10x(x-y)-(y-x)²

=10x(x-y)+(x-y).(y-x)

=(x-y).(10x+y-x)

=(x-y).(9x+y)

\(10x\left(x-y\right)-\left(y-x\right)^2\)

\(=10x\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left(10x-x+y\right)\)

\(=\left(x-y\right)\left(9x+y\right)\)

49x²+28x-5=49x²+28x+4-9=(7x+2)²-9=(7x+2-9)(7x+2+9)=(7x-7)(7x+11)=7(x-1)(7x+11)

còn cách nào nữa không giải ra hết luôn nha

=49x²-7x+35x-5

=7x(7x-1)+5(7x-1)

=(7x-1)(7x+5)

a) \(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)

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

\(x^4+8x^3+28x^2+48x-13\)

\(=x^4+4x^3+13x^2+4x^3+16x^2+52x-x^2-4x-13\)

\(=x^2\left(x^2+4x+13\right)+4x\left(x^2+4x+13\right)-\left(x^2+4x+13\right)\)

\(=\left(x^2+4x-1\right)\left(x^2+4x+13\right)\)

14x²y - 21xy² + 28x²y² = 7xy(2x - 3y + 4xy)

cảm ơn

\(f\left(x\right)=x^4+8x^3+28x^2+48x-13\)

\(=\left(x^4+4x^3+7x^2\right)+\left(4x^3+16x^2+28x\right)+\left(5x^2+20x+35\right)-48\)

\(=x^2\left(x^2+4x+7\right)+4x\left(x^2+4x+7\right)+5\left(x^2+4x+7\right)-48\)

\(=\left(x^2+4x+7\right)\left(x^2+4x+5\right)-48\)

đặt t=\(x^2+4x+6\)khi đó g(t)=(t-1)(t+1)-48=t²-49=(t-7)(y+7)

vậy f(x)=(x²+4x-1)(x²+4x+13)

Trả lời:

Thay \(f\left(x\right)=0\), ta có:

\(0=x^4+8x^3+28x^2+48x-13\)

\(\Leftrightarrow-x^4-8x^3-28x^2-48x+13=0\)

\(\Leftrightarrow-x^4-4x^3-4x^3+x^2-16x^2-13x^2+4x-56x+13=0\)

\(\Leftrightarrow\left(-x^4-4x^3+x^2\right)+\left(-4x^3-16x^2+4x\right)+\left(-13x^2-56x+13\right)=0\)

\(\Leftrightarrow-x^2.\left(x^2+4x-1\right)-4x.\left(x^2+4x-1\right)-13.\left(x^2+4x-1\right)=0\)

\(\Leftrightarrow\left(-x^2-4x-13\right).\left(x^2+4x-1\right)=0\)

Vì \(-x^2-4x-13=-x^2-4x-4-9\)

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

                                     \(=-\left(x+2\right)^2-9< 0\forall x\)

\(\Rightarrow x^2+4x-1=0\)

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

\(\Leftrightarrow\left(x+2\right)^2=5=\left(\pm\sqrt{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=\sqrt{5}\\x+2=-\sqrt{5}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2+\sqrt{5}\\x=-2-\sqrt{5}\end{cases}}\)

Vậy đa thức có 2 nghiêm \(x\in\left\{-2+\sqrt{5},-2-\sqrt{5}\right\}\)

a) \(5x^2y-20xy+20y=5y\left(x^2-4x+4\right)=5y\left(x-2\right)^2\)

b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)

c) \(3x^2y-12y=3y\left(x^2-4\right)=3y\left(x-2\right)\left(x+2\right)\)

d) \(7x^3-28x^2+28x=7x\left(x^2-4x+4\right)=7x\left(x-2\right)^2\)

a: \(5x^2y-20xy+20y\)

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

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

b: \(3x^3+6x^2+3x\)

\(=3x\left(x^2+2x+1\right)\)

\(=3x\left(x+1\right)^2\)

c: \(3x^2y-12y\)

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

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

d: \(7x^3-28x^2+28x\)

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

\(=7x\left(x-2\right)^2\)

a: \(12x^3-6x^2+3x\)

\(=3x\cdot4x^2-3x\cdot2x+3x\cdot1\)

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

b: \(\dfrac{2}{5}x^2+5x^3+x^2y\)

\(=x^2\cdot\dfrac{2}{5}+x^2\cdot5x+x^2\cdot y\)

\(=x^2\left(\dfrac{2}{5}+5x+y\right)\)

c: \(14x^2y-21xy^2+28x^2y^2\)

\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)

\(=7xy\left(2x-3y+4xy\right)\)