Chọn A.

Do 2^x + 2^-x  > 0

Nên 

1. A = 6x^3 - 3x^2 + 2.|x| + 4 với x = -23

Thay x = -23 vào biểu thức trên, ta có:

A = 6.(-23)^3 - 3.(-23)^2 + 2.|-23| + 4

A = -74539

2. B = 2.|x| - 3.|y| với x = 12; y = -3

Thay x = 12; y = -3 vào biểu thức trên, ta có:

B = 2.|12| - 3.|-3|

B = 15

3. |2 + 3x| = |4x - 3|

ta có: 2 + 3x = \(\hept{\begin{cases}4x-3\Leftrightarrow4x-3\ge0\Leftrightarrow x\ge\frac{3}{4}\\-\left(4x-3\right)\Leftrightarrow4x-3< 0\Leftrightarrow x< \frac{3}{4}\end{cases}}\)

Nếu x >= 3/4, ta có phương trình:

2 + 3x = 4x - 3

<=> 3x - 4x = -3 - 2

<=> -x = 5

<=> x = 5 (TM)

Nếu x < 3/4, ta có phương trình:

 2 + 3x = -(4x - 3)

<=> 2 + 3x = -4x + 3

<=> 3x + 4x = 3 - 2

<=> 7x = 1

<=> x = 1/7 (TM) 

Vậy: tập nghiệm của phương trình là: S = {5; 1/7}

a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)

a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)

= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)

= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)

Có vài bước mình làm tắc á nha :>

a,

23 x (7 - 4) = 23 x 3 = 69

23 x 7 - 23 x 4 = 161 - 92 = 69

Ta có: 23 x (7 - 4) = 23 x 7 - 23 x 4

b,

(8 - 3) x 9 = 5 x 9 = 45

8 x 9 - 3 x 9 = 72 - 27 = 45

Ta có: (8 - 3) x 9 = 8 x 9 - 3 x 9

a) (16,32 + 20,8) : 4 - 2,12 x 3

= 37,12 : 4 - 6,36

= 9,28 - 6,36

= 2,92

b) 23% : 4 + 31,5% x 2

= 0,23 : 4 + 0,315 x 2

= 0,0575 + 0,63

= 0,6875

a) (16,32 + 20,8) : 4 - 2,12 x 3

= 37,12 : 4 - 6,36

= 9,28 - 6,36

= 2,92

b) 23% : 4 + 31,5% x 2

= 0,23 : 4 + 0,315 x 2

= 0,0575 + 0,63

= 0,6875

                               Đây nhé bạn ###

a)

A=\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5x-5}\)

\(\Leftrightarrow\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5\left(x-1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+1\\x=0-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

MTC: 5(x-1)(x+1)

\([\dfrac{5\left(x+1\right)\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}-\dfrac{5\left(x-1\right)\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}]\div\dfrac{2x\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow[5\left(x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)]\div2x\left(x+1\right)\)

\(\Leftrightarrow[5\left(x+1\right)^2-5\left(x-1\right)^2]\div2x^2+2x\)

\(\Leftrightarrow[5\left(x^2+2x+1\right)-5\left(x^2-2x+1\right)]\div2x^2+2x\)

\(\Leftrightarrow(5x^2+10x+5-5x^2+10x-5)\div2x^2+2x\)

\(\Leftrightarrow20x\div\left(2x^2+2x\right)\)

\(\Leftrightarrow10x+10\)

 x*17 + x *23 = 120

x*(17+23)=120

x*40=120

x=120:40

x=3

A