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

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) \(x^2+36=12x\)

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

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

\(\Rightarrow x=6\)

a: Ta có: \(2-x=2\left(x-2\right)^3\)

\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2

c: Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^3=0\)

\(\Leftrightarrow\left(x-1.5\right)^6-2\left(x-1.5\right)^3=0\)

\(\Leftrightarrow\left(x-1.5\right)^3\cdot\left[\left(x-1.5\right)^3-2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1.5\\x=\sqrt[3]{2}+1.5\end{matrix}\right.\)

Bài 2: 

a: =>4x(x+5)=0

=>x=0 hoặc x=-5

b: =>(x+3)(x-3)=0

=>x=-3 hoặc x=3

`1)`

`a)3x^2-6xy+3y^2=3(x^2-2xy+y^2)=3(x-y)^2`

`b)(x-y)^2-4x^2=(x-y-2x)(x-y+2x)=(-x-y)(3x-y)`

`2)`

`a)2x(x-3)-x+3=0`

`<=>2x(x-3)-(x-3)=0`

`<=>(x-3)(2x-1)=0`

`<=>[(x=3),(x=1/2):}`

`b)x^2+5x+6=0`

`<=>x^2+2x+3x+6=0`

`<=>(x+2)(x+3)=0`

`<=>[(x=-2),(x=-3):}`

\(1,=8xy+14y^2-4xz-7yz\\ 2,=y\left(4x^2-12x+9\right)=y\left(2x-3\right)^2\\ 3,\Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

Câu 1: \(\left(2y-z\right)\left(4x+7y\right)=8xy-4xz+14y^2-7yz\)

câu 2: \(4x^2y-12xy+9y=y\left(4x^2-12x+9\right)\)

câu 3: \(\left(x-2\right)\left(x+3\right)+x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-2+x\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-2\right)=0\\ \Leftrightarrow2\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

Bài 3

a) 2x(x - 3) - x + 3 = 0

2x(x - 3) - (x - 3) = 0

(x - 3)(2x - 1) = 0

x - 3 = 0 hoặc 2x - 1 = 0

*) x - 3 = 0

x = 3

*) 2x - 1 = 0

2x = 1

x = 1/2

Vậy x = 1/2; x = 3

b) (3x - 1)(2x + 1) - (x + 1)² = 5x²

6x² + 3x - 2x - 1 - x² - 2x - 1 - 5x² = 0

(6x² - x² - 5x²) + (3x - 2x - 2x) = 0 + 1 + 1

-x = 2

x = -2

Bài 2

a) 5x² + 30y

= 5(x² + 6y)

b) x³ - 2x² - 4xy² + x

= x(x² - 2x - 4y² + 1)

= x[(x² - 2x + 1) - 4y²]

= x[(x - 1)² - (2y)²]

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

_{Chiều dài của hình chữ nhật sau khi tăng thêm 5% là:}

     _{100%+5%=105% (chiều dài ban đầu)}

_{Chiều rộng của hình chữ nhật sau khi tăng thêm 8% là:}

     _{100%+8%=108% (chiều rộng ban đầu)}

_{Diện tích của hình chữ nhật sau khi tăng là:}

     _{105%×108%=113,4% (diện tích ban đầu)}

              _{Đáp số:113,4$}

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

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