\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

M = x⁹ - x⁷ + x⁶ - x⁵ - x⁴ + x³ - x² + 1

= ( x⁹ - x⁷ ) + ( x⁶ - x⁴ ) - ( x⁵ - x³ ) - ( x² - 1 )

= x⁷( x² - 1 ) + x⁴( x² - 1 ) - x³( x² - 1 ) - ( x² - 1 )

= ( x² - 1 )( x⁷ + x⁴ - x³ - 1 )

= ( x - 1 )( x + 1 )[ x⁴( x³ + 1 ) - ( x³ + 1 ) ]

= ( x - 1 )( x + 1 )( x³ + 1 )( x⁴ - 1 )

= ( x - 1 )( x + 1 )( x + 1 )( x² - x + 1 )( x² - 1 )( x² + 1 )

= ( x + 1 )²( x - 1 )( x² - x + 1 )( x - 1 )( x + 1 )( x² + 1 )

= ( x + 1 )³( x - 1 )²( x² + 1 )( x² - x + 1 )

Bài 1:

$5x+10=5(x+2)$

Bài 2:

Tại $x=8$ thì $x^2+4x+4=(x+2)^2=(8+2)^2=10^2=100$

Bài 3:

$x^2-6x+9=x^2-2.3.x+3^2=(x-3)^2$

Bài 4:

Diện tích mảnh đất là:

$(x+5)(x-5)=24$

$\Leftrightarrow x^2-25=24$

$\Leftrightarrow x^2=49$

$\Rightarrow x=7$ (do $x>5$)

Chiều dài mảnh đất là: $x+5=7+5=12$ (m)

Nhớ mình nha mình âm diểm rồi:

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

M=(x²+3x+2x+6)(x²+5x+4x+20)-24

M=(x²+5x+6)(x²+9x+20)-24

M=x⁴+9x³+20x²+5x³ +14x+100x+6x²+54x+120-24

M=x⁴+14x³+26x²+168x+96

(x^m+2)+(x^m) = 2x^m+2 = 2(x^m+1)

(x^x+1)-(x^x)-1 = x^x+1-x^x-1 = 0

(m^4)-(n^4) = (m²)²-(n²)² = (m²-n²)(m²+n²)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(M=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+3\right)\left(x+4\right)\left(x+2\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+12\right)\left(x^2+7x+10\right)-24\)

\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25\)

\(=\left(x^2+7x+11\right)^2-25\)

\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+6\right)\)

Dòng cuối Đinh Đức Hùng viết nhầm kìa

Ờ đúng òy :)) dòng cuối viết nhầm

sửa lại là :\(\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\) nha