Len photomax

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

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

=>x+100=0

hay x=-100

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

=>x+100=0

hay x=-100

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

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

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

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

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

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........

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....

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)

7)    \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)

\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+36}{65}+1+\frac{x+40}{60}+1\)

\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)            (vì 1/75 + 1/70 - 1/65 - 1/60  \(\ne\)0)

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

Vậy.....

7)    \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)

\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1\)

\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)

\(\Leftrightarrow\)\(x+100=0\)    (1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)

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

Vậy...

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

\(\Rightarrow\)

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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