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....

Đáp án :C

là C.10

la c nha

\(\Leftrightarrow-x^2=-9\cdot\dfrac{4}{49}=\dfrac{-36}{49}\\ \Leftrightarrow x^2=\dfrac{36}{49}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{7}\\x=-\dfrac{6}{7}\end{matrix}\right.\)

a) = (x+3).(x-3)^2-(x-3)(x+3)^2

=(x^2-9)(x-3)-(x^2-9)(x+3)

=(x^2-9)(x-3-x-3)

=-6(x^2-9)

các câu còn lại tương tự

\(a,\left(x+3\right)\left(x^2-3x+9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

\(=x^3+3-\left(x^3-3\right)\)

\(=x^3+3-x^3+3\)

\(=6\)

\(b,\left(x-5\right)\left(x^2+5x+25\right)-\left(x+5\right)\left(x^2-5x+25\right)\)

\(=x^3-5^3-x^3-5^3\)

\(=-125-125\)

\(=-250\)

Khoang cach cua day so tren la 5

So cac so hang la:

[54-4]:5+1=11[so]

Tong day so tu 4 den 54 la

[54+4]x11:2=319

=>x x 11+319=1430

x x 11         =1430-319

x x 11         =1111

 x               =1111:11

x                 =101

x + (4 + 9 + 14 + ... + 49 + 54 ) = 1430

Số các số hạng của các dãy số trong ngoặc là:

(54 - 4 ) : 5 + 1 = 11 (số)

Tổng của dãy số đó là:

( 54 + 4 ) x 11 : 2 = 319

Ta có biểu thức:

x + 319 = 1430

       x = 1430 - 319

       x = 1111

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...

9: \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)

=>x-99=0

hay x=99

7: \(\Leftrightarrow\left(\dfrac{x+25}{75}+1\right)+\left(\dfrac{x+30}{70}+1\right)=\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+40}{60}+1\right)\)

=>x+100=0

hay x=-100

8:

Sửa đề: \(\dfrac{99-x}{101}+\dfrac{97-x}{103}+\dfrac{95-x}{105}+\dfrac{93-x}{107}=-4\) 

\(\Leftrightarrow\left(\dfrac{99-x}{101}+1\right)+\left(\dfrac{97-x}{103}+1\right)+\left(\dfrac{95-x}{105}+1\right)+\left(\dfrac{93-x}{107}+1\right)=0\)

=>200-x=0

hay x=200