5/12 = 10/24

2/3 = 16/24

7/24 giữ nguyên

\(\dfrac{5}{12}=\dfrac{5\times2}{12\times2}=\dfrac{10}{24}\\ \dfrac{2}{3}=\dfrac{2\times8}{3\times8}=\dfrac{16}{24}\)

7/24 giữ nguyên

10/24

16/24

7/24

Tham khảo!

https://olm.vn/hoi-dap/detail/10185285054.html

25/6

giúp mình với, và giải chi tiết

25/6

\(\dfrac{3}{x}=\dfrac{49}{24}-\dfrac{5}{3}\\ \dfrac{3}{x}=\dfrac{49}{24}-\dfrac{40}{24}\\ \dfrac{3}{x}=\dfrac{3}{8}\\ x=8\)

\(\dfrac{5}{3}+\dfrac{3}{x}=\dfrac{49}{24}\\ \dfrac{3}{x}=\dfrac{49}{24}-\dfrac{5}{3}\\ \dfrac{3}{x}=\dfrac{3}{8}\\ x=8\)

\(\Leftrightarrow12x^3-12x^3-8x=24\)

hay x=-3

a) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)-4=\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\)

Đặt \(t=x^2+6x+5\)

\(PT=t\left(t+3\right)-4=t^2+3t-4=\left(t-1\right)\left(t+4\right)\)

Thay t: \(PT=\left(x^2+6x+5-1\right)\left(x^2+6x+5+4\right)=\left(x^2+6x+4\right)\left(x^2+6x+9\right)=\left(x^2+6x+4\right)\left(x+3\right)^2\)

b)  Đặt \(t=\left(2x+1\right)^2\)

\(PT=t^2-3t+2=\left(t^2-3t+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(t+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(t+1\right)\left(t+2\right)\)

Thay t:

\(PT=\left[\left(2x+1\right)^2+1\right]\left[\left(2x+1\right)^2+2\right]=\left[4x^2+4x+2\right]\left[4x^2+4x+3\right]=2\left[2x^2+2x+1\right]\left[4x^2+4x+3\right]\)

32,25 đúng nha bạn

\(21+\left(4,5.2,5\right)\)

\(=21+11,25\)

\(=32,25\)

=>\(\text{Cái thứ 2 đúng}\)

Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=k\Leftrightarrow a=2k;b=3k\)

\(ab=24\Leftrightarrow6k^2=24\Leftrightarrow k^2=2\\ \Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=4;b=6\\a=-4;b=-6\end{matrix}\right.\)

Ta có :

\(\dfrac{a}{2}=\dfrac{b}{3}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\end{matrix}\right.\)

mà \(ab=24\)

\(\Rightarrow2k.3k=24\)

\(\Rightarrow6k^2=24\)

\(\Rightarrow k^2=2^2\)

\(\Rightarrow k=\left\{{}\begin{matrix}2\\-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{b}{3}=2\\\dfrac{a}{2}=\dfrac{b}{3}=-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=4;b=6\\a=-4;b=-6\end{matrix}\right.\)

Ta có:

(a.b)/(2.3)= 24/6=4(T/c dãy tỉ số = nhau )

=>a=4.2=8

b=4.3=12

p là số nguyên tố lớn hơn 3 nên p ko chia hết cho 3 và p lẻ

=>p^2-1=(p-1)(p+1) chia hết cho cho 2*4=8(1)

TH1: p=3k+1

p^2-1=9k^2+6k+1=9k^2+6k=3k(3k+2) chia hết cho 3(2)

TH2: p=3k+2

p^2-1=9k^2+12k+4-1

=9k^2+12k+3=3(3k^2+4k+1) chia hết cho 3(3)

Từ (1), (2), (3) suy ra p^2-1 chia hết cho BCNN(3;8)=24