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

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

tick cho mình rồi mình làm cho

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 \(\left[\begin{array}{nghiempt}x=0\\x=-5\end{array}\right.\)

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

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

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

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

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

Ta có: 

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

Nên pt vô nghiệm

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

xem lại đề hộ mk nhé 

mk sửa đề thế này 

x+ ( x+1)+(x+2)+....+(x+24)+(x+25) = 25 

=> x+ x+1 + x + 2 +...+ x+ 24 + x + 25 = 25 

=> 25x + 325 = 25 

từ đây tính ra x

x^2 -2x = 24

=> x^2 - 2x - 24=0

=>x^2 -8x+6x - 24 = 0

=> ( x^2- 8x)+( 6x-24) = 0

=> x(x-8) + 6(x-8) = 0

=> (x+6)(x-8)=0

=>\(\orbr{\begin{cases}x=-6\\x=8\end{cases}}\)

\(=\frac{\left(2.5\right)^4.3^4-2^4\left(3.5\right)^2}{2^8.5^2.3^3}=\frac{2^4.3^2.5^2\left(5^2.3^2-1\right)}{2^8.5^2.3^3}=\frac{255-1}{16.3}=\frac{14}{3}\)

xin lỗi mình làm lộn bài

x=6

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