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

b, 707 : [ ( 2^x - 5 ) + 74 ] = 4² - 3²

=> 707 : [ ( 2^x - 5 ) + 74 ] = 16 - 9 = 7

=> ( 2^x - 5 ) + 74 = 707 : 7 = 101

=> 2^x - 5 = 101 - 74 = 27

=> 2^x = 27 + 5 = 32

=> 2^x = 2⁵

=> x = 5

a) \(\left(-396\right)+2016+189+\left(-604\right)-109\)

\(=-396+2016+189-604-109\)

\(=-\left(396-2016\right)+189-604-109\)

\(=-\left(-1620\right)+189-604-109\)

\(=1620+189-604-109\)

\(=1809-604-109\)

\(=1205-109\)

\(=1096\)

b) \(10-\left|x+3\right|=-8-10\)

\(\Leftrightarrow10-\left|x+3\right|=-\left(8+10\right)\)

\(\Leftrightarrow10-\left|x+3\right|=-18\)

\(\Leftrightarrow\left|x+3\right|=10-\left(-18\right)\)

\(\Leftrightarrow\left|x+3\right|=10+18\)

\(\Leftrightarrow\left|x+3\right|=28\)

Xét trường hợp 1: \(x+3=28\)

\(\Rightarrow x=28-3\)

\(\Rightarrow x=25\)

Xét trường hợp 2: \(x+3=-28\)

\(\Rightarrow x=-28-3\)

\(\Rightarrow x=-\left(28+3\right)\)

\(\Rightarrow x=-31\)

Vậy \(x=25\) hoặc \(x=-31\)

3^x : 3 = 9

3^x = 9 . 3

3^x = 27

3^x = 3³

=> x = 3

c. 4^{2x - 1} : 2² = 16

    4^{2x - 1} = 16 . 2² 

    4^{2x - 1} = 64

    4^{2x - 1} = 4³

=> 2x - 1 = 3

=> 2x = 3 + 1

=> 2x = 4

=> x = 2

a,3

b,2

c,2

d,5

a, 3^{x :}3=9

3^x =9:3

3^x=3¹

=>x=1

\(b,2\left(x-1\right)^5+2=4\)

\(2\left(x-1\right)^5=4-2\)

\(2\left(x-1\right)^5=2\)

\(\left(x-1\right)^5=2:2\)

\(\left(x-1\right)^5=1\)

\(\Rightarrow x=2\)

\(c,4^{2x-1}:2^2=16\)

\(4^{2x-1}:4=16\)

\(4^{2x-1}=16.4\)

\(4^{2x-1}=64\)

\(4^{2x-1}=4^3\)

\(\Rightarrow2x-1=3\)

\(2x=3+1\)

\(2x=4\)

\(x=4:2\)

\(x=2\)

\(d,707:[\left(2^x-5\right)+74]=4^2-3^2\)

\(707:\left[\left(2^x-5\right)+74\right]=16-9\)

\(707:\left[\left(2^x-5\right)+74\right]=7\)

\(\left(2^x-5\right)+74=707:7\)

\(\left(2^x-5\right)+74=101\)

\(2^x-5=101-74\)

\(2^x-5=27\)

\(2^x=27+5\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

a: \(\Leftrightarrow4^{x-5}\cdot17=68\)

=>4^x-5=4

=>x-5=1

=>x=6

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

=>|2x-1|=1/3

=>2x-1=1/3 hoặc 2x-1=-1/3

=>x=2/3 hoặc x=1/3

c: =>|2x-2|=|3x+15|

=>3x+15=2x-2 hoặc 3x+15=-2x+2

=>x=-17 hoặc x=-13/5

a 

\(230+\left[16+\left(y-5\right)\right]=315\cdot1\)   

\(230+\left[16+\left(y-5\right)\right]=315\)   

\(16+\left(y-5\right)=315-230\)   

\(16+\left(y-5\right)=85\)   

\(y-5=85-16\)   

\(y-5=69\)   

\(y=69+5\)   

\(y=74\)   

b 

\(707:\left[\left(2^y-5\right)+74\right]=16-9\)   

\(707:\left[\left(2^y-5\right)+74\right]=7\)   

\(\left(2^y-5\right)+74=707:7\)  

\(\left(2^y-5\right)+74=101\)   

\(2^y-5=101-74\)   

\(2^y-5=27\)   

\(2^y=27+5\)   

\(2^y=32\)   

\(2^y=2^5\)   

\(\Rightarrow y=5\)

a) x = 74    

b) x = 5 

c) x = 3      

d) x = 14

a) 210 -5.x= 200

<=> 5.x    = 210 - 200

<=> 5.x     = 10

<=> x         = 10 : 5

<=> x         = 2

Vậy x = 2

b) 210 - 5 . ( x-10 ) = 200

<=> 5. (x-10)          = 210 - 200

<=> 5.(x-10)           = 10

<=> x -10                = 10 : 5

<=> x - 10              = 2

<=>x                     = 12

Vậy x = 12

d)350 : x + 10 = 20

<=> 350 : x       = 20 - 10

<=> 350 : x         = 10

<=> x                 = 350 : 10

<=> x               = 35

Vậy x = 35

e)36 : ( x - 5 ) = 2²

<=> 36 : ( x-5) = 4

<=> x - 5        = 36 : 4

<=> x - 5          = 9

<=> x                = 9 + 5

<=> x               = 14

Vậy x =14

e) [3 . (70-x) + 5 ] : 2 = 46

<=>  [3 . (70-x) + 5 ]  = 46 . 2 

<=>  [3 . (70-x) + 5 ]  = 92

<=> 210 - 3.x + 5        = 92

<=>210- 3.x              = 92 - 5

<=> 210 - 3.x            = 87

<=> 3.x                       = 210 - 87

<=> 3.x                    = 123

<=> x                      = 123 : 3

<=> x                     = 41

Vậy x = 41

\(a,6x-4=5x\\ \Leftrightarrow x-4=0\\ \Leftrightarrow x=4\\ b,\dfrac{2x+3}{3}=\dfrac{5-4x}{2}\\ \Leftrightarrow2\left(2x+3\right)=3\left(5-4x\right)\\ \Leftrightarrow4x+6=15-12x\\ \Leftrightarrow16x-9=0\\ \Leftrightarrow x=\dfrac{9}{16}\\ c,\left(x+7\right)\left(x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)

d, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\\ \Leftrightarrow\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{3x+5}{\left(x+3\right)\left(x-3\right)}=0\\ \Leftrightarrow\dfrac{2x+6+3x-9-3x-5}{\left(x+3\right)\left(x-3\right)}=0\\ \Rightarrow2x-8=0\\ \Leftrightarrow x=4\left(tm\right)\)

a.6x-4=5x <=> x=4

b.\(\dfrac{2x+3}{3}=\dfrac{5-4x}{2}\)

\(\Leftrightarrow\dfrac{2\left(2x+3\right)}{6}=\dfrac{3\left(5-4x\right)}{6}\)

\(\Leftrightarrow2\left(2x+3\right)=3\left(5-4x\right)\)

\(\Leftrightarrow4x+6=15-12x\)

\(\Leftrightarrow16x=11\)

\(\Leftrightarrow x=\dfrac{11}{16}\)

c.(x+7)(x-10)=0

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)

d.\(ĐK:x\ne\pm3\)

\(\Rightarrow\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{2\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+5}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow2\left(x+3\right)+3\left(x-3\right)=3x+5\)

\(\Leftrightarrow2x+6+3x-9-3x-5=0\)

\(\Leftrightarrow2x-8=0\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=4\left(tm\right)\)

a, \(6x-5x=4\Leftrightarrow x=4\)

b, \(4x+6=15-12x\Leftrightarrow16x=9\Leftrightarrow x=\dfrac{9}{16}\)

c, \(\left[{}\begin{matrix}x+7=0\\x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=10\end{matrix}\right.\)

d, đk : x khác -3 ; 3 

\(2x+6+3x-9=3x+5\Leftrightarrow2x=8\Leftrightarrow x=4\left(tmđk\right)\)