1.(2x-9)chia het cho (x-5)

suy ra 2x-9 chia het cho (x-5)

ta co (x-5) chia het cho (x-5)

suy ra 2.(x-5) chia het cho (x-5)

suy ra 2x-10 chia het cho (x-5)

suy ra (2x-10)-(2x-9) chia het cho (x-5)

suy ra 2x-10-2x+9 chia het cho (x-5)

suy ra -1 chia het cho (x-5) 

suy ra x-5 thuoc Ư(-1)

Ư(-1)=...

neu x-5=1 suy ra x=6

neu x-5=-1 ...

vay x=...

chào bạn

xin chào mọi người

219-7.(x-5)=100

<=> 7.(x-5)=119

<=> x-5=17

<=> x=22

Đáp số: x=22

219-7.(x-5)=100

        7.(x-5)=219-100

         7.(x-5)=119

              (x-5)=119:7

               (x-5)=17

                x=17+5

                x=22

vậy x=22

                              Bài làm :

Ta có :

\(219-7.\left(x-5\right)=100\)

\(\Leftrightarrow7. \left(x-5\right)=119\)

\(\Leftrightarrow x-5=17\)

\(\Leftrightarrow x=22\)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(\dfrac{5}{x+2}-\dfrac{x-1}{x-2}=\dfrac{12}{x^2-4}+1\left(x\ne-2;x\ne2\right)\)

\(< =>\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

suy ra

`5x-10-(x^2 +2x-x-2)=12+x^2 -4`

`<=>5x-10-x^2 -2x+x+2-12-x^2 +4=0`

`<=>-x^2 -x^2 +5x-2x+x-10+2+4=0`

`<=>-x^2 +4x-4=0`

`<=>x^2 -4x+4=0`

`<=>(x-2)^2 =0`

`<=>x-2=0`

`<=>x=2(ktmđk)`

vậy phương trình vô nghiệm

ĐKXĐ: \(x\ne\pm2\)

\(\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow5\left(x-2\right)-\left(x-1\right)\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow5x-10-\left(x^2+x-2\right)=12+x^2-4\)

\(\Leftrightarrow-x^2+4x-8=x^2+8\)

\(\Leftrightarrow2x^2-4x+16=0\)

\(\Leftrightarrow2\left(x-1\right)^2+14=0\)

Do \(\left\{{}\begin{matrix}2\left(x-1\right)^2\ge0\\14>0\end{matrix}\right.\) ;\(\forall x\)

\(\Rightarrow2\left(x-1\right)^2+14>0\)

Vậy phương trình đã cho vô nghiệm

x/y = 2/5 ⇒ x/2 = y/5

⇒ x/5 = 2y/10

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

x/2 = 2y/10 = (x + 2y)/(2 + 10) = 36/12 = 3

x/2 = 3 ⇒ x = 2 . 3 = 6

y/5 = 3 ⇒ y = 5 . 3 = 15

Vậy x = 6; y = 10

x+ ( 1/2 x3/4 )=5

x+ 3/8 =5

x=5-3/8

x=37/8

cảm ơn pẹn

2/3 x Y = 23/18
Y = 23/18 : 2/3
Y = 23/12

kq 

2/3 x Y = 69/54

Y = 69/54 : 2/3

=> Y = 207/108 = 23/12 =)

à giờ thì m hiểu r

2/3xy=8/18+15/18

2/3xy=23/18

y=23/18 : 2/3

y=23/18 x3/2

y=69/36

y=23/12 

a) \(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)

\(\Rightarrow\left(\frac{3}{5}-\frac{2}{3}-1\right).x=\frac{-5}{21}:\frac{1}{7}=\frac{-5}{3}\)

\(\Rightarrow\frac{-16}{15}.x=\frac{-5}{3}\Rightarrow x=\frac{-5}{3}:\frac{-16}{15}=\frac{25}{16}\)

b) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{36}\)

\(\Rightarrow\left(x-\frac{1}{4}\right)^2=\left(±\frac{1}{6}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{1}{6}\\x-\frac{1}{4}=\frac{-1}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)

c) \(14-\left|\frac{3x}{2}-1\right|=9\)

\(\Rightarrow\left|\frac{3x}{2}-1\right|=14-9=5\)

\(\Rightarrow\orbr{\begin{cases}\frac{3x}{2}-1=5\\\frac{3x}{2}-1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{3x}{2}=6\\\frac{3x}{2}=-4\end{cases}}\Rightarrow\orbr{\begin{cases}3x=12\\3x=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{-8}{3}\end{cases}}\)