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câu g)
\(G=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right).\)
\(=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{120}{121}\)
\(=\frac{3.\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{2.2.3.3.4.4.5.5....11.11}\)
\(=\frac{12}{3}=4\)
B= 333300
C=328350
D=(n+1) /( n nhân 2)
E=(1/3 trừ 1/3^100):2
1)=>3B=1.2.3+2.3.3+3.4.3+...+99.100.3
3B=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
3B=99.100.101
=>B=333300
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 \(\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}=0\)
Đề câu C sai nhé, sửa: ... < 1/2
\(C=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\\ 3C=1+\frac{1}{3}+...+\frac{1}{3^{98}}\\ 3C-C=1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{99}}\\ 2C=1-\frac{1}{3^{99}}\\ C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\left(đpcm\right)\)
Đề câu D sai nhé, sửa: ... > -1/2
\(D=\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)\cdot...\cdot\left(\frac{1}{100^2}-1\right)< \left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot...\cdot\left(\frac{1}{100}-1\right)\)
Mặt khác \(\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)\cdot...\cdot\left(\frac{1}{100}-1\right)\\ =\frac{-1}{2}\cdot\frac{-2}{3}\cdot\frac{-3}{4}\cdot...\cdot\frac{-99}{100}\\ =-\left(\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\right)\\ =\frac{-1}{100}\)
Mà \(\frac{1}{100}< \frac{1}{2}\Rightarrow\frac{-1}{100}>\frac{-1}{2}\)
Vậy \(D< \frac{-1}{2}\left(đpcm\right)\)
Cảm ơn bạn nhiều nhé.