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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=\(8a^3-3.2a.\left(a^2-b^2\right)-2a^3=6a\left[a^2-\left(a^2-b^2\right)\right]=6ab^2\)
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
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
a, <=>y2-32 <=> y2 -9 (hằng đẳng thức số 3)
b, <=> m3+n3 ( hằng đẳng thức số 6)
c, <=> 23-a3 (__________________số 7)
d, <=> (a-b-c-a+b-c )( a-b-c+a-b+c)
<=> -2c*2a= -4ac
e, <=> (a-x-y-a-x+y) [(a-x-y) 2+(a-x-y)(a+x-y)+(a+x-y)2]
(Nhân phá ngoặc) -)
d <=> (1-x2)[(1+x2)2-x2)
<=> (1-x2)(1+2x2)
<=> 1+2x2-x2-2x4
<=> 1+x2-2x4
a) \(\cdot\left(m+n\right)^2-\left(m-n\right)^2+\left(m+n\right)\left(m-n\right)\)
\(=\left(m+n+m-n\right)\left(m+n-m+n\right)+\left(m+n\right)\left(m-n\right)\)
\(=\left(2m\cdot2n\right)+m^2-n^2\)
\(=4mn+m^2-n^2\)
b) \(\left(a+b\right)^2-\left(a-b\right)^2-2a^3\)
\(=\left(a+b+a-b\right)\left(a+b-a+b\right)-2a^3\)
\(=2ab-2a^3\)
c) \(\left(2x+1\right)^2+\left(2x-1\right)^2+2\left(4x^2-1\right)\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2\)
\(=\left(4x\right)^2=16x^2\)
d) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(a+b+c-b-c\right)^2=a^2\)
xin lỗi mk ghi sai đề ở bài :d) (a+b+c)^2-2(a+b+c)(b+c)+(b+c)^2