a, thiếu đề 

b, -5 ( 2 - x ) + 4 ( x - 3 ) = 10x - 15 

<=> -10 + 5x + 4x - 12 = 10x - 15 

<=> -x = 7 <=> x = 7 

c,-7(5-x)-2(x-10)=15

<=> -35 + 7x - 2x + 20 = 15 <=> 5x = 30 <=> x = 6 

d, -4(x+1) + 89x - 3 = 24 

<=> 85x = 31 <=> x = 31/85 

e, 5(x-30-2(x+6)) = 9 

<=> 5 (-x-49) = 9 <=> -5x = 254 <=> x = -254/5 

b: =>-10+5x+4x-12=10x-15

=>9x-10x=-15+22

=>-x=7

hay x=-7

c: =>-35+7x-2x+20=15

=>5x-15=15

=>x=6

d: =>-4x-4+72x-24=24

=>68x-32=24

hay x=14/17

(Tớ sửa lại đề nhé.)

\(\dfrac{x-5}{x^2-9}-\dfrac{5}{3-x}=\dfrac{4}{x+3}\)

Điều kiện: \(x\ne\pm3\)

\(\Leftrightarrow\dfrac{x-5}{\left(x+3\right)\left(x-3\right)}+\dfrac{5\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(\Rightarrow x-5+5x+15=4x-12\)

\(\Leftrightarrow x+5x-4x=-12-15+5\)

\(\Leftrightarrow2x=-22\)

\(\Leftrightarrow x=-11\)

Sai môn học r bạn ơi

a)x\(\in\left\{-5;-4;-3;-2;-1\right\}\)

b)x\(\in\left\{-2;-1;0;1;2;3\right\}\)

c)x\(\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)

d)x\(\in\left\{1;2;3;4\right\}\)

e)x\(\in\left\{0;1;2;3;4\right\}\)

f)x\(\in\left\{-10;-9;-8;-7\right\}\)

`x/2+x+x/3+x+x+x/4=5 3/4`

`=>3x+x/2+x/3+x/4=23/4`

`=>49/12x=23/4`

`=>x=69/49`

Vậy `x=69/49`

giúp mik với xin các bạn đó mik đnag cần gấp

Soạn đề bằng thanh công cụ cho rõ nha bạn

giúp mik nhé

đi mà giúp đi

e)175:x=345:43

3,4:x=6,8:43

3,4:x=5,1

x=3,4:5,1

x=23

f)x:45 =258:54

x:45=52

x=52×45

x=2

g)(x×0,25+2012)×2013=(50+2012)×2013g)

(x×0,25+2012)×2013=2062×2013

x×0,25+2012=(2062×2013):2013

x×0,25+2012=2062

x×0,25=2062−2012

x×0,25=50

x=50:0,25

x=200

h)(x−12)×53=74−12h)

(x−12)×53=54

x−12=54:53

x−12=34

x=34+12

x=54

\(1,\)

\(2x\left(x-3\right)-\left(3-x\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

\(x^4-x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

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

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

x=-6/19 (^-^)b

cộng cả 2 vế với -1

x=105

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

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

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

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

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

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

<=> x = 105

Vậy nghiệm phương trình là x = 105

#muon roi ma sao con

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

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

( cả 2 vế trừ đi 3 và từng phân thức trừ đi 1 ) 

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

Vậy tập nghiệm của pt là S = { 105 }