
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


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 = \(\frac{20}{3}\)

`@` `\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}`

1)-(-3)3-(2x+1)3=152
27-(2x+1)3=152
(2x+1)3=-125
(2x+1)3=(-5)3
2x+1=-5
2x=-4
x=-2
2)-(x+6)-34=12+3x
-x-6-34=12+3x
-x-3x=12+6+34
-4x=52
x=-13

`@` `\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?

\(8x+2x=25\cdot2^2\Leftrightarrow8x+2x=25\cdot4\)
\(\Leftrightarrow8x+2x=100\Leftrightarrow x\left(8+2\right)=100\)
\(\Leftrightarrow x\cdot10=100\Leftrightarrow\orbr{\begin{cases}x=100:10\\x\cdot10=10\cdot10\end{cases}}\)
\(\Leftrightarrow x=10\)
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

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
\(8x+\left(-34\right)=-2x+6\)
\(\Leftrightarrow8x+2x=6+34\)
\(\Leftrightarrow10x=40\)
\(\Leftrightarrow x=4\)
\(8x+\left(-34\right)=-2x+6\)
\(\Leftrightarrow8x+2x=6+34\)
\(\Leftrightarrow10x=40\)
\(\Leftrightarrow x=\frac{40}{10}=4\)