1. \(\text{6 + 2.(x - 19) = 16}.\)

\(\Leftrightarrow2.\left(x-19\right)=10.\)

\(\Leftrightarrow x-19=5.\)

\(\Leftrightarrow x=24.\)

Vậy \(x=24.\)

2. \(\text{(-240) : x – 16 = 64}.\)

\(\Leftrightarrow\left(-240\right):x=80.\)

\(\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

3. \(2x^3=16.\)

\(\Leftrightarrow x^3=8.\)

\(\Leftrightarrow x=2.\)

Vậy \(x=2.\)

\(\left|x^2+x+16\right|=x^2+\left|x+16\right|\)( vì \(x^2\ge0\))

\(\left|x^2+x-6\right|=x^2+\left|x-6\right|\)(vì \(x^2\ge0\))

\(\left|x+16\right|+\left|x-6\right|=\left|x+16\right|=\left|-x+6\right|\ge\left|22\right|=22\)

dấu = xảy ra khi và chỉ khi \(\left(x+16\right).\left(-x+6\right)\ge0\Rightarrow-16\le x\le6\)(1)

\(x^2\ge0\Rightarrow x^2+x^2\ge0\)

dấu = xảy ra khi và chỉ khi x=0 (2)

=> \(x^2+\left|x+16\right|+x^2+\left|x-6\right|\ge22+0=22\)

dấu = xảy ra khi dấu = ở (1) và (2) đồng thời xảy ra 

=> x=0

Vậy min A=22 khi và chỉ khi x=0

p/s: ko chắc lắm, sai sót bỏ qua :))

3x - 6 = x² - 16

<=> x² - 3x - 10 = 0

<=> x² - 5x + 2x - 10 = 0

<=> x(x - 5) + 2(x - 5) = 0

<=> (x + 2)(x - 5) = 0

<=> \(\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

a: ĐKXĐ: x∉{2;-2;3}

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\right):\frac{16}{x^2-x-6}\)

\(=\left(\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\frac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right):\frac{16}{\left(x-3\right)\left(x+2\right)}\)

\(=\left(\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\frac{\left(x-3\right)\cdot\left(x+2\right)}{16}\)

\(=\frac{4x\left(x+2\right)-4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\cdot\frac{\left(x-3\right)\left(x+2\right)}{16}=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)}\cdot\frac{x-3}{16}\)

\(=\frac{-16\left(x-3\right)}{16\left(x+2\right)}=\frac{-x+3}{x+2}\)

b: A<0

=>\(\frac{-x+3}{x+2}<0\)

=>\(\frac{x-3}{x+2}>0\)

=>x>3 hoặc x<-2

c: A>=5

=>\(\frac{-x+3}{x+2}-5\ge0\)

=>\(\frac{-x+3-5x-10}{x+2}\ge0\)

=>\(\frac{-6x-7}{x+2}\ge0\)

=>\(\frac{6x+7}{x+2}\le0\)

=>-2<x<=-7/6

42-2(x+16) +12 :2=6(x-1)+ (-4)^3 :16

42- 2x- 32 +6 = 6x - 6 - 64: 16

(42 - 32 +6) - 2x = 6x - 6 - 4

16 = 2x = 6x- 10

16+10= 2x+6x

26= 8x 

x= 26/8

x= 13/4

42-2(x+16)+12:2=6(x-1)+(-4)³:16

42-2(x+16)+6=6(x-1)-64:16

42-2x-32+6=6x-6-4

-2x+16=6x-10

6x+2x=16+10

8x=26

x=3,25

16.(x-2) -6.(x-3)=6

=> 16x-32-6x+18=6

=>10x-14=6

=>10x=20

=>x=2

Vậy x=2

hok tốt

bạn ơi x =2 nè bn k cho mk nha

    \(\left(x-4\right)\left(x^2+4x+16\right)-x\left(x^2-6\right)=2\)

\(\Rightarrow x^3-64-x^3+6x=2\)

\(\Rightarrow-64+6x=2\Rightarrow6x=66\Rightarrow x=11\)

\(\left(\dfrac{1}{16}\right)^{x+2}=\left(\dfrac{1}{32}\right)^6\)

\(\Leftrightarrow\dfrac{1}{16^{x+2}}=\dfrac{1}{32^6}\)

\(\Leftrightarrow\dfrac{1}{2^{4x+8}}=\dfrac{1}{2^{30}}\)

\(\Leftrightarrow4x+8=30\Leftrightarrow4x=22\Leftrightarrow x=\dfrac{11}{2}\)

b) (x – 4). (x2 + 4x + 16) – x. (x2 - 6) = 2

⇔ x3 + 4x2 + 16x – 4x2 – 16x – 64 – (x3 - 6x ) – 2= 0

⇔ x3 + 4x2 + 16x – 4x2 – 16x – 64 – x3 + 6x – 2= 0

⇔ 6x – 66 =0

⇔ 6x = 66

⇔ x = 66 : 6

⇔ x = 11

Vậy x = 11

a x * 4 + [1/2 + 1/4 + 1/8 + 1/6] = 23/16

x * 4 +             25/24             = 23/16

x * 4                                     = 23/16 - 25/24

x * 4                                     =   38/96 = 19/48

x                                          = 19/48 / 4

x                                          = 19/172

b x *         1              = 425

x                              = 425 / 1

x                              =    425