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đề cần chứng minh nhỏ hơn 1 hay 11
nếu 1 thì
\(\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{100^2}\)
\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.......+\frac{1}{99\cdot100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrowđcm\)
nếu nhỏ hơn 11 thì làm như thế thêm câu
vì đẳng thức trên <1<11
=>đcm
\(2+\frac{1}{1+\frac{1}{1+\frac{1}{3+\frac{1}{4}}}}=2+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{13}{4}}}}=2+\frac{1}{1+\frac{1}{1+\frac{4}{13}}}=2+\frac{1}{1+\frac{1}{\frac{17}{13}}}\)
\(=2+\frac{1}{1+\frac{13}{17}}=2+\frac{1}{\frac{30}{17}}=2+\frac{17}{30}=\frac{77}{30}\)
vậy_
a) \(\frac{3x-6}{x+4}=\frac{2\left(x+5\right)+\left(x-3\right)}{x-2}\)
\(\frac{3\left(x-2\right)}{x+4}=\frac{2\left(x+5\right)+x-3}{x-2}\)
\(\frac{3\left(x-4\right)}{x+4}=\frac{3x+7}{x-2}\)
\(3\left(x-2\right)\left(x-2\right)=\left(3x+7\right)\left(x+4\right)\)
\(3\left(x-2\right)^2=\left(3x+7\right)\left(x+4\right)\)
\(3x^2-12x+12=3x^2+12x+7x+28\)
\(3x^2-12x+12=3x^2+19x+28\)
\(-12x+12=19x+28\)
\(12=19x+28+12x\)
\(19x+28+12x=12\) (chuyển vế)
\(31x+28=12\)
\(31x=12-28\)
\(31x=-16\)
\(x=-\frac{16}{31}\)
\(\Rightarrow x=-\frac{16}{31}\)
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+2015}\)
\(=\frac{2}{1.2}+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+2+3\right).3}{2}}+.....+\frac{1}{\frac{\left(2015+1\right).2015}{2}}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+....+\frac{2}{2015.2016}\)
a) Ta có :
\(\frac{a}{5}=\frac{-3}{b}\)
\(\Rightarrow a.b=-3.5=-15\)
\(\Rightarrow a;b\in\left(-15\right)=\left(\pm1;\pm3;\pm5;\pm15\right)\)
Ta có bảng sau :
| a | 1 | 15 | -15 | -1 | 3 | 5 | -5 | -3 |
| b | 15 | 1 | -1 | -15 | 5 | 3 | -3 | -5 |
Vậy................
bn từ phần đầu đến phàn -15 đúng rùi
a.b=-15=3.(-5);(-15).1;(-3).5;5.(-3);1.(-15)
=> a,b là các cặp trên
=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+............+\frac{1}{18.19.20}\)
=\(\frac{2}{1.2.3.2}+\frac{2}{2.3.4.2}+............+\frac{2}{18.19.20.2}\)
=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}............+\frac{1}{18.19}-\frac{1}{19.20}\)
=\(\frac{1}{1.2}-\frac{1}{19.20}\)
=\(\frac{189}{380}\)
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
1/3+1/4+4/5-1/2=4/12+3/12+4/5-1/2=7/12+4/5-1/2=35/60+48/60-1/2=83/60-1/2=83/60-30/60
=53/60
MTC;60
20/60+15/60+12/60-30/60
=20+15+12-30
=17 nha bn