\(a)x=\dfrac{1}{4}+\dfrac{5}{13}=\dfrac{33}{52}.\\ b)\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}.\\ \Leftrightarrow\dfrac{x}{3}=\dfrac{11}{21}.\\ \Leftrightarrow\dfrac{7x}{21}=\dfrac{11}{21}.\\ \Rightarrow7x=11.\\ \Leftrightarrow x=\dfrac{11}{7}.\\ c)\dfrac{x}{3}=\dfrac{16}{24}+\dfrac{24}{36}=\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}.\\ \Rightarrow x=4.\\ d)\dfrac{x}{15}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{13}{15}.\\ \Rightarrow x=13.\)

Giải

a) 24 + (15-x) = 16

<=> 15-x=16-24

<=> 15-x=(-8)

<=> x = 15-(-8) => x = 23

b) 11 - (24-x) = 

<=> 24-x= 11 - (-27)

<=> 24 - x = 38

<=> x = 24 - 38 => x = -14

c) |x-16| = 42

<=> x-16=42 và  x-16=-42

<=> x = 42 + 16 ; x =-42+16

<=> x = 58 ; x = -26

d) 10-2|x| = (-2) . (-3)

<=> 10 - (-2) . (-3) = 2|x|

<=> 10 - 6 = 2|x|

<=> 2|x| = 4

<=> |x| = 4 : 2 = 2

=> x = 2 hoặc x = (-2)

A/ 24+(15-x)=16                                                  

          (15-x)= 16- 24

          (15-x)= - 18

               x = 5 - (-18)

               x = 23

B/  11-(24-x)=(-3)³

          (24-x)=11- 27

          (24-x)= -16

               x = 24 -(- 16)

               x= 40

C/ 

\(\left|x-16\right|=42\hept{\begin{cases}x-16=42\\x-16=-42\end{cases}}\hept{\begin{cases}x=42+16\\x=-42+16\end{cases}}\hept{\begin{cases}x=58\\x=-26\end{cases}}\)

  D/ 10-2|x|=(-2)x(-3)

       10-2|x|=6

          2|x|=10-6

         2|x|=4

           x=4:2

           x=2

a)\(\dfrac{4}{x}=\dfrac{x}{16}\)

<=>\(x^2=4.16=64\)

<=>\(x=\pm8\)

<=>x=-8(vì x<0)

b)\(\dfrac{x}{-24}=\dfrac{-6}{x}\)

<=>\(x^2=\left(-24\right)\left(-6\right)=144\)

<=>\(x=\pm12\)

<=>x=12(Vì x>0)

\(B=\frac{1+x^2+x^4+...+x^{26}}{1+x^4+x^8+...+x^{24}}\)

\(=\frac{\frac{\left(x^2-1\right)\left(1+x^2+x^4+...+x^{26}\right)}{x^2-1}}{\frac{\left(x^4-1\right)\left(1+x^4+x^8+...+x^{24}\right)}{x^4-1}}\)

\(=\frac{\frac{x^{28}-1}{x^2-1}}{\frac{x^{28}-1}{x^4-1}}=\frac{x^4-1}{x^2-1}=x^2+1\)

1: \(\Leftrightarrow x=UCLN\left(24;36;150\right)=6\)

2: \(\Leftrightarrow x\in\left\{24;48;72;...\right\}\)

mà 16<=x<=50

nên \(x\in\left\{24;48\right\}\)

3: \(\Leftrightarrow x\inƯ\left(6\right)\)

mà x>-10

nên \(x\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

4: \(\Leftrightarrow x\in BC\left(4;5;8\right)\)

\(\Leftrightarrow x\in\left\{...;-40;0;40;80;120;160;200;...\right\}\)

mà -20<x<180

nên \(x\in\left\{0;40;80;120;160\right\}\)

2.

a, x-13=-46

=>x=(-46)+13

=>x=33

b, 4x-6=22

=>4x=22+6

=>4x=28

=>x=28:4

=>x=7

3.

a, 32=2⁵

48=2⁴.3

=>ƯCLN(32,48)=2⁴=16

16=2⁴

72=2³.3²

=>ƯCLN(16,72)=2³=8

b,

24=2³.3

60=2².3.5

=>BCNN(24,60)=2³.3.5=120

72=2³.3²

180=2².3².5

=>BCNN(72,180)=2³.3².5=360

a) x - 13  =  -   46 

    x         =  - ( 46 + 13 )                                

    x         =  - 59   

Chắc vậy

2a/ -33

2b/  7

3a/ 16

       8

3b/ 120

       360