Bài 1:

a) \(8xy^2+24x^2y-32x^3y^2=8xy\left(y+3x-4x^2y\right)\)

b) \(x^2-16x-y^2+64=\left(x-8\right)^2-y^2=\left(x-8-y\right)\left(x-8+y\right)\)

Bài 2:

\(\left(x-4\right)^2-\left(12x+x^2\right)=6\)

\(\Rightarrow x^2-8x+16-12x-x^2=6\)

\(\Rightarrow20x=10\Rightarrow x=\dfrac{1}{2}\)

\(1,\\ =8xy\left(y+3x-4x^2y\right)\\ =\left(x-8\right)^2-y^2=\left(x-y-8\right)\left(x+y-8\right)\)

\(2,\Leftrightarrow x^2-8x+16-12x-x^2=6\\ \Leftrightarrow-20x=-10\\ \Leftrightarrow x=2\)

a. 523 – x = 301

x = 523 – 301

x = 222

b. x : 642 = 6

x = 642 x 6

x = 3 852

mình cần gấp

a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)

\(\Leftrightarrow2x^2-2x-2x^2=-6\)

\(\Leftrightarrow x=3\)

b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

0=(2x+1)²

4x² + 4x + 1 = 0

4x² = 0 hay 4x + 1 = 0

x = 2 hay x= \(-\dfrac{1}{4}\)

(2x+1)=(2x+1)

=> (2x+1)^4 - (2x+1)^6 = 0

=> (2x+1)^4 * [1 - (2x+1)^2] = 0

=> \(\left[{}\begin{matrix}\left(2x+1\right)^4=0\\\left[1-\left(2x+1\right)^2\right]=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}2x+1=0\\\left(2x+1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\\left[{}\begin{matrix}2x+1=1\\2x+1=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x=0\\2x=-2\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\\x=-1\end{matrix}\right.\)Vậy x\(\in\){0;-1;\(\dfrac{1}{2}\)}

\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)

a) \(\Leftrightarrow x^3-8-x^3+9x=1\)

\(\Leftrightarrow9x=9\Leftrightarrow x=1\)

b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)

\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)

Xét cấp số cộng 1, 6, 11, ..., 96.

Ta có: 96 = 1 + 5(n − 1) ⇒ n = 20

Suy ra

Và 2x.20 + 970 = 1010

Từ đó x = 1

(x+1)(2x-6)(2x^2+1)

=> 

(x+1)(2x-6)(2x^2+1) = 0

=> x + 1 = 0  => x = -1

2x - 6 = 0 => x = 3

2x² + 1 = 0 => x = -\(\sqrt{0,5}\)

làm bậy thui!!!

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