1.

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x-1}=\frac{98}{99}\)

\(1-\frac{1}{x-1}=\frac{98}{99}\)

\(\frac{1}{x-1}=1-\frac{98}{99}\)

\(\frac{1}{x-1}=\frac{1}{99}\)

\(\Rightarrow x-1=99\)

\(\Rightarrow x=99+1=100\)

b) \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}\right)+10.\left(\frac{1}{13}-\frac{1}{15}\right)+10.\left(\frac{1}{15}-\frac{1}{17}\right)+...+10.\left(\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-\left[10.\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(x-10.\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

\(\Rightarrow x=\frac{3}{11}+\frac{8}{11}=1\)

c) 5^x + 2 . 5^x + 2³ = 83

5^x . ( 1 + 2 ) + 8 = 83

5^x . 3 = 83 - 8

5^x . 3 = 75

5^x = 75 : 3

5^x = 25

\(\Rightarrow\)5^x = 5²

\(\Rightarrow\)x = 2

2.

Ta thấy \(2016^{2016}>2016^{2016}-3\)

\(\Rightarrow B=\frac{2016^{2016}}{2016^{2016}-3}>\frac{2016^{2016}+2}{2016^{2016}-3+2}=\frac{2016^{2016}+2}{2016^{2016}-1}=A\)

\(\Rightarrow A< B\)

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)

Ta có \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{99}\)

      = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{98}{99}\)(áp dụng công thức)

      = \(1-\frac{1}{x+1}=\frac{98}{99}\)

      = \(\frac{1}{x+1}=1-\frac{98}{99}\)(quy tắc tìm số trừ)

      = \(\frac{1}{x+1}=\frac{1}{99}\Rightarrow\frac{1}{x+1}=\frac{1}{98+1}\Rightarrow x=98\)

Vậy x = 98 :)

Còn nữa, công thức mà mình áp dụng là: \(\frac{a}{b.c}=\frac{1}{b}-\frac{1}{c}\)nếu \(a=c-b\)

cảm ơn nhé các bạn

các bạn phại đổi kiểu chữ để làm bài này (VNI) , (TELEX)

vni

x - \(\dfrac{20}{11.13}-\dfrac{20}{13.15}-\dfrac{20}{15.17}-........-\dfrac{20}{53.55}=\dfrac{3}{11}\)

x - \(\left(\dfrac{20}{11.13}+\dfrac{20}{13.15}+\dfrac{20}{15.17}+...+\dfrac{20}{53.55}\right)=\dfrac{3}{11}\)

Đặt A = \(\dfrac{20}{11.13}+\dfrac{20}{13.15}+\dfrac{20}{15.17}+...+\dfrac{20}{53.55}\)

A = \(\dfrac{20}{2}\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+\dfrac{2}{15.17}+....+\dfrac{2}{53.55}\right)\)

A = \(\dfrac{20}{2}\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)

A = \(10.\left(\dfrac{1}{11}-\dfrac{1}{55}\right)\)

A = 10 . 4/55

A = 8/11

TA Có : x - 8/11 = 3/11

x = 3/11 + 8/11

x = 11/11

x = 1

Vậy x = 1

a) \(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

x = 1

b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\) ( nhân cho cả tử và mẫu của các số hạng trên ( ngoại trừ 2/x.(x+1) ) là 2)

\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

=> x + 1 = 18

x = 17

\(a,x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)

\(\Rightarrow x=1\)

\(b,\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{18}=\frac{1}{x+1}\)

\(\Rightarrow x+1=18\Leftrightarrow x=17\)

a) Vì \(\dfrac{x+5}{3}\)= \(\dfrac{x-6}{7}\) nên 7(x+5) = 3(x-6)

=> 7x+ 35 = 3x - 18

7x - 3x = -18 -35

4x = -53

x = -53:4

x = \(\dfrac{-53}{4}\)

\(x-\frac{20}{11.13}-\frac{20}{23.15}-....-\frac{20}{53.55}=\frac{3}{11}\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+....+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

=> \(x=\frac{3}{11}+\frac{8}{11}=1\)

thank you nha !

 -Nếu |x-2013|-2014=2015->|x-2013| = 4029

                                                  + Nếu x-2013 =4029 -> x= 6042

                                                  + Nếu x-2013 = -4029 -> x = -2016

  - Nếu |x-2013|-2014= -2015     ->  |x-2013| = -1 (loại vì  |x-2013| \(\ge\)0 )

Vậy x= 6042 ; x=-2016  là các giá trị cần tìm.

   \(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

=> \(x-\left(20\times\frac{1}{11.13}+20\times\frac{1}{13.15}+20\times\frac{1}{15.17}+...+20\times\frac{1}{53.55}\right)=\frac{3}{11}\)

      \(x-20\times\left(\frac{1}{11.13}+\frac{1}{13.15}+\frac{1}{15.17}+...+\frac{1}{53.55}\right)=\frac{3}{11}\)

       \(x-20\times\left(\frac{1}{2}\times\left(\frac{1}{11}-\frac{1}{13}\right)+\frac{1}{2}\times\left(\frac{1}{13}-\frac{1}{15}\right)+\frac{1}{2}\times\times+...+\frac{1}{2}\times\left(\frac{1}{53}-\frac{1}{55}\right)\right)=\frac{3}{11}\)

\(x-20\times\frac{1}{2}\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-...-\frac{1}{53}+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10\times\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10\times\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{10}{11}=\frac{3}{11}\)

=> \(x=\frac{3}{11}+\frac{10}{11}=\frac{13}{11}\)

Vậy x=\(\frac{13}{11}\)

cả 2012 cũng đc đó nhé thử xem

ý b sai rồi Phạm Ngọc Thạch ak

1/2 đào đâu ra ma nhân cơ chứ

Trả lời nhanh giúp mình nhé

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câu a: x=1

câu b x=17