1)\(x^3-2x^2y+x-xz^2\)

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

\(x^2y+xy^2-x-y\)

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

\(=\left(x+y\right)\left(xy-1\right)\)

2) \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)

\(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\)

\(\left(x-2\right)\left[x^2+2x+7+2.\left(x+2\right)-5\right]=0\)

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

Ta có: \(x^2+6>0\forall x\)

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

\(\Rightarrow x-2=0\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

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

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

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

\(-2x\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}-2x=0\\x-2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

P/S: Câu 1 nghi sai đề

Ta có: \(A=\frac{3x^2+6x+11}{x^2+2x+3}=3+\frac{2}{x^2+2x+3}=3+\frac{2}{\left(x+1\right)^2+2}\)

Đặt \(B=\frac{2}{\left(x+1\right)^2+2}\),để A đạt giá trị lớn nhất thì B lớn nhất.

Mà B lớn nhất khi \(\left(x+1\right)^2+2\) bé nhất. 

Lại có: \(\left(x+1\right)^2\ge0\forall x\Rightarrow\left(x+1\right)^2+2\ge2\) (1)

Từ (1) suy ra: \(B\le\frac{2}{2}=1\Rightarrow A=3+B\le3+1=4\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy \(A_{max}=4\Leftrightarrow x=-1\)

Bài 2:

a) x + 2a.(x-y) - y

= 2a.(x-y) + (x-y)

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

b) 5a² - 5ax - 7a + 7x

= 5a.(a-x) - 7.(a-x)

= (a-x).(5a-7)

Bài 1:

a) 5x.(10x+7) - 25x.(2x-3) = 40

50x² + 35x - 50x² + 75x = 40

110x = 40

x = 4/11

b) (3x+2).(x-2) - (x-1).(x-3) = 4

3x² - 6x + 2x - 4 - x² + 3x + x - 3 = 4

2x² - 7 = 4

...

bn tự làm tiếp nha

Đặt \(A=\left(x^2-2x+3\right)\left(x^2-2x+5\right)-8\). Rút gọn A,ta được:

\(A=x^4-4x^3+12x^2-16x+7\)

\(=x^4-2x^3+x^2-2x^3+4x^2-2x+7x^2-14x+7\)

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

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

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

Ok chứ?

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=24\)

\(\Rightarrow\left(x-1\right)\left(x-4\right)\left(x-2\right)\left(x-3\right)=24\)

\(\left(x^2-5x+4\right)\left(x^2-5x+6\right)=24\)

Đặt \(x^2-5x+5=a,\)ta có

\(\left(a-1\right)\left(a+1\right)=24\Rightarrow a^2=25\Rightarrow a=\pm5\)

Theo cánh đặt,ta có

+,\(x^2-5x+5=5\Rightarrow x\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

+\(x^2-5x+5=-5\Rightarrow x^2-2\cdot\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\)

\(\Rightarrow\left(x-\frac{5}{2}\right)^2+\frac{15}{4}=0\)(vô lí)

Vậy 

1, \(2016^2+16^2-32\cdot2016=2016^2-2\cdot16\cdot2016+16^2\)

\(=\left(2016-16\right)^2=2000^2=4000000\)

2,\(A\left(x\right)=x^3-3x^2+2x+a=x^2\left(x-3\right)+2\left(x-3\right)+6+a\)

\(=\left(x-3\right)\left(x^2+2\right)+6+a\)

Vì \(6+a\)bậc bé hơn \(x-3\)nên

Để \(A\left(x\right)⋮B\left(x\right)\)thì \(6+a=0\Rightarrow a=-6\)