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

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

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

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

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

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

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

=> \(x+205=0\)

=> \(x=-205\)

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

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

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

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

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

\(=>x+205=0\)

\(=>x=-205\)

\(a;\)

\(=0-1\)

\(=-1\)

\(b;\)

\(=0-4\)

\(=-4\)

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

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

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

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

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

Ta có : \(\frac{x+4}{200}+\frac{x+3}{201}=\frac{x+2}{202}+\frac{x+1}{203}\)

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

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

=> \(\frac{x+204}{200}+\frac{x+204}{201}-\frac{x+204}{202}-\frac{x+204}{203}=0\)

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

=> \(x+204=0\)

=> \(x=-204\)

Vậy phương trình có tập nghiệm là S = { -204 }

Thank nha bạn

2.Áp dụng tc dãy tỉ số bằng nhau ta có:

\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)

\(\Rightarrow a+b+c=a+b-c\)

\(\Rightarrow a+b+c-a-b+c=0\)

\(\Rightarrow2c=0\)

\(\Rightarrow c=0\)

Vậy c=0

BT5: Ta có: f(1)=1.a+b=1 =>a+b=1 (1)

f(2)=2a+b=4 (2)

Trừ (1) cho (2) ta có: 2a+b-a-b=4-1 => a=3

Với a=3 thay vào (1) ta có: 3+b=1 => b=-2

Vậy a=3, b=-2