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

tick cho mình rồi mình làm cho

a: =>6/x=x/24

=>x^2=144

=>x=12 hoặc x=-12

b: =>x(1-7/12+3/8)=5/24

=>x*19/24=5/24

=>x=5/24:19/24=5/19

c: =>(x-1/3)^2=1+3/4+1/2=9/4

=>x-1/3=3/2 hoặc x-1/3=-3/2

=>x=11/6 hoặc x=-7/6

d: =>(x-3)^2=16

=>x-3=4 hoặc x-3=-4

=>x=-1 hoặc x=7

e: =>9/x=-1/3

=>x=-27

f: =>x-1/2=0 hoặc -x/2-3=0

=>x=1/2 hoặc x=-6

A x=12

B x=2

a) 4(2x+7)-3(3x-2)=24

8x+28-9x-6=24

(8x-9x)+(28-6)=24

(-1)x+22=24

(-1)x=24-22=2

x=2:(-1)=-2

b) -2(x+3)+(-4)²=3(1-x)

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

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

[(-2)x+3x]+[(-6)+16]=3

x+10=3

x=3-10=-7

`1/2-(1/3+3/4)<=x<=1/24-(1/8-1/3)`

`<=>6/12-4/12-9/12<=x<=1/24-3/24+8/24`

`<=>-7/12<=x<=1/4`

`<=>-14/24<=x<=3/12`

`=>-14<=x<=3`

`=>x\in{-14;-13;-12;...;3}` do `x\inZZ`

Ta có:\(\left(x+3\right)^2=\left(x+3\right)\left(x-3\right)\)

Xét \(x+3=0\Rightarrow x=-3\)

Xét \(x+3\ne0\) ta có:

\(x+3=x-3\)

\(\Rightarrow0=6\left(VL\right)\)

Vậy \(x=-3\)

a) 

(x + 3)² = (x + 3)(x – 3)

⇔ (x + 3)² - (x + 3)(x - 3) = 0

⇔ (x + 3)(x + 3 - x + 3) = 0

⇔ 6(x + 3) = 0

⇔ x = -3

Vậy: x = -3

b) Ta có A = (x + 1)(x + 2)(x + 3)(x + 4) – 24

= (x + 1)(x + 4)(x + 2)(x + 3) - 24

= (x² + 5x + 4)(x2 + 5x + 6) - 24(*)

Đặt x² + 5x + 5 = t

Thay x² + 5x + 5 = t vào (*) ta được:

A = (t - 1)(t + 1) - 24

= t² - 25

= (t - 25)(t + 25)

= (x² + 5x + 5 + 5)(x² + 5x + 5 - 5)

= (x² + 5x + 10)(x² + 5x)

(x² + 5x + 10).x(x + 5) chia hết cho x (Với x ≠ 0)

Vậy: A chia hết cho x (Với x ≠ 0)

a) (x + 3)² = (x + 3)(x - 3)

<=> x² + 6x + 9 = x² - 3²

<=> x² + 6x + 9 = x² - 9

<=> 6x + 9 = -9

<=> 6x = -9 - 9

<=> 6x = -18

<=> x = -3

=> x = -3

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

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

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