\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)

Từ đó bạn tìm những giá trị của x nha!

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{-13}{12}\)

Ta có ( x - 5 )( y + 3 ) = -9

Vì x; y ϵ Z nên x - 5; y + 3 ϵ Z

Vậy x - 5; y + 3 ϵ Ư_{( -9 )} = { -1; 1; -3; 3; -9; 9 }

Lập bảng giá trị

Vậy các cặp số nguyên ( x; y ) cần tìm là ( -9; -12 ) ; ( 9; 6 ) ; ( -3; -6 ) ; ( 3; 0 ) ; ( -1; -4 ) ; ( 1; -2 )

(x-5)(y+3)=-1x9=-3x3=-9x1(x,y ϵ z)

=>

Vậy (x,y)=(4,6)=(2,0)=(-4,-2)

20,23*20,24-20,22*8,25-20,22*5,9-2,022*60,9

=20,23*20,24-20,22(8,25+5,9+6,09)

=20,23*20,24-20,22*20,24

=20,24*(-0,01)

=-0,2024

240.(x-5) = 2².5-20

\(\Rightarrow\)240.(x-5) = 4 . 5 - 20

\(\Rightarrow\)240.(x-5) = 20 - 20

\(\Rightarrow\)240.(x-5) = 0

\(\Rightarrow\)x-5           = 0 : 240

\(\Rightarrow\)x-5           = 0

\(\Rightarrow\)x              = 0 + 5

\(\Rightarrow\)x              =  5

\(\Rightarrow\)Vậy x = 5

\(\Rightarrow\)240.(x-5)=0

\(\Rightarrow\)x-5= -240

\(\Rightarrow\)x= -240/-5

\(\Rightarrow\)x=48 

vậy x=48 k mik nha !

Nhân chứ ko phải trừ đâu Minh êi

Chọn B

B

B

`x-1/9 =8/3`

`=>x=8/3 +1/9`

`=> x= 24/9 +1/9`

`=>x= 25/9`

Vậy `x=25/9`

__

`x-2/20=-5/2-x`

`=>x+x=-5/2 +2/20`

`=> 2x= -50/20 +2/20`

`=> 2x= -48/20`

`=> x= -12/5:2`

`=>x=-12/5 xx1/2`

`=>x= -12/10`

`=>x= -6/5`

Vậy `x=-6/5`

`@` `\text {Ans}`

`\downarrow`

`a)`

Đề là \(\dfrac{x-1}{9}=\dfrac{8}{3}\) phải hongg bạn?

\(\dfrac{x-1}{9}=\dfrac{8}{3}\)

`=>` `(x-1)3 = 8*9`

`=> (x-1)3=72`

`=> x-1=72 \div 3`

`=> x-1=24`

`=> x=25`

`b)`

\(\dfrac{x-2}{20}=\dfrac{-5}{2-x}\)

`=>` `(x-2)(2-x)=20*(-5)`

`=> (x-2)(2-x)=-100`

`=> -[(x-2)(x-2)]=-100`

`=> -(x-2)^2 = -100`

`=> (x-2)^2 = 100`

`=> (x-2)^2 = (+-10)^2`

`=>`\(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

\(\left(x-2\right)\left(4x-20\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\4x-20=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\4x=20\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ \left(x-5\right)\left(25-5x?\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\25-5x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\5x=25\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=5\end{matrix}\right.\\ \left(x-4\right)\left(2x-8\right)\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\2x-8=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\2x=8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=4\end{matrix}\right.\)

a,(x-2)(4x-20)=0

=>x-2=0 hoặc 4x-20=0

=>x=2 hoặc x=5

b,(x-5)(25-5)=0

=>x-5=0  ( vì 25-5 ≠0)

=>x=5

c,(x-4)(2x-8)=0

=>x-4=0 hoặc 2x-8=0

=>x=4 

\(\left(x-2\right)\left(4x-20\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\4x-20=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\4x=20\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy x={2;5}

________________________

\(\left(x-5\right)\left(25-5\right)=0\\ \Rightarrow\left(x-5\right)20=0\\ \Rightarrow x-5=0\\ \Rightarrow x=5\)

Vậy x=5

_____________________

\(\left(x-4\right)\left(2x-8\right)=0\\ \Rightarrow2x-8=x-4\\ \Rightarrow2x-8-x=-4\\ \Rightarrow x-8=-4\\ \Rightarrow x=-4+8\\ \Rightarrow x=4\)