`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

Ta có: \(2x\left(8x-1\right)^2\cdot\left(4x-1\right)=9\)

\(\Leftrightarrow\left(8x-1\right)^2\cdot\left(8x^2-2x\right)=9\)

\(\Leftrightarrow\left(64x^2-16x+1\right)\left(8x^2-2x\right)-9=0\)

\(\Leftrightarrow512x^4-128x^3-128x^3+32x^2+8x^2-2x-9=0\)

\(\Leftrightarrow512x^4-256x^3+40x^2-2x-9=0\)

\(\Leftrightarrow256x^3\left(2x-1\right)+40x^2-20x+18x-9=0\)

\(\Leftrightarrow256x^3\left(2x-1\right)+20x\left(2x-1\right)+9\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(256x^3+20x+9\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(256x^3+64x^2-64x^2-16x+36x+9\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left[64x^2\left(4x+1\right)-4x\left(4x+1\right)+9\left(4x+1\right)\right]=0\)

\(\Leftrightarrow\left(2x-1\right)\left(4x+1\right)\left(64x^2-4x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\4x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{1}{2};-\dfrac{1}{4}\right\}\) 

Tên cậu hay đó chứ

Tên thiệt là Lê Vâng Lời ah?

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

f: Ta có: \(2x^2+7x+3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

ko phải "-" đâu mà là nhân đấy

giữa 2 ngoặc đó

https://olm.vn/hoi-dap/detail/55678450227.html

a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3