x=-100

\(\frac{x+5}{95}+\frac{x+6}{94}+\frac{x+7}{93}+\frac{x+8}{92}+\frac{x+9}{91}=-5\)

\(\left(\frac{x+5}{95}+1\right)+\left(\frac{x+6}{94}+1\right)+\left(\frac{x+7}{93}+1\right)+\left(\frac{x+8}{92}+1\right)+\left(\frac{x+9}{91}+1\right)=-5+5=0\)

x=100

k hộ với

\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)

=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)

=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)

=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)

=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)

=> x =  - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)

Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)

\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)

\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)

\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)

mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)

nên x+100=0

hay x=-100

Vậy: S={-100}

a)\(\frac{x}{108}=\frac{-7}{9}.\frac{5}{6}\)

\(\frac{x}{108}=\frac{-35}{54}\)

\(\frac{x}{108}=\frac{-70}{108}\)

\(x=-70\)

b) 

Ta có : 

\(\frac{x+1}{65}+\frac{x+3}{63}< \frac{x+5}{61}+\frac{x+7}{59}\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)< \left(\frac{x+5}{61}+1\right)+\left(\frac{x+7}{59}+1\right)\)

\(\Leftrightarrow\)\(\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+5}{61}-\frac{x+7}{59}< 0\)

\(\Leftrightarrow\)\(\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)< 0\)

Vì \(\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)< 0\)

\(\Rightarrow\)\(x+66>0\)

\(\Rightarrow\)\(x>-66\)

Vậy \(x>-66\)

mình nhầm câu 2 phải là <=0

Ta có : 

\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4\le0\)

\(\Leftrightarrow\)\(\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)+4\le0\)

\(\Leftrightarrow\)\(\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}+4\le0\)

\(\Leftrightarrow\)\(\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)+4\le0\)

Hết biết giải, mk mới lớp 7 :')

\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{97}+\frac{x+100}{96}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

Dễ thấy \(\left(\frac{1}{99}< \frac{1}{98}< \frac{1}{97}< \frac{1}{96}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)\ne0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

Vậy x = -100

\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)

\(\Rightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)

\(\Rightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)

\(\Rightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)=0\)

Dễ thấy \(\left(\frac{1}{91}>\frac{1}{93}>\frac{1}{95}>\frac{1}{97}\right)\)nên \(\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)\ne0\)

\(\Rightarrow200-x=0\Rightarrow x=200\)

Vậy x = 200

<=> (1927-x/91 + 1)+(1925-x/93 + 1) + (1923-x/95 + 1) + (1921-x/97 + 1) = 0

<=> 2018-x/91 + 2018-x/93 + 2018-x/95 + 2018-x/97 = 0

<=> (2018-x).(1/91 + 1/93 + 1/95 + 1/97) = 0

<=> 2018-x = 0 ( vì 1/91 + 1/93 + 1/95 + 1/97 > 0 )

<=> x = 2018

Vậy x = 2018

Tk mk nha

\(\frac{x}{108}=-\frac{7}{9}.\frac{5}{6}\)

\(\frac{x}{108}=-\frac{35}{54}\)

\(\Rightarrow x=\frac{108.-35}{54}=-70\)
bài dưới để làm tiếp cho

\(\frac{x+5}{95}\)+\(\frac{x+6}{94}\)+\(\frac{x+7}{93}\)+\(\frac{x+8}{92}\)+\(\frac{x+9}{91}\)= -5

\(\frac{x+5}{95}\)+\(\frac{x+6}{94}\)+\(\frac{x+7}{93}\)+\(\frac{x+8}{92}\)+\(\frac{x+9}{91}\)+ 5 = 0

\(\left(\frac{x+5}{95}+1\right)\)+ \(\left(\frac{x+6}{94}+1\right)+\left(\frac{x+7}{93}+1\right)+\left(\frac{x+8}{92}+1\right)+\left(\frac{x+9}{91}+1\right)=0\)

\(\frac{x+5+95}{95}+\frac{x+6+94}{94}+\frac{x+7+93}{93}+\frac{x+8+92}{92}+\frac{x+9+91}{91}=0\)

\(\frac{x+100}{95}+\frac{x+100}{94}+\frac{x+100}{93}+\frac{x+100}{92}+\frac{x+100}{91}\)

\(\left(x+100\right)\left(\frac{1}{95}+\frac{1}{94}+\frac{1}{93}+\frac{1}{92}+\frac{1}{91}\right)=0\)

                          SUY RA         \(x+100=0\)

                                                \(x=-100\)

                                    VẬY:   \(x=-100\)

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