a) \(\left(m+n\right)^2-\left(m-n\right)^2+\left(m+n\right)\left(m-n\right)\)

\(=\left(m+n+m-n\right)\left(m+n-m+n\right)+m^2-n^2\)

\(=m^2-n^2+4mn\)

b) \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3\)

\(=\left(a+b-a+b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]-2a^3\)

\(=2b\left[a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right]-2a^3\)

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

\(=2a^2b+6b^3-2a^3.\)

Tương tự áp dụng các HĐT.

a) \(\left(m+n\right)^2-\left(m-n\right)^2=\left[\left(m+n\right)-\left(m-n\right)\right]\left[\left(m+n\right)+\left(m-n\right)\right]=\left(2n\right)\left(2m\right)=4mn\)\(\left(m+n\right)\left(m-n\right)=m^2-n^2\)

A=\(4mn+m^2-n^2\) tối giản rồi

b)

\(\left(a+b\right)^3+\left(a-b\right)^3=\left[\left(a+b\right)+\left(a-b\right)\right]^3-3\left(a+b\right)\left(a-b\right)\left[\left(a+b\right)+\left(a-b\right)\right]=8a^3-3.2a.\left(a^2-b^2\right)\)B=\(8a^3-3.2a.\left(a^2-b^2\right)-2a^3=6a\left[a^2-\left(a^2-b^2\right)\right]=6ab^2\)

Câu c :

\(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2=\left(2x+1+2x-1\right)^2\)

Câu d :

\(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2=\left[\left(a+b+c\right)-\left(b+c\right)\right]^2=\left(a+b+c-b-c\right)^2\)

Bài 12:

1) A = x² - 6x + 11

= (x² - 6x + 9) + 2

= (x - 3)² + 2

Ta có: (x - 3)² ≥ 0 ∀ x

Dấu ''='' xảy ra khi x - 3 = 0 ⇔ x = 3

Do đó: (x - 3)² + 2 ≥ 2

Hay A ≥ 2

Dấu ''='' xảy ra khi x = 3

Vậy Min A = 2 tại x = 3

2) B = x² - 20x + 101

= (x² - 20x + 100) + 1

= (x - 10)² + 1

Ta có: (x - 10)² ≥ 0 ∀ x

Dấu ''='' xảy ra khi x - 10 = 0 ⇔ x = 10

Do đó: (x - 10)² + 1 ≥ 1

Hay B ≥ 1

Dấu ''='' xảy ra khi x = 10

Vậy Min B = 1 tại x = 10

Sao bạn KO tách ra cho dễ nhìn

a) \(\dfrac{2x-6}{x^2-x-6}\)

\(=\dfrac{2\left(x-3\right)}{x^2-3x+2x-6}\)

\(=\dfrac{2\left(x-3\right)}{x\left(x-3\right)+2\left(x-3\right)}\)

\(=\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+2\right)}\)

\(=\dfrac{2}{x+2}\)

b) \(\dfrac{6x^2-x-2}{4x^2-1}\)

\(=\dfrac{6x^2+3x-4x-2}{\left(2x\right)^2-1^2}\)

\(=\dfrac{3x\left(2x+1\right)-2\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\)

\(=\dfrac{\left(2x+1\right)\left(3x-2\right)}{\left(2x-1\right)\left(2x+1\right)}\)

\(=\dfrac{3x-2}{2x-1}\)

\(c,\dfrac{x^3-x^2+3x-3}{x^3+2x^2+3x+6}\)

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

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

d,Sửa đề :

\(\dfrac{a^2-b^2+c^2+2ac}{a^2+b^2-c^2+2ab}\)

\(=\dfrac{\left(a^2+2ac+c^2\right)-b^2}{\left(a^2+2ab+b^2\right)-c^2}\)

\(=\dfrac{\left(a+c\right)^2-b^2}{\left(a+b\right)^2-c^2}\)

\(=\dfrac{\left(a-b+c\right)\left(a+b+c\right)}{\left(a+b-c\right)\left(a+b+c\right)}\)

\(=\dfrac{a-b+c}{a+b-c}\)

e,g Đề ko rõ

\(\dfrac{2x-6}{x^2-x-6}=\dfrac{2\left(x-3\right)}{\left(x^2+2x\right)-\left(3x+6\right)}\\ =\dfrac{2\left(x-3\right)}{x\left(x+2\right)-3\left(x+2\right)}\\ =\dfrac{2\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{2}{x+2}\)

b.

\(\dfrac{6x^2-x-2}{4x^2-1}\\ =\dfrac{\left(6x^2+3x\right)-\left(4x+2\right)}{\left(2x-1\right)\left(2x+1\right)}\\ =\dfrac{3x\left(2x+1\right)-2\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\\ =\dfrac{3x-2}{2x-1}\)

c.

\(\dfrac{x^3-x^2+3x-3}{x^3+2x^2+3x+6}\\ =\dfrac{x^2\left(x-1\right)+3\left(x-1\right)}{x^2\left(x+2\right)+3\left(x+2\right)}\\ =\dfrac{\left(x-1\right)^2}{\left(x+2\right)^2}\)

mk chỉ đưa ra kết quả thui nghen chứ lm thì dài lm, bn coi kết quả r đối chiếu bài lm của bn ấy

a/ = x³ - 16x² + 25x

b/ = -2ab + a² + 2a

c/ = 2a²

A) \(\left(x-3\right)^2-\left(x+2\right)^2\)

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

\(=-5.\left(2x-1\right)\)

B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)

\(=8x^3-y^3-8x^3-y^3\)

\(=-2y^3\)

C) \(x^2+6x+8\)

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

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

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

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

bài 3 A) \(x^2-16=0\)

\(\left(x-4\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

B) \(x^4-2x^3+10x^2-20x=0\)

\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\left(x^3+10x\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

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

x=0

x=2

a)

A= (-m+n-p)-(-m-n-p)

A= -m+n-p+m+n+p

A= (-m+m) +(n+n) + (-p+p)

A= 0+2n+0

A = 2n

Bài 1: 

A = (-m + n - p) - (-m - n - p)

A = -m + n - p + m + n + p

A = (-m + m) + (n + n) - (p - p)

A = 2n

Với n = -1 => A = 2(-1) = -2

Bài 2: 

A = (-2a + 3b - 4c) - (-2a -3b - 4c)

A = -2a + 3b - 4c + 2a + 3b + 4c

A = (-2a + 2a) + (3b + 3b) - (4c - 4c)

A = 6b

Với b = -1 => A = 6(-1) = -6

Bài 3:

a) A = (a + b) - (a - b) + (a - c) - (a + c)

A= a + b - a + b + a - c - a - c

A = (a - a + a - a) + (b + b) - (c + c)

A = 2(b - c)

b) B = (a + b - c) + (a - b + c) - (b + c - a) - (a - b - c)

B = a + b - c + a - b + c - b - c + a - a + b + c

B = (a + a + a - a) + (b - b - b + b) - (c - c + c - c)

B = 2a