x² + 6x - y² + 9 = 

= ( x² + 6x + 9 ) - y² 

= ( x + 3 )² - y² 

= ( x + 3 - y ) . ( x + 3 + y )

Hok tốt!!!!!!!!!!!

ta có 

a.\(N=\left(10x\right)^2-2.\left(10x\right)+1=\left(10x-1\right)^2=\left(3-1\right)^2=4\text{ khi }x=\frac{3}{10}\)

b.\(P=\left(5x\right)^2-2.5xy^2+y^4=\left(5x-y^2\right)^2=\left(5.5-4^2\right)=9^2=81\)

1+1=3-1=2 nha bạn

\(A=\left(x^2-5x-4\right)^2+7\left(x^2-5x-4\right)+10\)

Đặt \(t=x^2-5x-4\).

\(A=t^2+7t+10=t^2+2t+5t+10=\left(t+2\right)\left(t+5\right)\)

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

\(=\left(x^2-5x-2\right)\left(x^2-5x+1\right)\)

\(x^2-16+2\left(4-x\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)-2\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4-2\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2\right)=0\)

Trường hợp 1: \(x-4=0\Leftrightarrow x=4\)

Trường hợp 2: \(x+2=0\Leftrightarrow x=-2\)

Vậy: \(S=\left\{4;-2\right\}\)

ko k cho mình nha team mình có người k cho mình 3 k nên bạn ko k cho mình 3 câu trả lời 

a.(−4xy)(2xy^2–3x^2y)

=(−4xy)(2xy^2)+(−4xy)(−3x^2y)

=−8x^2y^3+12x^3y^2

nha bạn chúc bạn học tốt ạ 

a) = \(\left(-4xy\right)\left(2xy^2\right)\)\(+\left(-4xy\right)\left(-3x^2y\right)\)

    = \(-8x^2y^3\)\(+12x^3y^2\)

b) = \(\left(-5x\right)\left(3x^3\right)\)\(+\left(-5x\right)\left(7x^2\right)\)\(+\left(5x\right)\left(-x\right)\)

    = \(-15x^4-35x^3+5x^2\)

a.\(3x^2-4x+1=0\Leftrightarrow3x^2-3x-x+1=0\Leftrightarrow\left(3x-1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}\)

b.\(x\left(2x^2-4x+3x-6\right)=0\Leftrightarrow x\left(2x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\text{ hoặc }x=-\frac{3}{2}\)

c.\(2x^3-x^2-13x-6=0\Leftrightarrow\left(x+2\right)\left(2x+1\right)\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}\text{ hoặc }x=3}\)

\(a,3x^2-4x+1=0\)

\(\Leftrightarrow3x^2-3x-x+1=0\)

\(\Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)

Trường hợp 1: \(x-1=0\Leftrightarrow x=1\)

Trường hợp 2: \(3x-1=0\Leftrightarrow x=\frac{1}{3}\)

Vậy: \(S=\left\{1;\frac{1}{3}\right\}\)

\(b,2x^3-x^2-6x=0\)

\(\Leftrightarrow x\left(2x^2-x-6\right)=0\)

\(\Leftrightarrow x\left(2x^2-4x+3x-6\right)=0\)

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

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

Trường hợp 1: \(x=0\)

Trường hợp 2: \(x-2=0\Leftrightarrow x=2\)

Trường hợp 3: \(x+3=0\Leftrightarrow x=-3\)

Vậy: \(S=\left\{0;2;-3\right\}\)

\(c,2x^3-x^2-13x-6=0\)

\(\Leftrightarrow2x^3-5x^2-6x^2+2x-15x-6=0\)

\(\Leftrightarrow\left(2x^3+5x^2+2x\right)-\left(6x^2+15x+6\right)=0\)

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

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

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

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

\(\Leftrightarrow\left(x+2\right)\left(2x+1\right)\left(x-3\right)=0\)

Trường hợp 1: \(x+2=0\Leftrightarrow x=-2\)

Trường hợp 2: \(2x+1=0\Leftrightarrow x=-\frac{1}{2}\)

Trường hợp 3: \(x-3=0\Leftrightarrow x=3\)

Vậy: \(S=\left\{-2;-\frac{1}{2};3\right\}\)

a)(x - 3)(x² + 3x + 9) + x(x + 2)(2 - x) = 1;

<=> x^3-27+x(4-x^2) = 1

<=> x^3-27+4x-x^3=1

<=> 4x=28

<=> x=7

Vậy x=7

b) (x + 1)³ - (x - 1)³ - 6(x - 1)² = -19.

<=> [(x+1)-(x-1)][(x+1)^2+(x+1)(x-1)+(x-1)^2]-6(x^2-2x+1)=-19

<=> 2(x^2+2x+1+x^2-1+x^2-2x+1)-6x^2+12x-6=-19

<=> 2(3x^2+1)-6x^2+12x=-13

<=> 6x^2+2-6x^2+12x=-13

<=> 12x=-15

<=> x=-15/12=-5/4

Vậy x=-5/4

Trả lời:

a, \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)

\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-4x+1-7\left(x^2-9\right)=36\)

\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\)

\(\Leftrightarrow8x+76=36\)

\(\Leftrightarrow8x=-40\)

\(\Leftrightarrow x=-5\)

Vậy x = - 5 là nghiệm của pt.

b, \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)

\(\Leftrightarrow x^3-27+x\left(4-x^2\right)=1\)

\(\Leftrightarrow x^3-27+4x-x^3=1\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

Vậy x = 7 là nghiệm của pt.

c, \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-19\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-19\)

\(\Leftrightarrow12x-4=-19\)

\(\Leftrightarrow12x=-15\)

\(\Leftrightarrow x=-\frac{5}{4}\)

Vậy x = - 5/4 là nghiệm của pt.

a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=\)

\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16-9=-12x-24=0\)

\(\Leftrightarrow12x+24=0\Leftrightarrow12x=24\Leftrightarrow x=2\)

mình biết câu a và b rồi cảm ơn