230 + [2² + (x - 5)²] = 315 . 2021⁰

230 + 4 + (x - 5)² = 315.1

234 + (x - 5)² = 315

(x - 5)² = 315 - 234

(x - 5)² = 81

x - 5 = 9 hoặc x - 5 = -9

*) x - 5 = 9

x = 9 + 5

x = 14

*) x - 5 = -9

x = -9 + 5

x = -4

Vậy x = -4; x = 14

Cứucứu

a,

13[x-9] = 169

=> x - 9 = 169/13

=> x - 9 = 13

=> x = 13+9

=> x = 22

b,

Viết lại đề:

7^x+3 = 343

<=> 7^x+3 = 7³

=> x + 3 = 3 

=> x = 3-3

=> x = 0

c,

230 + [16 + [x-5]] = 315 . 2³

=> 230 + [16 + x - 5] = 315 . 8

=> 230 + 16 + x - 5 = 2520

=> 230 + 16 + x = 2520 + 5 = 2525

=> x = 2525 - 230 - 16 = 2279

d,

13.x - 3².x = 2017¹ - 1²⁰¹⁸

=> 13x - 9x = 2017 - 1

=> 4x = 2016

=> x = 504

a) 13 ( x-9 )=169

=> x-9 =169 : 13 =13

=> x=13+9 =22

b)\(7^{x+3}=343\)

\(7^x.7^3=343\)

\(7^x=343:7^3\)

\(7^x=1\Rightarrow x=1\)

c)230 + 16 +x -5 =315.8

241 +x =2520

x=2520-241=2279

d) 13x -\(3^2.x\)=2017-1

x(13-9)=2016

x.4=2016

x=2016:4

x=504

d , ( x : 7 + 15 ) . 23 + 391 => Đề thiếu

e , ( 19 . x + 2 . 5² ) : 14 = ( 13 - 8 ) ² - 4²

=> ( 19 . x + 2 . 5² ) : 14 = 5² -  16

=> ( 19 . x + 2 . 5² ) : 14 = 25 - 16 = 9

=> 19 . x + 2 . 52 = 9 x 14 = 126

=> 19 . x + 2 . 25 = 126

=> 19 . x + 50 = 126

=> 19 . x = 126 - 50 = 76

=> x = 76 : 19 = 4

a, 230 + [ 2⁴ + ( x - 5 ) ] = 315 . 2020⁰

=> 230 + [16 + ( x - 5 ) ] = 315 . 1

=> 230 + [ ( 16 - 5 ) + x ) = 315

=> 11 + x = 315 - 230 = 85

=> x = 85 - 11 = 74

a) |x| = 4

\(\left[ {_{x =  - 4}^{x = 4}} \right.\)

Vậy \(x \in \{ 4; - 4\} \)

b) |x| = \(\sqrt 7 \)

\(\left[ {_{x =  - \sqrt 7 }^{x = \sqrt 7 }} \right.\)

Vậy \(x \in \{ \sqrt 7 ; - \sqrt 7 \} \)

c) ) |x+5| = 0

x+5 = 0

x = -5

Vậy x = -5

d) \(\left| {x - \sqrt 2 } \right|\) = 0

x - \(\sqrt 2 \) = 0

x = \(\sqrt 2 \)

Vậy x =\(\sqrt 2 \)

b: \(5x^2+3x-2-4x^2+x+5=0\)

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

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

a) \(\Rightarrow x^8-2x^4-8=0\Rightarrow\left(x^4-4\right)\left(x^4+2\right)=0\)

\(\Rightarrow\left(x^2-2\right)\left(x^2+2\right)\left(x^4+2\right)=0\)

\(\Rightarrow x^2=2\Rightarrow x=\pm\sqrt{2}\)(do \(x^2+2\ge2>0,x^4+2\ge2>0\))

b) \(\Rightarrow x^2+4x+3=0\Rightarrow\left(x+1\right)\left(x+3\right)=0\)

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

a)\(=>2x=-10=>x=-5\)

b)\(=>-2x=-5=>x=\dfrac{-5}{-2}=\dfrac{5}{2}\)

c)\(4-x=0=>x=4-0=4\)

d)\(=>2x=-1=>x=-\dfrac{1}{2}\)

e)\(=>x^2=-2\)=> x ko tồn tại

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

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 40x = 15 <=> x = 3/8

Sorry, cái này mình nhầm

\(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)

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

Vậy \(S=\varnothing\)

b: \(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|\ge0\forall x\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)