right)\left(a^3+b^3\right)\)

\(=\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)

=.= hok tốt!!

\(n^3-n=n\left(n^2-1\right)\)

\(=\left(n-1\right)n\left(n+1\right)⋮2\)

\(\left(n-1\right)n\left(n+1\right)⋮3\)

\(\Rightarrow\left(n-1\right)n\left(n+1\right)⋮2.3=6\)

\(\Rightarrow n^3-n⋮6\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)

\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)

\(a,x^2+y^2-4x-2y+6\)

\(=\left(x^2-4x+4\right)+\left(y^2-2y+1\right)+1\)

\(=\left(x-2\right)^2+\left(y-1\right)^2+1\)

Ta có: \(\left(x-2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2\right)^2+\left(y-1\right)^2+1\ge1\forall x,y\)

Hay: \(x^2+y^2-4x-2y+6\ge1\)

\(b,x^2+4y^2+z^2-4x+4y-8z+25\)

\(=\left(x^2-4x+4\right)+\left(4y^2+4y+1\right)+\left(z^2-8z+16\right)+4\)

\(=\left(x-2\right)^2+\left(2y+1\right)^2+\left(z-4\right)^2+4\)

Vì: \(\left(x-2\right)^2+\left(2y+1\right)^2+\left(z-4\right)^2\ge0\forall x,y,z\)

\(\Rightarrow\left(x-2\right)^2+\left(2y+1\right)^2+\left(z-4\right)^2+4\ge4\forall x,y,z\)

Hay: \(x^2+4y^2+z^2-4x+4y-8z+25\ge4\)

=.= hok tốt !!

Chúc bạn có 1 ngày vui vẻ!!!

\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

\(4x^2-5\left(x-1\right)=-21x-2\)

\(\Leftrightarrow4x^2-5x+5+21x+2=0\)

\(\Leftrightarrow4x^2+16x+7=0\)

\(\Leftrightarrow\left(4x^2+16x+16\right)-9=0\)

\(\Leftrightarrow\left(2x+4\right)^2=9\)

\(\Leftrightarrow\left(2x+4\right)^2=3^2=\left(-3\right)^2\)

TH1 : \(2x+4=3\Rightarrow x=-\frac{1}{2}\)

TH2: \(2x+4=-3\Rightarrow x=-\frac{7}{2}\)

Vậy..............................................................

=.= hok tốt!!

\(4x^2-5x+5=-21x-2\)

\(\Rightarrow4x^2+16x=-7\)

\(\Rightarrow4x\left(x+4\right)=-7\)

Lập bảng