A=\(\frac{\left(1+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)

A=\(\frac{\left(1+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).0}{1-2+3-4+...+99-100}\)

A= 0

KẾT QUẢ ĐÚNG 100%

THÔNG MINH

63.1,2=75,6

21.3,6=75,6

=>63.1,2-21.3,6=0

nên biểu thức trên bằng 0

\(A=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right)\left(63.1,2-21.3,6+1\right)}{1-2+3-4+....+99-100}\)

\(=\frac{\frac{100\left(100+1\right)}{2}\left(\frac{3+2-6}{12}\right)\left[63\left(1,2-1,2\right)+1\right]}{\left(1-2\right)+\left(3-4\right)+....+\left(99-100\right)}\)

\(=\frac{5050.\left(-\frac{1}{12}\right).1}{-1+\left(-1\right)+\left(-1\right)+...+\left(-1\right)}\)

\(=\frac{2525.\left(-\frac{1}{6}\right)}{-50}=\frac{101}{12}\)

101/12 bạn nha

CHÚC BẠN HỌC GIỎI

A=(1/100- 1^2). (1/100-(1/2)^2).....(1/100- (1/510)^2).....(1/100-(1/20)^2)

A=(1/100- 1^2). (1/100-(1/2)^2).....(1/100- 1/100).....(1/100-(1/20)^2)

A=(1/100- 1^2). (1/100-(1/2)^2).....0.....(1/100-(1/20)^2)

A=0

Mình ko biết gõ ngoặc vuông bạn thông cảm nha! Chúc bạn học tốt!!!

345,345678

Xét : \(\frac{1}{100}-\frac{1}{n^2}=\frac{n^2-100}{100n^2}=\frac{\left(n-10\right)\left(n+10\right)}{100n^2}\)

Áp dụng , đặt biểu thức cần tính là A , ta có : 

\(A=\left(\frac{1}{100}-\frac{1}{1^2}\right)\left(\frac{1}{100}-\frac{1}{2^2}\right)\left(\frac{1}{100}-\frac{1}{3^2}\right)...\left(\frac{1}{100}-\frac{1}{20^2}\right)\)

\(=\frac{\left(1-10\right)\left(1+10\right)}{100.1^2}.\frac{\left(2-10\right)\left(2+10\right)}{100.2^2}.\frac{\left(3-10\right)\left(3+10\right)}{100.3^2}...\frac{\left(10-10\right)\left(10+10\right)}{100.10^2}...\frac{\left(20-10\right)\left(20+10\right)}{100.20^2}\)

Nhận thấy trong A có một nhân tử (10-10) = 0 nên A = 0

làm thế thì hơi dài đấy Hoàng Lê Bảo Ngọc

ta nhận thấy trong biểu thức chứa thừa số \(\frac{1}{100}-\left(\frac{1}{10}\right)^2=\frac{1}{100}-\frac{1}{100}=0\)

=>biểu thức ấy =0

Nguyễn Thiều Công Thành Ừ , tại mình quên không để ý :)

Ta có:

\(P\left(1\right)=7=7.1^2\); \(P\left(2\right)=28=7.2^2\); \(P\left(3\right)=63=7.3^2\)

\(\Rightarrow\)Đặt \(g\left(x\right)=7x^2\).

Đặt \(Q\left(x\right)=P\left(x\right)-g\left(x\right)\).

Ta có:

\(Q\left(1\right)=Q\left(2\right)=Q\left(3\right)=0\)

\(\Rightarrow x=1;x=2;x=3\)là các nghiệm của đa thức Q(x)

\(\Rightarrow Q\left(x\right)⋮\left(x-1\right);\left(x-2\right);\left(x-3\right)\)

Do Q(x) là đa thức bậc 4 có hệ số cao nhất bằng 1 nên

\(Q\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right).\)

\(\Rightarrow P\left(x\right)=Q\left(x\right)+g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right)+7x^2\)

Ta có:

\(P\left(100\right)=\left(100-1\right)\left(100-2\right)\left(100-3\right)\left(100-m\right)+7.100^2\)

\(=99.98.97\left(100-m\right)+7.100^2==97.98.99.100-97.98.99m+7.100^2\)

\(P\left(-96\right)=\left(-96-1\right)\left(-96-2\right)\left(-96-3\right)\left(-96-m\right)+7.\left(-96\right)^2\)

\(=\left(-97\right).\left(-98\right).\left(-99\right).\left(-96-m\right)+7.96^2\)

\(=\left(-96\right).\left(-97\right).\left(-98\right).\left(-99\right)-\left(-97\right).\left(-98\right).\left(-99\right).m+7.96^2\)

\(=96.97.98.99+97.98.99m+7.96^2\)

\(A=\frac{P\left(100\right)+P\left(-96\right)}{8}\)

\(=\frac{97.98.99.100-97.98.99m+7.100^2+96.97.98.99+97.98.99m+7.96^2}{8}\)

\(=\frac{97.98.99\left(100+96\right)+7.\left(100^2+96^2\right)}{8}=112244867\)

Ta có:

\(P\left(1\right)=7=7.1^2\); \(P\left(2\right)=28=7.2^2\); \(P\left(3\right)=63=7.3^2\)

\(\Rightarrow\)Đặt \(g\left(x\right)=7x^2\).

Đặt \(Q\left(x\right)=P\left(x\right)-g\left(x\right)\).

Ta có:

\(Q\left(1\right)=Q\left(2\right)=Q\left(3\right)=0\)

\(\Rightarrow x=1;x=2;x=3\)là các nghiệm của đa thức Q(x)

\(\Rightarrow Q\left(x\right)⋮\left(x-1\right);\left(x-2\right);\left(x-3\right)\)

Do Q(x) là đa thức bậc 4 có hệ số cao nhất bằng 1 nên

\(Q\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right).\)

\(\Rightarrow P\left(x\right)=Q\left(x\right)+g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-m\right)+7x^2\)

Ta có:

\(P\left(100\right)=\left(100-1\right)\left(100-2\right)\left(100-3\right)\left(100-m\right)+7.100^2\)

\(=99.98.97\left(100-m\right)+7.100^2==97.98.99.100-97.98.99m+7.100^2\)

\(P\left(-96\right)=\left(-96-1\right)\left(-96-2\right)\left(-96-3\right)\left(-96-m\right)+7.\left(-96\right)^2\)

\(=\left(-97\right).\left(-98\right).\left(-99\right).\left(-96-m\right)+7.96^2\)

\(=\left(-96\right).\left(-97\right).\left(-98\right).\left(-99\right)-\left(-97\right).\left(-98\right).\left(-99\right).m+7.96^2\)

\(=96.97.98.99+97.98.99m+7.96^2\)

\(A=\frac{P\left(100\right)+P\left(-96\right)}{8}\)

\(=\frac{97.98.99.100-97.98.99m+7.100^2+96.97.98.99+97.98.99m+7.96^2}{8}\)

\(=\frac{97.98.99\left(100+96\right)+7.\left(100^2+96^2\right)}{8}=112244867\)

\(3A=100+\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}\)

\(3A-A=2A=100-\frac{100}{3^4}\)

\(A=50-\frac{\frac{100}{3^4}}{2}\)

\(\text{Đặt }A=\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}+\frac{100}{3^4}\)

\(3A=100+\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}\)

\(3A-A=2A=100-\frac{100}{3^4}\)

\(2A=100-\frac{100}{81}=\frac{8000}{81}\)

\(A=\frac{8000}{81}\text{ : }2\)

\(A=\frac{4000}{81}\)

Đặt \(A=\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}+\frac{100}{3^4}\)

\(\Rightarrow A=100\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)\)

Đặt \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}\)

\(\Rightarrow3B-B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}\)\(-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}\)

\(\Rightarrow2B=1-\frac{1}{3^4}\)

\(\Rightarrow B=\frac{1}{2}\left(1-\frac{1}{3^4}\right)\)

\(\Rightarrow B=\frac{1}{2}.\frac{80}{81}=\frac{40}{81}\)

\(\Rightarrow A=100.\frac{80}{81}=\frac{8000}{81}\)

\(\left[\frac{100}{3}\right]+\left[\frac{100}{3^2}\right]+\left[\frac{100}{3^3}\right]+\left[\frac{100}{3^4}\right]\)

\(=33+11+3+1\)

\(=48\)

\(\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}+\frac{100}{3^4}\)

\(=\frac{100.3^3}{3^4}+\frac{100.3^2}{3^4}+\frac{100.3}{3^4}+\frac{100}{3^4}\)

\(=\frac{100.3^3+100.3^2+100.3+100}{3^4}\)

\(=\frac{100.\left(3^3+3^2+3+1\right)}{3^4}\)

\(=\frac{100.\left(27+9+3+1\right)}{81}\)

\(=\frac{100.40}{81}\)

\(=\frac{4000}{81}\)

Tính ko cần nhanh ak zZz Cool Kid zZz