\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

<=>  \(\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)

<=>  \(\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)

<=>  \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

Nhận thấy:   \(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\)

=>  \(x-105=0\)

<=>  \(x=105\)

\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

\(\Leftrightarrow\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}-\frac{x-100}{5}-\frac{x-101}{4}-\frac{x-102}{3}=0\)

\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)-\left(\frac{x-100}{5}-1\right)-\left(\frac{x-101}{4}-1\right)-\left(\frac{x-102}{3}-1\right)=0\)

\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

\(\Leftrightarrow x-105=0\left(Vì\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)

\(\Leftrightarrow x=105\)

Đề bài: Tìm x

Ta có : \(\frac{3}{x-5}=\frac{-4}{x+2}\)

\(\Rightarrow3\left(x+2\right)=-4\left(x-5\right)\)

\(3x+6=-4x+20\)

\(3x+4x=-6+20\)

\(7x=14\)

\(x=14\div7\)

\(x=2\)

Vậy \(x=2\).

\(\frac{3}{x-5}=\frac{-4}{x+2}\)

\(\Rightarrow3x+6=-4x+20\)

\(\Rightarrow7x=14\)

\(\Rightarrow x=2\)

  3/x-5= - 4/x+2

 =>3.(x+2)=-( - 4).(x-5)

=> 3x+3.2= - 4x-(-20)

=>3x+6= - 4x+20

=>4x+3x=20-6

=> 7x=14

=>x=14:7

=>x=2

7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x + 2) - (x + 4)

=> 14x - 35 - 35x + 10 + 10x -14 = x + 2 - x - 4

=> (14x - 35x + 10x) + (-35 + 10 - 14) = -2

=> -11x + (-39) = -2

=> -11x = -2 - (-39)

=> -11x = 37

=> x = \(\frac{-37}{11}\)

7(2x - 5) - 5(7x - 2) + 2(5x - 7)             = (x + 2) - (x + 4)

=> 14x - 35 - 35x + 10 + 10x -14         = x + 2 - x - 4

=> (14x - 35x + 10x) + (-35 + 10 - 14) = -6
=> -11x - 39                                         = -6
=> -11x                                                = -6+39
=> -11x                                                = 33
=>      x                                                = 33:(-11)
=>      x                                                = -3

5^x+3(5-1*3)=5^11*2

5^x+3*2=5^11*2

5^x+3=5^11

x+3=11

x=8

BÀI \(1\):

\(\left(-1005\right).\left(x+2\right)=0\)

\(x+2=0:\left(-1005\right)\)

\(x+2=0\)

\(x=0-2\)

\(x=-2\)

BÀI \(2\):

\(x+x+x+91=-2\)

\(3x=\left(-2\right)-91\)

\(3x=-93\)

\(x=\left(-93\right):3\)

\(x=-31\)

BÀI \(3\):

\(\left|5x+1\right|=11\)

Có hai trường hợp:

\(TH^{ }1:_{ }5x+1=11\)

\(5x=11-1\)

\(5x=10\)

\(x=10:5\)

\(x=2\)

\(TH2:^{ }5x+1=-11\)

\(5x=\left(-11\right)-1\)

\(5x=-12\)

\(x=\left(-12\right):5\)

\(x=-2,4\)

\(k\)\(minh\)\(nhe\)\(.\)

a, \(\left(-2\right)^2-2x^2=-50.\)

\(\Leftrightarrow4-2x^2=-50\)

\(\Leftrightarrow2x^2-54=0\)

\(\Leftrightarrow x^2-27=0\)

\(\Leftrightarrow\left(x-3\sqrt{3}\right)\left(x+3\sqrt{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\sqrt{3}\\x=-3\sqrt{3}\end{cases}}\)

\(\left(-2\right)^2-2x^2=-50\)

\(\Leftrightarrow x^2=\frac{-50-\left(-2\right)^2}{-2}\)

\(\Leftrightarrow x^2=27\)

\(\Leftrightarrow x=\pm\sqrt{27}\)

\(\Leftrightarrow x=\pm3\sqrt{3}\)

\(\frac{2}{x}-\frac{5}{x}=-9\)

\(\Leftrightarrow\frac{2-5}{x}=-9\)

\(\Leftrightarrow-3=-9x\)

\(\Leftrightarrow x=\frac{-3}{-9}=\frac{1}{3}\)

\(2|x|-5|x|=-9\)

*Với x < 0 ta có phương trình: 

\(-2x+5x=-9\)

\(\Leftrightarrow3x=-9\)

\(\Leftrightarrow x=-3\)(TM)

*Với \(x\ge0\)ta có phương trình: 

\(2x-5x=-9\)

\(\Leftrightarrow-3x=-9\)

\(\Leftrightarrow x=3\)(TM)

Vậy ..........

a 24x-15x=75+33

9x=108

x=108:9=12

b 3-2/5x=2/3

2/5x=3-2/3=1/3

x=1/3:2/5=5/6

Tk & kết bạn với mik nha

24x+75=15x-33

24x-15x= -33-75

9x       = -108

x        = -108:9= -12

CÂU A PHẢI LÀ -12
 

\(\frac{x^3-x^2-x-2}{x^5-3x^4+4x^3-5x^2+3x-2}\)

\(=\frac{x^3-2x^2+x^2-2x+x-2}{x^5-2x^4-x^4+2x^3+2x^3-4x^2-x^2+2x+x-2}\)

\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)

\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)

\(=\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(x^4-x^3+2x^2-x+1\right)}=\frac{x^2+x+1}{x^4-x^3+2x^2-x+1}\)

\(\left(x-3\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)

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

\(\Leftrightarrow3x-24=0\Leftrightarrow x=8\)