`#3107.101107`

`(19 - x)^3 * 2 - 20 = 34`

`=> (19 - x)^3 * 2 = 34 + 20`

`=> (19 - x)^3 * 2 = 54`

`=> (19 - x)^3 = 54 \div 2`

`=> (19 - x)^3 = 27`

`=> (19 - x)^3 = 3^3`

`=> 19 - x = 3`

`=> x = 19 - 3`

`=> x = 16`

Vậy, `x = 16.`

(19-x)³.2-20=34 

<=> (19-x)³=27

<=> 19-x=3

=> x=16

a)3+x=7

x=7-3

x=4

b)-3x+5=-22

-3x=-22-5

-3x=-27

x=-27:(-3)

x=9

d)(-x+3)(2x+4)=0

-x+3=0 hoặc 2x+4=0

x=3 hoặc x=-2

Phần a;b tính dễ dàng nên bạn tự làm nha

c)(-3).|x-1|=-9

=>|x-1|=(-9):(-3)

=>|x-1|=3

=>x-1 =3 hoặc x-1=-3

=>x =3+1 =>x =-3+1

=>x =4 =>x =-2

Vậy \(x\in\left\{4;-2\right\}\)

d)\(\left(-x+3\right).\left(2x+4\right)=0\)

=>-x+3=0 hoặc 2x+4=0

\(\Rightarrow-x=-3\) \(\Rightarrow2x=-4\)

\(\Rightarrow x=3\) \(\Rightarrow x=-4:2\)

\(\Rightarrow x=-2\)

Vậy \(x\in\left\{3;-2\right\}\)

Chúc bn học tốt

`x+17=3^5:3^2`

`=>x+17=3^3`

`=>x+17=27`

`=>x=27-17`

`=>x=10`

__

`5.6^(x+1)-2.3^2=12`

`=>5.6^(x+1)-2.9=12`

`=>5.6^(x+1)-18=12`

`=>5.6^(x+1)=12+18=30`

`=>6^(x+1)=30:5`

`=>6^(x+1)=6`

`=>x+1=1`

`=>x=0`

A. \(\text{x + 17 = 3⁵ : 3²}\)

    \(x+17=3^{5-2}\)

    \(x+17=3^3\)

     \(x+17=27\)

     \(x=27-17\)

     \(x=10\)

B.\(5\cdot6^{x+1}-2\cdot3^2=12\)

  \(5\cdot6^{x+1}-2\cdot9=12\)

 \(5\cdot6^{x+1}-18=12\)

\(5\cdot6^{x+1}=18+12\)

\(5\cdot6^{x+1}=30\)

\(6^{x+1}=\dfrac{30}{5}\)

\(6^{x+1}=6\)

\(x+1=1\) 

\(x=0\)

\(x+1=1\) vì \(6=6^1\)

Yx,2-3,45=4,68

Yx1,2=4,68+3,45

Yx1,2=8,13

Y=8,13:1,2

Y=6,775

Xx1,2-3,45=4,68

=>Xx1,2=4,68+3,45=8,13

=>x=8,13:1,2=6,775

Đánh giá:  \(\left|x-2\right|^2\ge0\)      \(\forall x\)

\(\Rightarrow\)\(-\left|x-2\right|^2\le0\)   \(\forall x\)

\(\Rightarrow\)\(N=-\left|x-2\right|^2-18\le-18\)

Dấu  '='    xảy ra    \(\Leftrightarrow\)\(x-2=0\)   \(\Leftrightarrow\)\(x=2\)

Vậy  MAX  \(N=-18\)   \(\Leftrightarrow\)\(x=2\)

Thanks bạn

có nè

đỗ trà my(tag hộ)

nguyễn trúc giang(tag hộ)

HISINOMA KINIMADO

\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)

\(x^8=x^{12}:x^5\)

\(x^8=x^7\)

=> x⁸ - x⁷ = 0

x⁷.(x-1) = 0

=> x⁷ = 0=> x = 0

x-1 = 0 => x = 1

KL:  x = 1 hoặc x = 0

\(\frac{x}{\left(x^4\right)^2}=\frac{x^{12}}{x^5}\)

=>\(\frac{x}{x^8}=x^7\)

=>\(\frac{1}{x^7}=x^7\)

=>\(1=x^7.x^7\)

=>\(1^{14}=x^{14}\)

=>\(x=1\)

Bài này \(x^5\rightarrow x^4\) mới đúng chứ?

\(\frac{x^{12}}{x^4}=x^{\left(12-4\right)}=x^8=\left(x^4\right)^2\)

Đánh giá:   \(\left|x-2\right|^2\ge0\)

\(\Rightarrow\)\(-\left|x-2\right|^2\le0\)

\(\Rightarrow\)\(N=-\left|x-2\right|^2-18\le-18\)

Dấu  '='  xảy ra  \(\Leftrightarrow\)\(x-2=0\) \(\Leftrightarrow\)\(x=2\)

Vậy MAX   \(N=-18\) \(\Leftrightarrow\)\(x=2\)