gọi số lẻ đầu tiên là : 2n-3

vậy 4 số lẻ đó là 2n-3, 2n-1, 2n+1,2n+3

ta có : \(\left(2n+1\right)\left(2n+3\right)-\left(2n-1\right)\left(2n-3\right)=160\Leftrightarrow16n=160\)

vậy n=10 và bốn số đó là 17 ,19,21,23

Mình đã trả lời câu này rồi nhé, bạn vào tham khảo UwU

https://olm.vn/hoi-dap/detail/53010159395.html

(x+2)(x+3)(x+4)(x+5)=24

x=-6,

x=-1;

x = -(căn bậc hai(3)căn bậc hai(5)i+7)/2

;x = (căn bậc hai(3)căn bậc hai(5)i-7)/2;

nha bạn chúc bạn học tốt nha 

(x + 2)(x + 3)(x + 4)(x + 5) = 24

<=> [(x + 2)(x + 5][(x + 3)(x + 4] = 24

<=> (x² + 7x + 10)(x² + 7x + 12) - 24 = 0

<=> (x² + 7x + 11 - 1)(x² + 7x + 11 + 1) - 24 = 0

<=> (x² + 7x + 11)² - 25 = 0 

<=> (x² + 7x + 16)(x² + 7x + 6) = 0

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left[\left(x+\frac{7}{2}\right)^2+\frac{15}{4}\right]=0\)

<=> (x + 1)(x + 6) = 0 

<=> \(\orbr{\begin{cases}x+1=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-6\end{cases}}\)

Vậy \(x\in\left\{-1;-6\right\}\)

a,\(\frac{1}{4}\left(a+b\right)^2-1=\left(\frac{a+b}{2}\right)^2-1^2=\left(\frac{a+b}{2}-1\right)\left(\frac{a+b}{2}+1\right)\)

b,\(9\left(x-y\right)^2-4\left(x-y\right)^2=\left(x-y\right)^2\left(9-4\right)=5\left(x-y\right)^2\)

c,\(\left(p-2q\right)^2-4\left(p+q\right)^2=\left(p-2q-2p-2q\right)\left(p-2q+2p+2q\right)\)

\(=\left(-p-4q\right)3p\)

d, \(25p^2m^4-\frac{1}{36}p^4=\left(5pm^2\right)^2-\left(\frac{p^2}{6}\right)^2=\left(5pm-\frac{p^2}{6}\right)\left(5pm+\frac{p^2}{6}\right)\)

a, \(\frac{1}{4}\left(a+b\right)^2-1=\left(\frac{1}{2}a+\frac{1}{2}b\right)^2-1=\left(\frac{a}{2}+\frac{b}{2}-1\right)\left(\frac{a}{2}+\frac{b}{2}+1\right)\)

b, \(9\left(x-y\right)^2-4\left(x-y\right)^2=\left(3x-3y\right)^2-\left(2x-2y\right)^2\)

\(=\left(3x-3y-2x+2y\right)\left(3x-3y+2x-2y\right)=5\left(x-y\right)^2\)

c, \(\left(p-2q\right)^2-4\left(p+q\right)^2=\left(p-2q\right)^2-\left(2p+2q\right)^2\)

\(=\left(p-2q-2p-2q\right)\left(p-2q+2p+2q\right)^2=9p^2\left(-p-4q\right)\)

d, \(25p^2m^4-\frac{1}{36}p^4=\left(5pm^2\right)^2-\left(\frac{1}{6}p^2\right)^2=\left(5pm^2-\frac{1}{6}p^2\right)\left(5pm^2+\frac{1}{6}p^2\right)\)

\(=p^2\left(5m^2-\frac{1}{6}p\right)\left(5m^2+\frac{1}{6}p\right)\)

271​x3−32​x2+4x−8

 x  -  x  +  4x  -  8

=  4x  -  8

=  4( x  -  8)

b, \(\left(x^2-9\right)\left(x^2+9\right)-\left(x^4-18x^2+81\right)=\left(x^4-81\right)-x^4+18x^2-81\)=\(18x^2-162\)

a, \(5x^2\)\(-3x-x^3+x^2+x^3-6x^2\)\(-10+3x\)=\(-10\)

b,\(x^3+x^2+x-x^3-x^2-x+5=5\)