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

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

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

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

\(\Rightarrow x\left(x+3\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Rightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=24\)

Đặt \(x^2+3x+1=t\)

\(\Rightarrow\left(t-1\right)\left(t+1\right)=24\)

\(\Rightarrow t^2-1=24\Rightarrow t^2=25\Rightarrow t=5;-5\)

Xét t=5 thì \(x^2+3x+1=5\Rightarrow x^2+3x-4=0\)

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

\(\Rightarrow x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x-1\right)=0\Rightarrow x=-4;1\)

Xét t=-5 ta có

\(x^2+3x+1=-5\Rightarrow x^2+3x+6=0\)

\(\Rightarrow x_1=\frac{-3+\sqrt{15}i}{2};x_2=\frac{-3-\sqrt{15}i}{2}\)

mà \(x\in Z\)nên x=-4;1

\(5x\left(x-2\right)-3\left(x-1\right)=20x^2-15x\left(2x+1\right)-24\)

\(\Rightarrow5x^2-10x-3x+3=20x^2-30x^2-15x-24\)

\(\Rightarrow5x^2-13x+3=-10x^2-15x-24\)

\(\Rightarrow5x^2+10x^2-13x+15x+3+24=0\)

\(\Rightarrow15x^2+2x+27=0\)

Ta có: 

\(\Delta=2^2-4\cdot15\cdot27==-1616< 0\)

Nên pt vô nghiệm

\(5x\left(x-2\right)-3\left(x-1\right)=20x^2-15x\left(2x+1\right)-24\\ \Leftrightarrow5x^2-10x-3x+3=20x^2-30x^2-15x-24\\ \Leftrightarrow5x^2-20x^2+30x^2-10x-3x+15x+3+24=0\\ \Leftrightarrow15x^2+2x+27=0\\ \Leftrightarrow15x^2-2.x.\sqrt{15}+\dfrac{2}{15}+\dfrac{403}{15}=0\\ \Leftrightarrow\left(\sqrt{15}x+\dfrac{\sqrt{30}}{15}\right)^2+\dfrac{403}{15}=0\left(Vô.lí\right)\\ Vậy:Không.có.x.thoả\)

tick cho mình rồi mình làm cho

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

<=> x⁴ + 2x³ - x² - 2x - 24 = 0

<=> (x⁴ - 2x³) + (4x³ - 8x²) + (7x² - 14x) + (12x - 24) = 0

<=> (x - 2)(x³ + 4x² + 7x + 12) = 0

<=> (x - 2)[(x³ + 3x²) + (x² + 3x) + (4x + 12)] = 0

<=> (x - 2)(x + 3)(x² + x + 4) = 0

<=> x = 2 or x = - 3

xin lỗi mình ko biết nha bạn

xin lỗi mình ko biết nha bạn

xin lỗi mình ko biết nha bạn

đề sai

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

Ta có:

(x + 1)(x + 4) = x² + 5x + 4

(x + 2)(x + 3) = x² + 5x + 6

Đặt x² + 5x + 5 = t, lúc này ta có:

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

=> t² - 1 = 24

=> t² = 25

\(\Rightarrow t=\pm5\)

Thay vào x² + 5x + 5 = t ta được \(\left[\begin{array}{nghiempt}x=0\\x=-5\end{array}\right.\)