Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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%
Ta có: \(63.1,2-21.3,6=0,9.7.10.1,2-21.3,6\)
\(=6,3.12-21.3,6\)
\(=0,9.7.4.3-7.3.0,9.4\)
\(=6,3.12-6,3.12\)
\(=0\)
\(\Rightarrow\frac{\left(1+2+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)0}{1-2+3-4+...+99-100}=0\)
Vậy \(\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}=0\)
Bài giải
\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\cdot\left(63\cdot1,2-21\cdot3,6\right)}{1-2+3-4+...+99-100}\)
\(=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\cdot\left(75,6-75,6\right)}{1-2+3-4+...+99-100}\)
\(=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\cdot0}{1-2+3-4+...+99-100}\)
\(=\frac{0}{1-2+3-4+...+99-100}\)
\(=0\)
\(A=\frac{\left(1+2+3+...+99+100\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63\cdot1,2-21\cdot3,6\right)}{1-2+3-4+...+99-100}\)
\(=\frac{\left(1+2+3+...+99+100\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(75,6-75,6\right)}{1-2+3-4+...+99-100}\)
\(=\frac{\left(1+2+3+...+99+100\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\cdot0}{1-2+3-4+...+99-100}\)
\(=0\)
Ta thấy : \(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
nên kết quả dãy trên bằng 0
câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
cho mik hỏi có gì đặc biệt ko!!
Ta có : \(6,3.1,2-2,1.3,6=0\)
Thay vào ta được C=0