\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)

=> \(\left(1+\frac{x+5}{200}\right)+\left(1+\frac{x+4}{201}\right)=\left(1+\frac{x+3}{202}\right)+\left(1+\frac{x+2}{203}\right)\)

=> \(\frac{x+205}{200}+\frac{x+205}{201}=\frac{x+205}{202}+\frac{x+205}{203}\)

=> \(\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)

=> \(\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)

Do \(\frac{1}{200}>\frac{1}{202};\frac{1}{201}>1-\frac{1}{203}\)

=> \(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)

=> \(x+205=0\)

=> \(x=-205\)

\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)

\(=>\frac{x+5+200}{200}+\frac{x+4+201}{201}-\frac{x+3+202}{202}-\frac{x+2+203}{203}=0\)

\(=>\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)

\(=>\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)

\(Do:\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)

\(=>x+205=0\)

\(=>x=-205\)

\(a;\)

\(=0-1\)

\(=-1\)

\(b;\)

\(=0-4\)

\(=-4\)

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{15}\)

\(\frac{181\left(x+1\right)}{660}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\frac{181\left(x+1\right)}{660}=\frac{17\left(x+1\right)}{52}\)

\(2353\left(x+1\right)=2805\left(x+1\right)\)

\(2353x+2353=2805x+2805\)

\(2353=2805x+2805-2353x\)

\(2353=452x+2805\)

\(2353-2805=452x\)

\(-452=452x\)

\(x=-1\)

<=> (2-x/201 + 1) + (x/203 - 1) = (1-x/202 + 1) + (1-1)

<=> 203-x/201 + x-203/203 = 203-x/202

<=> 203-x/201 - 203-x/203 - 203-x/202 = 0

<=> (203-x).(1/201-1/203-1/202) = 0

<=> 203-x = 0 ( vì 1/201-1/203-1/202 khác 0 )

<=> x=203

Vậy x=203

k mk nha

\(\frac{x+199}{200}+\frac{x+198}{201}+\frac{x+197}{202}=-3\)

\(\frac{x+199}{200}+1+\frac{x+198}{201}+1+\frac{x+197}{202}+1=0\)

\(\frac{x+399}{200}+\frac{x+399}{201}+\frac{x+399}{202}=0\)

\(\left(x+399\right)\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)=0\)

Mà \(\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)\ne0\)

=> x + 399 = 0

=> x = -399 

BT1: 2015²⁰¹⁴ có tận cùng là 5

    2014²⁰¹⁵=2014.(2014²)¹⁰⁰⁷=2014.4056196¹⁰⁰⁷=2014.(...6) => Có tận cùng là ...4

=> 2015²⁰¹⁴-2014²⁰¹⁵ có tận cùng là ...5-...4=...1 

BT2: f(1)=a.1+b=1  (1)

       f(2)=a.2+b=4    (2)

Trừ (2) cho (1) => a=3

Thay a=3 vào (1) => b=-2

ĐS: a=3; b=-2

Sao ko ai trả lời vậy?! Bộ câu của mình khó quá ak???

\(\frac{x+1}{203}+\frac{x+2}{202}+\frac{x+3}{201}+\frac{x+4}{200}+\frac{x+5}{199}+5=0\)

\(\Leftrightarrow\frac{x+1}{203}+1+\frac{x+2}{202}+1+\frac{x+3}{201}+1+\frac{x+4}{200}+1+\frac{x+5}{199}+1=0\)

\(\Leftrightarrow\frac{x+204}{203}+\frac{x+204}{202}+\frac{x+204}{201}+\frac{x+204}{200}+\frac{x+204}{199}=0\)

\(\Leftrightarrow\left(x+204\right)\left(\frac{1}{203}+\frac{1}{203}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\right)=0\)

\(\Leftrightarrow x+204=0\).Do \(\frac{1}{203}+\frac{1}{203}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\ne0\)

\(\Leftrightarrow x=-204\)

Ta có :

\(\frac{x+1}{203}+\frac{x+2}{202}+\frac{x+3}{201}+\frac{x+4}{200}+\frac{x+5}{199}+5=0\)

\(\Leftrightarrow\left(\frac{x+1}{203}+1\right)+\left(\frac{x+2}{202}+1\right)+\left(\frac{x+3}{201}+1\right)+\left(\frac{x+4}{200}+1\right)+\left(\frac{x+5}{199}+1\right)=0\)

\(\Leftrightarrow\left(\frac{x+204}{203}\right)+\left(\frac{x+4}{202}\right)+\left(\frac{x+4}{201}\right)+\left(\frac{x+204}{200}\right)+\left(\frac{x+204}{199}\right)=0\)

\(\Leftrightarrow\left(x+204\right)\left(\frac{1}{203}+\frac{1}{202}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\right)=0\)

Dễ thấy \(\left(\frac{1}{203}+\frac{1}{202}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\right)\ne0\)

=> x + 204 = 0

<=> x = - 204

Vậy pt có nghiệm x = - 204

a/\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)

=\(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)

=2.5

=10

= 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/x - 1/x + 1

= 1/2 - 1 /x + 1 =199/200

=1/x+1 = 1/2 - 199/200

1/ x + 1 = ....

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x.\left(x+1\right)}=\frac{199}{200}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{199}{200}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{199}{200}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{199}{200}\)

\(\Rightarrow\frac{1}{x+1}=-\frac{99}{200}\)

\(\Rightarrow\frac{-99}{-\left(x+1\right).99}=\frac{-99}{200}\)

\(\Rightarrow-99x+-99=200\)