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

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

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

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

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

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

\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+14}{83}+\frac{x+116}{4}=0\)

\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+14}{83}+1+\frac{x+116}{4}-4=0\)

\(\frac{x+14+86}{86}+\frac{x+15+85}{85}+\frac{x+16+84}{84}+\frac{x+14+83}{83}+\frac{x+116-16}{4}=0\)

\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

Vì \(\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)\ne0\)

\(\Rightarrow x+100=0\)

\(\Rightarrow x=-100\)

Vậy........

a) (x-1)x(x+1)(x+2) = 24

<=> [(x-1)(x+2)][x(x+1) = 24

<=> (x^2+x-2)(x^2+x) = 24     (1)

Đặt t=x^2+x-1 = (x+1/2)^2 - 5/4    (*)

(1) trở thành (t-1)(t+1) = 24

<=> t^2 - 1 - 24 = 0

<=> t^2 - 25 = 0

<=> t^2 = 25

<=> t=5 hoặc t=-5

Mà t >= -5/4 ( từ *) => t = (x+1/2)^2-5/4 = 5

<=> (x+1/2)^2 = 25/4

Đến đây dễ r`

c) x^4 + 3x^3 + 4x^2 + 3x + 1 = 0

<=> x^4 + x^3 + 2x^3 + 2x^2 + 2x^2 + 2x + x + 1 = 0

<=> (x+1)(x^3 + 2x^2 + 2x + 1) = 0

<=> (x +1)(x^3 + x^2 + x^2 + x + x + 1) = 0

<=> (x+1)^2.(x^2+x+1) = 0

Mà x^2+x+1 = (x+1/2)^2 + 3/4 > 0

Nên x+1=0 <=> x=-1

Vậy ...

b, cộng 1 vào 4 phân thức đầu,trừ 4 ở pt cuối ,rồi đặt đc NTC (x+100)

b) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=0\)

\(\Rightarrow\)

số đó là

0 nhé bn 

mình chỉ

chăc thế thôi

hihi

câu a kq =-100, lấy x+14/86+z+x+15/85+1....=0

x+100/86+x+100/85+...=0

(x+100).(1/86+1/85...)=o

suy ra x+100=0, x=-100

10)   \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)

\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)

\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)   (vì  1/86 + 1/85 + 1/84 + 1/83 + 1/4  \(\ne\)0)

\(\Leftrightarrow\)\(x=-100\)

Vậy....

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

     \(\frac{x+200}{99}+\frac{x+200}{98}=\frac{x+200}{97}+\frac{x+200}{96}\)

   \(\frac{x+200}{99}+\frac{x+200}{98}-\frac{x+200}{97}-\frac{x+200}{96}=0\)

\(\left(x+200\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

mà\(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\ne0\)

==> x+200=0

<=>x=-200

Vậy nghiệm của phương trình là x=-200

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

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

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

mà  \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)

==>200-x=0

<=>x=200

vậy nghiệm của pt là x=200

a, \(2\left(\frac{11x}{12}+\frac{1}{3}\right)=2-\frac{x}{6}\)

\(2\left(\frac{11x+4}{12}\right)-2+\frac{x}{6}=0\)

\(\frac{44x+8}{12}-2+\frac{x}{6}=0\)

\(\frac{44x+8}{12}-\frac{24}{12}+\frac{2x}{12}=0\)

\(\frac{44x+8-24+2x}{12}=\frac{46x-16}{12}=0\)

\(\Leftrightarrow46x-16=0\)

\(\Leftrightarrow46x=16\Rightarrow x=\frac{8}{23}\)

Vậy nghiệm của pt là x=8/23

k mk

Ta có: \(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)

\(\Leftrightarrow\frac{99-x}{101}-1+\frac{97-x}{103}-1+\frac{95-x}{105}-1+\frac{93-x}{107}-1=-4+4\)

\(\Leftrightarrow\frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)

\(\Leftrightarrow\left(200-x\right).\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)

=> 200 - x = 0

=> x          = 200

Vậy x = 200

\(x=200\)

\(\left(\frac{99-x}{101}+1\right)+\left(\frac{97-x}{103}+1\right)+\left(\frac{95-x}{105}+1\right)+\left(\frac{93-x}{107}+1\right)=-4+4\)

\(\frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)

\(\left(200-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)  mà \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)

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

k nha

\(\left(\frac{99-x}{101}+1\right)+\left(\frac{97-x}{103}+1\right)+\left(\frac{95-x}{105}+1\right)+\left(\frac{93-x}{107}+1\right)=-4+4\)

 \(\frac{110-x}{101}+\frac{110-x}{103}+\frac{110-x}{105}+\frac{110-x}{107}=0\)

\(\left(110-x\right).\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

\(\Rightarrow110-x=0\)( vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)  )

\(\Rightarrow x=110\)

vậy x=110

Châu sai

\(\Leftrightarrow\left(\frac{x+14}{86}+1\right)+\left(\frac{x+15}{85}+1\right)+\left(\frac{x+16}{84}+1\right)+\left(\frac{x+17}{83}+1\right)+\left(\frac{166}{4}-4\right)=0\)

\(\Leftrightarrow\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)

\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(x+100\right)=0\Rightarrow x=-100\left(\text{vì }\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)\ne0\)