c: \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\\x=-5\\x=5\end{matrix}\right.\)

\(\frac{1}{x}=\frac{2}{y}\Rightarrow\frac{x}{1}=\frac{y}{2}=\frac{x+y}{1+2}=\frac{4}{3}\)

=> x=4/3.1=4/3

y=4/3.2=8/3

(x + 3) + (x + 7) + (x + 11) + ... + (x + 79) = 860

=> x + 3 + x + 7 + x + 11 + ... + x + 79 = 860

=> (x + x + x + ... + x) + (3 + 7 + 11 + ... + 79) = 860

=> 20x + (79 + 3).20 : 2 = 860

=> 20x + 82.20 : 2 = 860

=> 20x + 82.10 = 860

=> 20x + 820 = 860

=> 20x = 40

=> x = 2

vậy_

#)Giải :

\(\left(x+3\right)+\left(x+7\right)+...+\left(x+79\right)=860\)

\(\left(x+x+...+x\right)+\left(3+7+...+79\right)=860\)(trong mỗi ngoặc có 20 số hạng)

\(x\times20+\frac{\left(79+3\right)\times20}{2}=860\)

\(x\times20+820=860\)

\(x\times20=860-820\)

\(x\times20=40\)

\(x=40\div20\)

\(x=2\)

Lời giải 

\(\left(x+x+...+x\right)+\left(3+7+...+79\right)=860\)

\(x.20+\frac{\left(79+3\right).20}{2}=860\)

\(x.20+820=860\)

\(x.20=40\)

\(x=2\)

NHANH NHANH GIÚP MÌNH VỚI NHÉ MÌNH ĐANG CẦN GẤP 

Bài 7:

a, \(x\) = \(\dfrac{1}{5}\) + \(\dfrac{2}{11}\)

    \(x\) = \(\dfrac{11}{55}\) + \(\dfrac{10}{55}\)

     \(x=\dfrac{21}{55}\)

b, \(\dfrac{x}{15}\) = \(\dfrac{3}{5}\) - \(\dfrac{2}{3}\)

    \(\dfrac{x}{15}\) = \(\dfrac{9}{15}\) - \(\dfrac{10}{15}\)

      \(\dfrac{x}{15}\) = \(\dfrac{1}{15}\)

      \(x\) = 1

c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\)= \(\dfrac{85}{x}\)

     \(\dfrac{33}{24}\) + \(\dfrac{52}{24}\) = \(\dfrac{85}{x}\)

       \(\dfrac{85}{24}\) = \(\dfrac{85}{x}\)

        24 = \(x\)

EM CẢM ƠN CÔ Ạ

a) Ta có: \(x-12=\left(-9\right)-15\)

\(\Rightarrow x-12=-24\)

\(\Rightarrow x=-12\)

b) \(2-x=17-\left(-5\right)\)

\(\Rightarrow2-x=22\)

\(\Rightarrow x=2-22\)

\(\Rightarrow x=-20\)

c) \(11-\left(15+11\right)=x-\left(25-9\right)\)

\(\Rightarrow-15=x-16\)

\(\Rightarrow x=1\)

d) \(x-\left(17-x\right)=x-7\)

\(\Rightarrow x-17+x=x-7\)

\(\Rightarrow x+x-x=-7+17\)

\(\Rightarrow x=10\)

e) \(9-25=\left(7-x\right)-\left(25+7\right)\)

\(\Rightarrow9-25=7-x-25-7\)

\(\Rightarrow-16=-25-x\)

\(\Rightarrow x=-9\)

g) \(x-13=21-x\)

\(\Rightarrow x+x=21+13\)

\(\Rightarrow2x=34\)

\(\Rightarrow x=17\)

a, x - 12 = ( -9 ) - 15 

x - 12 = -24

x = -24 + 12

x = -12

b, 2 - x = 17 - ( -5 )

2 - x = 22

x = 2 - 22

x = -20

c, 11 - ( 15 + 11 ) = x - ( 25 - 9 )

11 - 26 = x - 16

-15 = x - 16

x = -15 + 16 

x = 1

d, x - ( 17 - x ) = x - 7

x - 17 + x = x - 7

x + x - 17 = x - 7

x + x - x = -7 + 17

x = 10

e.9 - 25 = ( 7- x ) - ( 25 + 7 )

-16 = ( 7 - x ) -32

7 - x = -32 + 16 

7 - x = 16

x = 7 - 16

x= -9

g, x - 13 = 21 - x

x + x = 21 + 13

2x = 44

x = 44 : 2

x=22

A= -12

B=-20

C=1

D=17

E=-9

G=17

a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)

b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)

c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)

d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)

a) Ta có: 4x-20=0

$\iff 4x=20$

hay x=5

Vậy: S={5}

b) Ta có: $2x+x+12=0$

$\iff 3x+12=0$

$\iff 3x=-12$

hay x=-4

a.

\(A=B\)

\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\);ĐK:\(x\ne\pm2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)

\(\Leftrightarrow x^2+4x+4-x^2+4x-4+16=0\)

\(\Leftrightarrow8x+16=0\)

\(\Leftrightarrow8\left(x+2\right)=0\)

\(\Leftrightarrow x=-2\left(ktm\right)\)

Vậy không có giá trị x thỏa mãn A=B

b.

\(A:B=\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{-16}{\left(x-2\right)\left(x+2\right)}< 0\)

\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+4x-4}{-16}< 0\)

\(\Leftrightarrow\dfrac{8x}{-16}< 0\)

\(\Leftrightarrow\dfrac{8x}{16}>0\)

\(\Leftrightarrow\dfrac{x}{2}>0\)

\(\Leftrightarrow x>0\)

\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}\cdot\left(x-7\right)^{10}=0\)

\(\left(x-7\right)^{x+1}\cdot\left[1-\left(x-7\right)^{10}\right]=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=7\\\left(x-7\right)^{10}=1\end{cases}}\)

Bạn xét nốt TH \(\left(x-7\right)^{10}=1\)nhé

\(\Rightarrow\hept{\begin{cases}x=7\\\hept{\begin{cases}x-7=1\\x-7=-1\end{cases}}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=7\\\hept{\begin{cases}x=7\\x=6\end{cases}}\end{cases}}\)

( x - 7 )^{x + 1} - ( x - 7 )^{x + 11} = 0

=> ( x - 7 )^{x + 1} .[ 1 - ( x - 7 )¹⁰] = 0

=>( x - 7 )^{x + 1} = 0 hoặc 1 - ( x - 7 )¹⁰ = 0

x - 7 = 0 hoặc ( x - 7 )¹⁰ = 0

x - 7 = 0 hoặc / x - 7 / = 1

x - 7 = 0 hoặc x - 7  = 1  hoặc  x - 7 = -1

=> x = 7 hoặc x = 8 hoặc x = 6

Hok tốt !

(x-7)^x+1 - (x-7)^x+11 =0

=> (x-7)^x+1. [ 1- (x-7)^10 ]

=> (x-7)^x+1 = 0 hoặc 1-(x-7)^10 = 0

=> x-7=0 hoặc (x-7)^10=1

=> x =7 hoặc x-7 =1 hoặc x-7 = -1

=> x= 7 hoặc x= 8 hoặc x=6