câu b nha

B= 1/100 - (1/2.1 + 1/3.2 + ... + 1/98.97 + 1/99.98 + 1/100.99)

B=1/100 - (1 - 1/2 + 1/2 - 1/3 + 1/3 - ... - 1/99 + 1/99 - 1/100)

B=1/100-(1-1/100)

B=1/100-99/100

B= - 98/100

B= - 49/50

đ ú g nha

- Câu a ~> http://olm.vn/hoi-dap/question/183158.html

A=\(\frac{-9}{16}\)

B= -1

Vì \(B=\frac{2014^{11}+2}{2014^{12}+2}<1\)

\(\Rightarrow B=\frac{2014^{11}+2}{2014^{12}+2}<\frac{2014^{11}+2+4026}{2014^{12}+2+4026}=\frac{2014^{11}+4028}{2014^{12}+4028}=\frac{2014.\left(2014^{10}+2\right)}{2014\left(2014^{11}+2\right)}=\frac{2014^{10}+2}{2014^{11}+2}=A\)

Vậy B<A hay A<B

ta chứng minh bài toán phụ:

nếu ta có b<d \(\frac{a}{b}\)>\(\frac{c}{d}\) thì ad>bc

dễ thây \(\frac{ad}{bd}>\frac{cb}{bd}\)

 => ad>bd

áp dụng:

dat 2014=a ta co

\(A=\frac{a^{10}+2}{a^{11+2}}\)

 \(B=\frac{a^{11}+2}{a^{12}+2}\)

 ta có 

\(A=\frac{a^{10}+2.a^{12}+2}{a^{11}+2.a^{12}+2}\)

 \(B=\frac{a^{11}+2.a^{11}+2}{a^{12}+2.a^{11}+2}\)=\(\frac{a^{10}+2a^{12}+2}{a^{12}+2a^{11}+2}\)

 => A=B

mk hok chắc đâu nha

\(A>\frac{196}{197+198}+\frac{197}{198+197}=\frac{196+197}{198+197}=B\)

\(\Leftrightarrow A>B\)

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}.\)

\(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)(cộng 2 vế cho 3)

\(\frac{x+1}{2009}+\frac{2009}{2009}+\frac{x+2}{2008}+\frac{2008}{2008}+\frac{x+3}{2007}+\frac{2007}{2007}=\frac{x+10}{2000}+\frac{2000}{2000}+\frac{x+11}{1999}+\frac{1999}{1999}+\frac{x+12}{1998}+\frac{1998}{1998}.\)

\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}.\)

\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

x+2010=0

x=-2010

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

\(\Leftrightarrow\left(1+\frac{x+1}{2009}\right)+\left(1+\frac{x+2}{2008}\right)+\left(1+\frac{x+3}{2007}\right)\)

\(=\left(1+\frac{x+10}{2000}\right)+\left(1+\frac{x+11}{1999}\right)+\left(1+\frac{x+12}{1998}\right)\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x=2010}{1998}\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}\)

\(=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

\(\Leftrightarrow x+2010=0\)

\(\Leftrightarrow x=-2010\)

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

\(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\right)=0\Leftrightarrow x=-2010\)

Câu 2:

\(a,\Rightarrow x=-23+15=-8\\ b,\Rightarrow5\left(x+4\right)=85\\ \Rightarrow x+4=17\Rightarrow x=13\\ c,\Rightarrow5^{x+2}=24+1=25=5^2\\ \Rightarrow x+2=2\Rightarrow x=0\\ d,\Rightarrow x+4+x+5⋮9\\ \Rightarrow2x+9⋮9\\ \Rightarrow2x⋮9\Rightarrow x\in\left\{0;9\right\}\left(0< x< 10\right)\)

|1,5-x|+|2,5-x|=0

=> |1,5-x| = 0 và |2,5-x| = 0

=> x = 1,5 và x = 2,5

Không thể tồn tại 2 giá x cung một lúc.

Vậy không tồn tại x.

bạn tự làm đi bài này dễ mà 

1/2 . P = 1/2.6 + 1/6.10 + 1/10.14 + ... + 1/198.202

4.1/2. P= 4/2.6 + 4/6.10 + 4/10.14 + ... + 4/198.202

2P=1/2-1/6+1/6-1/10+1/10-1/14+...+1/198-1/202

2P=1/2-1/202=50/101

P=50/101:2=50/101.1/2=25/101