$\frac{x-11}{89}+\frac{x-13}{87}+\frac{x-15}{85}+\frac{x-17}{83}=4$$=>\left(\frac{x-11}{89}-1\right)+\left(\frac{x-13}{87}-1\right)+\left(\frac{x-15}{85}-1\right)+\left(\frac{x-17}{83}-1\right)=0$$=>\frac{x-100}{89}+\frac{x-100}{87}+\frac{x-100}{85}+\frac{x-100}{83}=0$$=>\left(x-100\right)\left(\frac{1}{89}+\frac{1}{87}+\frac{1}{85}+\frac{1}{83}\right)=0$=> x-100 =0 => x=100Vậy nghiệm là 100 

ta có 0=\(\frac{x-11}{89}-1+\frac{x-13}{87}-1+\frac{x-15}{85}-1+\frac{x-17}{83}-1=0\)

0=\(\frac{x-11-89}{89}+\frac{x-13-87}{87}+\frac{x-15-85}{85}+\frac{x-17-83}{83}\)

0=\(\frac{x-100}{89}+\frac{x-100}{87}+\frac{x-100}{85}+\frac{x-100}{83}\)

0=(x-100)(\(\frac{1}{89}+\frac{1}{87}+\frac{1}{85}+\frac{1}{83}\))

vì \(\frac{1}{89}+\frac{1}{87}+\frac{1}{85}+\frac{1}{83}\)khác 0      suy ra x-100=0 

                                                                                    x=100

x−11/89+x−13/87+x−15/85+x−17/83=4

=>(x−11/89−1)+(x−13/87−1)+(x−15/85−1)+(x−17/83−1)=0

=>x−100/89+x−100/87+x−100/85+x−100/83=0

=>(x−100)(1/89+1/87+1/85+1/83)=0

=> x-100 =0 

Vậy nghiệm là 100

a) (x + 5) / 95 + (x +10)/90 + (x + 15)/85 + (x + 20)/80 = -4

<=> (x + 5)/95 + (x + 5)/90 + 5/90 + (x + 5)/85 + 10/85+ (x + 5)/80 + 15/80 = -4

<=> (x + 5)(1/95+1/90+1/85+1/80) =-4 -5/90-10/85-15/85

<=> (x + 5)(1/95+1/90+1/85+1/80)= -1-(1 + 5/90 )-(1 + 10/85)  - (1 + 15/80)

<=>(x + 5)(1/95+1/90+1/85+1/80) = -1 - 95/90 - 95/85 - 95/80

<=>(x + 5)(1/95+1/90+1/85+1/80) = -95 (1/95+1/90+1/85+1/80)

<=> x + 5 = -95 => x = -100

b) Làm tương tự phần a => Kết quả x = 100

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)

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

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

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

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

c)Ta có: \(x^4+3x^3+4x^2+3x+1=0\)

\(\Leftrightarrow x\left(x^3+2x^2+2x+1\right)+1\left(x^3+2x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+2x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)

Ta có: \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\) nên vô nghiệm

Suy ra x + 1 =0 hay x = -1

X=0 hoặc -1

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

\(\dfrac{x+11}{89}+\dfrac{x+13}{87}=\dfrac{x+15}{85}+\dfrac{x+17}{83}\\\Leftrightarrow\left(\dfrac{x+11}{89}+1\right)+\left(\dfrac{x+13}{87}+1\right)=\left(\dfrac{x+15}{85}+1\right)+\left(\dfrac{x+17}{83}+1\right)\\ \Leftrightarrow\dfrac{x+100}{89}+\dfrac{x+100}{87}-\dfrac{x+100}{85}-\dfrac{x+100}{83}=0\\ \Leftrightarrow\left(x+100\right)\left(\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}\right)=0\\ \Leftrightarrow x+100=0\left(vì\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}\ne0\right)\\ \Leftrightarrow x=-100\)

\(\dfrac{x-11}{89}\) +\(\dfrac{x-13}{87}+\dfrac{x-15}{85}+\dfrac{x-17}{83}\)=4

⇔\(\dfrac{x-11}{89}+\dfrac{x-13}{87}+\dfrac{x-15}{85}+\dfrac{x-17}{83}-4=0\)

⇔\(\dfrac{x-11}{89}-1+\dfrac{x-13}{87}-1+\dfrac{x-15}{85}-1+\dfrac{x-17}{83}-1=0\)

⇔\(\dfrac{x-100}{89}+\dfrac{x-100}{87}+\dfrac{x-100}{85}+\dfrac{x-100}{83}=0\)

⇔\(\left(x-100\right)\left(\dfrac{1}{89}+\dfrac{1}{87}+\dfrac{1}{85}+\dfrac{1}{83}\right)=0\)

\(Do\) \(\dfrac{1}{89}+\dfrac{1}{87}+\dfrac{1}{85}+\dfrac{1}{83}\)≠\(0\) nên x-100=0 nên x=100

KL........

⇔(\(\dfrac{x-11}{89}\)-1)+(\(\dfrac{x-13}{87}\)-1)+(\(\dfrac{x-15}{85}\)-1)+(\(\dfrac{x-17}{83}\)-1)=0

⇔\(\dfrac{x-100}{89}\)+\(\dfrac{x-100}{87}\)+\(\dfrac{x-100}{85}\)+\(\dfrac{x-100}{83}\)=0

⇔(x-100)(\(\dfrac{1}{89}\)+\(\dfrac{1}{87}\)+\(\dfrac{1}{85}\)+\(\dfrac{1}{83}\))=0 (1)

Do 1/89+1/87+1/85+1/83≠0 nên (1)⇔x-100=0 ⇔x=100

Vậy tập nghiệm của PT là S=\(\left\{100\right\}\)

Phương trình 1:
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\(\Rightarrow\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}-10=0\)
\(\Rightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
\(\Rightarrow\frac{x-85-15}{15}+\frac{x-74-26}{13}+\frac{x-67-33}{11}+\frac{x-64-36}{9}=0\)
\(\Rightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Do \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.

Phương trình 3:
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
\(\Rightarrow\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)=0\)
\(\Rightarrow\frac{1909-x+91}{91}+\frac{1907-x+93}{93}+\frac{1905-x+95}{95}+\frac{1903-x+97}{97}=0\)
\(\Rightarrow\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
\(\Rightarrow\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Do \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow2000-x=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000.

Phương trình 4:
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Rightarrow\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}-15=0\)
\(\Rightarrow\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
\(\Rightarrow\frac{x-90-10}{10}+\frac{x-76-24}{12}+\frac{x-58-42}{14}+\frac{x-36-64}{16}+\frac{x-15-85}{17}=0\)
\(\Rightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
Do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.