Ta có: 8x+(-34)=2x+6

\(\Rightarrow\)8x-34-2x-6=0

hay 6x-40=0

\(\Rightarrow\)6x=40

hay x=\(\frac{40}{6}=\frac{20}{3}\)

Vậy: \(x=\frac{20}{3}\)

8x+(-34) = 2x+6

8x -34 = 2x+6

8x -2x=6+34

6x=40

x=40:6

=> x=\(\frac{40}{6}\)=\(\frac{20}{3}\)

Vậy x = \(\frac{20}{3}\)

Ta có:

x²y + xy - x = 6

x²y + xy - x -1 = 5

xy.(x + 1) - (x + 1) = 5

(x = 1).(xy - 1) = 1.5 = (-1).(-5) = 5.1 = (-5).(-1)

Ta có bảng giá trị;

Vậy (x;y) = (-2;2) ; (-6;0)

Cảm ơn bạn ạ!

`8x+2x=25 . 2^2`

`=>(8+2)x=25 . 4`

`=>10x=100`

`=>x=100:10`

`=>x=10`

<=>x+127=43

<=>x=43-127

<=>x=-84

`@` `\text {Ans}`

`\downarrow`

`c)`

`(34 - 2x)(2x - 6) = 0`

`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)

Vậy, `x \in {17; 3}`

`d)`

`(2019 - x)(3x - 12) = 0`

`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=12\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)

Vậy, `x \in {2019; 4}.`

`@` `\text {Kaizuu lv uuu}`

\(6\cdot x+9=15\)

\(\Rightarrow6\cdot x=15-9\)

\(\Rightarrow6\cdot x=6\)

\(\Rightarrow x=\dfrac{6}{6}=1\)

_______________

\(43+2x=49\)

\(\Rightarrow2x=49-43\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=\dfrac{6}{2}=3\)

____________

\(x:2+5=11\)

\(\Rightarrow x:2=11-5\)

\(\Rightarrow x:2=6\)

\(\Rightarrow x=6\cdot2=12\)

______________

\(77-11x=0\)

\(\Rightarrow11x=77\)

\(\Rightarrow x=\dfrac{77}{11}\)

\(\Rightarrow x=7\)

_______________

\(12-4:x=8\)

\(\Rightarrow4:x=12-8\)

\(\Rightarrow4:x=4\)

\(\Rightarrow x=\dfrac{4}{4}=1\)

_____________

\(x:3+8=11\)

\(\Rightarrow x:3=11-8\)

\(\Rightarrow x:3=3\)

\(\Rightarrow x=3\cdot3=9\)

a: 6x+9=15

=>6x=6

=>x=1

b: 2x+43=49

=>2x=6

=>x=3

c: x:2+5=11

=>x:2=6

=>x=12

d: 77-11x=0

=>7-x=0

=>x=7

e: 12-4:x=8

=>4:x=4

=>x=1

f: x:3+8=11

=>x:3=3

=>x=9

\(8x+2x=25\cdot2^2\Leftrightarrow8x+2x=25\cdot4\)

\(\Leftrightarrow8x+2x=100\Leftrightarrow x\left(8+2\right)=100\)

\(\Leftrightarrow x\cdot10=100\Leftrightarrow\orbr{\begin{cases}x=100:10\\x\cdot10=10\cdot10\end{cases}}\)

\(\Leftrightarrow x=10\)

Vậy giá trị x cần tìm để thỏa mãn yêu cầu của đề bài là \(x=10\)

8x + 2x = 25 . 4

10x = 100

=> x = 100 : 10 = 10

`@` `\text {Ans}`

`\downarrow`

`c)`

`( 34 - 2x ) * ( 2x - 6 ) = 0`

`=>`\(\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=34-0\\2x=0+6\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=34\div2\\x=6\div2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)

Vậy, `x \in {17; 3}`

`d)`

\(\left(2019-x\right)\left(3x-12\right)=0\)

`=>`\(\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019-0\\3x=0+12\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019\\x=12\div3\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2019\\x=4\end{matrix}\right.\)

Vậy,` x \in {2019; 4}`

^{p/s: Bài này hnhu mk làm r mà ạ?}

a/ \(2x^3=8x\)

\(2.8=2x^3\)

\(16=2x^3\)

\(x^3=16:2\)

\(x^3=8\)

\(x=2\)

phần b mk chưa nghiên cứu dc