1) x² -7x + 10 = x² - 2x - 5x + 10 = x(x - 2) - 5(x - 2) = (x - 5)(x - 2)

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

3) x² - 7x + 12 = x² - 3x - 4x + 12 = x(x - 3) - 4(x - 3) = (x - 3)(x - 4)

4) x² + 7x + 12 = x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)

5) 16x - 5x² - 3 = 15x - 5x² + x - 3 = -5x(x - 3) + (x - 3) = (x - 3)(1 - 5x) 

6) 6x² + 7x - 3 = 6x² - 2x + 9x - 3 = 2x(3x - 1) + 3(3x - 1) = (2x + 3)(3x - 1)  

7) 3x² - 3x - 6 = 3x² - 6x + 3x - 6 = 3x(x - 2) + 3(x - 2) = (x - 2)(3x + 3) = 3(x - 2)(x + 1)

8) 3x² + 3x - 6 = 3x² - 3x + 6x - 6 = 3x(x - 1) + 6(x - 1) = (x - 1)(3x + 6) = 3(x - 1)(x + 2)

9) 6x² - 13x + 6 = 6x² - 9x -  4x + 6 = 3x(2x - 3) - 2(2x - 3) = (3x - 2)(2x - 3) 

10) 6x² + 15x  + 6 = 6x² + 12x + 3x + 6 = 6x(x + 2) + 3(x + 2) = (x + 2)(6x + 3) = 3(x + 2)(3x + 1)

11) 6x² - 20x + 6 = 6x² - 18x - 2x + 6 = 6x(x -3) - 2(x - 3) = (6x - 2)(x - 3) = 2(3x - 1)(x - 3)

12) 8x² + 5x - 3 = 8x² + 8x - 3x - 3 = 8x(x + 1) - 3(x + 1) = (x + 1)(8x - 3)

Thu gọn đa thức được:

  \(2x^5-4x^4-3x^2+1\)

Vậy bậc của đa thức là 5.(Nếu sai do tính toán thì đừng cmr nha)

dù sao thì cũng nên l i k e bài nào phải làm đủ hơn chứ vì đã nói là chỉ viết đấp số ngay từ đầu đâu

(5x^7-5x^7)+(-7x^6+7x^6)+(5x^5-3x^5)-4x^4-3x^2+1

=(5-5)*x^7+(-7+7)*x^6+(5-3)*x^5-4x^4-3x^2+1

=0+0+2x^5-4x^4-3x^2+1

=2x^5-4x^4-3x^2+1

a)x²-5x+6=(x²-2x)-(3x-6)=x(x-2)-3(x-2)(=(x-2)(x-3)

b)x²+5x+6=(x²+2x)+(3x+6)=x(x+2)+3(x+2)=(x+2)(x+3)

c)x²-7x+12=(x²-3x)-(4x-12)=x(x-3)-4(x-3)=(x-3)(x-4)

d)x²+7x+12=(x²+3x)+(4x+12)=x(x+3)+4(x+3)=(x+3)(x+4)

\(x^2+5x-6\\ =x^2-x+6x-6\\ =x\left(x-1\right)+6\left(x-1\right)\\ =\left(x+6\right)\left(x-1\right)\\ ---\\ 5x^2+5xy-x-y\\ =x\left(5x-1\right)+y\left(5x-1\right)\\ =\left(5x-1\right)\left(x+y\right)\\ ----\\ 7x-6x^2-2\\ =-6x^2+3x+4x-2\\ =-3x\left(2x-1\right)+2\left(2x-1\right)\\ =\left(2-3x\right)\left(2x-1\right)\)

\(x^2+5x-6=\left(x-1\right)\left(x+6\right)\\ 5x^2+5xy-x-y=\left(5x-1\right)\left(x+y\right)\\ 7x-6x^2-2=\left(3x-2\right)\left(2x-1\right)\)

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

2.|x - 6| + 7x - 2 = |x - 6| + 5x

<=> |x - 6| + 7x - 2 = 5x

<=> |x - 6| + 2x - 2 = 0

<=> |x - 6| = 2 - 2x

\(\Leftrightarrow\orbr{\begin{cases}x-6=2-2x\\x-6=-2+2x\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=8\\-x=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-4\end{cases}}\)

a) \(5x\left(\frac{1}{5}x-2\right)+3\left(6-\frac{1}{3}x^2\right)=12\)

=> \(x^2-10x+18-x^2=12\)

=> -10x + 18 = 12

=> -10x = -6

=> -5x = -3

=> x = 3/5

b) 7x(x - 2) - 5(x - 1) = 7x² + 3

=> 7x² - 14x - 5x + 5 = 7x² + 3

=> 7x² - 14x - 5x + 5 - 7x² - 3 = 0

=> -19x + 2 = 0

=> -19x = -2

=> x = \(\frac{2}{19}\)

c) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11

=> 10x - 16 - 12x + 15 = 12x - 16 + 11

=> 10x - 16 - 12x + 15 - 12x + 16 - 11 = 0

=> (10x - 12x - 12x) + (-16 + 15 + 16 - 11) = 0

=> -14x + 4 = 0

=> -14x = -4

=> -7x = -2

=> x = 2/7

ĐKXĐ: ...

\(\Leftrightarrow x^2+5x+8=3\sqrt{\left(2x+3\right)\left(x^2+x+2\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3}=a\\\sqrt{x^2+x+2}=b\end{matrix}\right.\)

\(\Rightarrow2a^2+b^2=3ab\Leftrightarrow2a^2-3ab+b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\Rightarrow\left[{}\begin{matrix}a=b\\b=2a\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=2x+3\\x^2+x+2=8x+12\end{matrix}\right.\)