\(x^4-2x^3-12x^2+12x+36=x^4+x^2+36-2x^3+12x-12x^2-x^2\)

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

Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn.

a) Ta có: \(4x^2-28xy+49y^2\)

\(=\left(2x\right)^2-2\cdot2x\cdot7y+\left(7y\right)^2\)

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

b) Ta có: \(x^2+8xy+16y^2\)

\(=x^2+2\cdot x\cdot4y+\left(4y\right)^2\)

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

c) Ta có: \(x^2-12x+36\)

\(=x^2-2\cdot x\cdot6+6^2\)

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

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

\(\left(6-x\right)^2\)

4x² - 28xy + 49y²

= (2x)² - 2.2x.7y + (7y)²

= (2x - 7y)²

x² + 8x²y + 16x²y² đề có bị thiếu không , nên mình bổ sung nhé

= x² + 2.x.4xy + (4xy)²

= (x + 4xy)²

36 - 12x + x²

= 6² + 2.6.x + x²

= (6 + x)²

 Chúc bạn học tốt

Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

Bài 4:

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

\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)

\(x^2-12x+36=\left(x-6\right)^2\)

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

c) \(36-12x+x^2=x^2-12x+36=x^2-6x-6x+36\)

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

\(x^4-y^4\)

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

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

\(4x^2+12x+9\)

\(=\left(2x\right)^2+2.2x.3+9\)

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

\(36-12x+x^2\)

\(=6^2-2.6.x+x^2\)

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

\(a.\) \(ax^2-a^2x-x+a\)

\(=\left(ax^2-a^2x\right)-\left(x-a\right)\)

\(=ax\left(x-a\right)-\left(x-a\right)\)

\(=\left(ax-1\right)\left(x-a\right)\)

\(b.\) \(18x^3-12x^2+2x\)

\(=2x\left(9x^2-6x+1\right)\)

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

\(c.\) \(x^3-5x^2-4x+20\)

\(=\left(x^3-5x^2\right)-\left(4x-20\right)\)

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

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

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

\(d.\) \(\left(x+7\right)\left(x+15\right)+15\)

\(=x^2+15x+7x+105+15\)

\(=x^2+22x+120\)

\(=\left(x+10\right)\left(x+12\right)\)

huhu , ai rảnh thì làm giúp mk vs ạ !!!

a)  \(x^3+8x^2+17x+10\)

\(=x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+7x+10\right)\)

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

b)  \(=x^4-2x^3-12x^2+12x+36\)

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

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

a) \(8x\left(x-3\right)+x-3=0\)

\(\Rightarrow8x\left(x-3\right)+\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(8x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{8}\end{matrix}\right.\)

b) \(x^2+36=12x\)

\(\Rightarrow x^2-12x+36=0\)

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

\(\Rightarrow x=6\)