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1.
Ta có:
1/2 < 2/3
3/4 < 4/5
.............
99/100 < 100/101
=> 1/2*3/4*5/6*...*99/100 < 2/3*4/5*6/7*...*100/101
=> A < B
2.
\(A\cdot B=\left[\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right]\cdot\left[\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right]\)
\(A\cdot B=\frac{\left[1\cdot3\cdot5\cdot7\cdot...\cdot99\right]\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]}{\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]\left[3\cdot5\cdot7\cdot9\cdot...\cdot101\right]}=\frac{1\cdot3\cdot5\cdot...\cdot99}{3\cdot5\cdot7\cdot...\cdot101}=\frac{1}{101}\)
3.
Vì A < B => A.A < A.B => A2 < 1/101 < 1/100
Mà A2 < 1/100 <=> A2 < \(\frac{1}{10}^2\)=> A < 1/10
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\frac{1}{2.4}-\frac{1}{4.6}-\frac{1}{6.8}-\frac{1}{8.10}\)
\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}-\frac{1}{2}+\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\frac{1}{2}.\left[\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\right]\)
\(S=\frac{1}{2}.\left[\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\right]\)
\(S=\frac{1}{2}.\left[\left(1-\frac{1}{9}\right)-\left(\frac{1}{2}-\frac{1}{10}\right)\right]\)
\(S=\frac{1}{2}.\left(\frac{8}{9}-\frac{2}{5}\right)\)
\(S=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
1/ chứng tỏ 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/101^2 < 100/101
Nhận xét : 1/2^2 = 1/2.2 < 1/1.2
1/3^2 = 1/3.3 < 1/2.3
.....
1/101^ 2 = 1/101 . 101 < 1/100 . 101
=> 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/101^2 < 1/1.2 + 1/2.3 + .... + 1/100 . 101
1/1.2 + 1/2.3 + .... + 1/100 . 101 = 1 - 1/2 +1/2 - 1/3 + .... + 1/100 - 1/101 = 1 - 1/101 = 100 / 101
=> 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/101^2 < 100/101
3/ x756y chia 2,5,9 đều dư 1
Để x756y chia 5 dư 1 => \(y\in\left\{1;6\right\}\)
Vì x756y chia 2 dư 1 => y lẻ => y = 1
=> x7561 chia 9 dư 1 \(\Leftrightarrow\)( x + 7 + 5 + 6 + 1 ) chia 9 dư 1 => x + 19 chia 9 dư 1 => x + 19 - 1 chia hết cho 9 => x + 18 chia hết cho 9 => x chia hết cho 9 => \(x\in\left\{0;9\right\}\)
Bài 1 :
\(4\left(x-1\right)^{100}-3^{100}=3^{101}\)
\(\Leftrightarrow\)\(4\left(x-1\right)^{100}=3^{101}+3^{100}\)
\(\Leftrightarrow\)\(4\left(x-1\right)^{100}=3^{100}\left(3+1\right)\)
\(\Leftrightarrow\)\(4\left(x-1\right)^{100}=3^{100}.4\)
\(\Leftrightarrow\)\(\left(x-1\right)^{100}=3^{100}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-1\right)^{100}=3^{100}\\\left(x-1\right)^{100}=\left(-3\right)^{100}\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3+1\\x=-3+1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)
Vậy \(x=-2\) hoặc \(x=4\)
Chúc bạn học tốt ~
- ( 1 + (-2) ) + ( 3 + (-4) ) + . . . + (19 + (-20))
= -1 + -1 +....+-1
= -10 - (1 – 2) + (3 – 4) + . . . + (99 – 100)
= -1 + -1 +...+-1
= -50 - (2 – 4) + (6 – 8) + . . . + (48 – 50)
= -2 + -2 +...+-2
= -50