\(a,\)\(x^2+2x-15=0\Rightarrow\left(x+1\right)^2-16=0\Rightarrow\left(x+1\right)^2=16\)

\(\hept{\begin{cases}x+1=4\\x+1=-4\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\x=-5\end{cases}}\)\(\)vậy \(S=\left\{3;-5\right\}\)

\(b,\)\(2x^2-7x+6=0\Rightarrow2x\left(x-1\right)-5\left(x-1\right)+1=0\)

\(\Rightarrow\left(x-1\right)\left(2x-5\right)=-1=1.-1=-1.1\)

\(x-1=1;2x-5=-1\)và \(x-1=-1;2x-5=1\)

\(\Rightarrow x=2\)và\(x=0;x=3\)vậy \(S=\left\{2;0;3\right\}\)

Mình làm nốt nhé

c) \(x^3-4x^2+5x=x\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x\left(x^2-4x+4+1\right)=0=x\left[\left(x-2\right)^2+1\right]=0\)

Vì \(\left(x-2\right)^2+1>0\forall x\Leftrightarrow x=0\)

d) \(x^3+5x^2+3x-9=0\Leftrightarrow x^3+3x^2+2x^2+6x-3x-9=0\)

\(\Leftrightarrow x^2\left(x+3\right)+2x\left(x+3\right)-3\left(x+3\right)=\left(x+3\right)\left(x^2+2x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left[\left(x+1\right)^2-4\right]=\left(x+3\right)\left(x+1-2\right)\left(x+1+2\right)=0\)

\(\Leftrightarrow\left(x+3\right)^2\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}}\)

Tự KL

x³ + 9x = 0 <=> x( x² + 9 ) = 0 <=> x = 0 ( vì x² + 9 > 0 )

Vậy pt có nghiệm x = 0

x³ + 9x = 0

<=> x(x²+9) = 0

=> x = 0 ( vì x²+9>0 với mọi x )

Vậy x = 0

a) 5x² -20

=  5(x² -4)

=5 (x² -2²)

= 5(x-2)(x+2)

b) 16 - (x+y)²

=4² -(x+y)²

= (4-x-y)(4+x+y) 

a, \(5\left(x^2-4\right)=5\left(x-2\right)\left(x+2\right)\)

b, \(16-\left(x+y\right)^2=\left(4-x-y\right)\left(4+x+y\right)\)

mấy bài này áp dụng hđt là được nhé 

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)