1) biến đổi vế trái:

= a²+2ab+b² -a² +2ab -b²

=4ab = vế phải ( đpcm)

3;5 tuong tu

1) (a + b)² - (a - b)² = a² + 2ab + b² - a² + 2ab - b² = 4ab

3) (a + b)² - 4ab = a² + 2ab + b² - 4ab = a² - 2ab + b² = (a - b)²

5) a³ + b³ = a³ + 3a²b + 3ab² + b³ - 3a²b - 3ab² = (a + b)³ - 3ab(a + b)

1) (a+b)^2 - (a-b)^2 = 4ab

VT= (a + b - a +b) (a+b + a-b) 

= 2b * 2a 

= 4ab = VP

Vậy (a+b)^2 - (a-b)^2 = 4ab

2) (a+b)^2 - 4ab = (a-b)^2 

VT= (a+b)^2 - 4ab 

= a^2 + 2ab + b^2 - 4ab

= a^2 - 2ab + b^2 

= (a-b)^2  = VP

Vậy (a+b)^2 - 4ab = (a-b)^2 

a) Ta có: \(\left(a-1\right)^2\ge0\forall a\)

\(\Leftrightarrow a^2-2a+1\ge0\forall a\)

\(\Leftrightarrow a^2+2a+1\ge4a\forall a\)

\(\Leftrightarrow\left(a+1\right)^2\ge4a\)(đpcm)

b) Áp dụng bất đẳng thức Cosi ta có:

\(a+1\ge2\sqrt{a};b+1\ge2\sqrt{b};c+1\ge2\sqrt{c}\\ \Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\)

Dấu = xảy ra khi và chỉ khi a=b=c=1

Ta có : 

\(\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)( luôn đúng ) 

\(\left(đpcm\right)\)

Chứng minh phản chứng :

Giả sử : \(\left(a+b\right)^2< 4ab\)

\(\Rightarrow a^2+2ab+b^2< 4ab\)

\(\Rightarrow a^2-2ab+b^2< 0\)

\(\Rightarrow\left(a-b\right)^2< 0\) (vô lí )

Vậy cần có :

\(\left(a+b\right)^2\ge4ab\)

a, \(\left(a+b+c\right)^2=\left[\left(a+b\right)+c\right]^2=\left(a+b\right)^2+2c\left(a+b\right)+c^2=a^2+b^2+c^2+2ab+2ac+2bc\)

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

c, \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b-a+b\right)\left(a+b+a-b\right)=2b.2a=4ab\)

\(\left(a+b+c\right)^2=\left[\left(a+b\right)+c\right]^2=\left(a+b\right)^2+2\cdot\left(a+b\right)\cdot c+c^2\\ =a^2+2ab+b^2+2ac+2bc+c^2\\ =a^2+b^2+c^2+2ab+2ac+2bc\)

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

\(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)\\ =2a\cdot2b=4ab\)

a) (a+b+c)² = (a+b)² + 2(a+b)c + c² = a² + 2ab +b² + 2ac+ 2bc+ c²

b) (a+b)² + (a-b)² = a²+ 2ab+ b²+ a²- 2ab +b²= 2a² + 2b²

c) (a+b)²- (a-b)² = a²+ 2ab+ b²- a²+ 2ab- b² = 4ab

\(\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\left(đpcm\right)\)

\(\left(a+b\right)^2\ge4ab\)

<=>  \(a^2+2ab+b^2\ge4ab\)

<=>  \(a^2+2ab+b^2-4ab\ge0\)

<=>  \(a^2-2ab+b^2\ge0\)

<=>  \(\left(a-b\right)^2\ge0\)  luôn đúng

Dấu "=" xảy ra <=> a=b

Mình không biết làm.

a) (a+b)² + (a-b)² = 2(a² +b²)

VT = (a² + 2ab + b²) + (a² - 2ab + b²) = a² +2ab + b² + a² - 2ab + b² = 2a² + 2b² = 2(a² + b²) = VP (đpcm)

b) (a+b)² = (a-b)² + 4ab

VP = (a-b)² + 4ab = a² - 2ab + b² + 4ab = a² + 2ab +b² = (a+b)² = VT (đpcm)

Ta có 12 ≥ ( a + b ) 3 + 4 a b ≥ 2 a b 3 + 4 a b . Đặt t = a b , t > 0  thì

12 ≥ 8 t 3 + 4 t 2 ⇔ 2 t 3 + t 2 − 3 ≤ 0 ⇔ ( t − 1 ) ( 2 t 2 + 3 t + 3 ) ≤ 0  

Do 2 t 2 + 3 t + 3 > 0 , ∀ t nên t − 1 ≤ 0 ⇔ t ≤ 1 . Vậy 0 < a b ≤ 1  

Chứng minh được 1 1 + a + 1 1 + b ≤ 2 1 + a b , ∀ a , b > 0  thỏa mãn a b ≤ 1  

Thật vậy, BĐT 1 1 + a − 1 1 + a b + 1 1 + b − 1 1 + a b ≤ 0  

a b − a ( 1 + a ) ( 1 + a b ) + a b − b ( 1 + b ) ( 1 + a b ) ≤ 0 ⇔ b − a 1 + a b a 1 + a − b 1 + b ⇔ ( b − a ) 2 ( a b − 1 ) ( 1 + a b ) ( 1 + a ) ( 1 + b ) ≤ 0  

Do 0 < a b ≤ 1  nên BĐT này đúng

Tiếp theo ta sẽ CM 2 1 + a b + 2015 a b ≤ 2016 , ∀ a , b > 0  thỏa mãn  a b ≤ 1

Đặt t = a b , 0 < t ≤ t  ta được 2 1 + t + 2015 t 2 ≤ 2016  

2015 t 3 + 2015 t 2 − 2016 t − 2014 ≤ 0 ⇔ ( t − 1 ) ( 2015 t 2 + 4030 t + 2014 ) ≤ 0  

BĐT này đúng  ∀ t : 0 < t ≤ 1  

Vậy  1 1 + a + 1 1 + b + 2015 a b ≤ 2016.  Đẳng thức xảy ra a = b = 1

a) Sửa đề :

\(x^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

\(x^4=\left(a^4+3a^3b+3a^2b^2+ab^3\right)+\left(a^3b+3a^2b^2+3ab^3+b^4\right)\)

\(x^4=a\left(a^3+3a^2b+3ab^2+b^3\right)+b\left(a^3+3a^2b+3ab^2+b^3\right)\)

\(x^4=\left(a+b\right)\left(a^3+3a^2b+3ab^2+b^3\right)\)

\(x^4=\left(a+b\right)\left[\left(a^3+2a^2b+ab^2\right)+\left(a^2b+2ab^2+b^3\right)\right]\)

\(x^4=\left(a+b\right)\left[a\left(a^2+2ab+b^2\right)+b\left(a^2+2ab+b^2\right)\right]\)

\(x^4=\left(a+b\right)^2\left(a+2ab+b^2\right)\)

\(x^4=\left(a+b\right)^4\)

b) Sửa đề:

 \(x^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\)

\(x^5=\left(a^5+4a^4b+6a^3b^2+4a^2b^3+ab^4\right)+\left(a^4b+4a^3b^2+6a^2b+4ab^4+b^5\right)\)

\(x^5=a\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)+b\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)\)

\(x^5=\left(a+b\right)\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)\)

\(x^5=\left(a+b\right)\left[\left(a^4+3a^3b+3a^2b^2+ab^3\right)+\left(a^3b+3a^2b^2++3ab^3+b^4\right)\right]\)

\(x^5=\left(a+b\right)\left[a\left(a^3+3a^2b+3ab^2+b^3\right)+b\left(a^3+3a^2b+3ab^2+b^3\right)\right]\)

\(x^5=\left(a+b\right)^2\left(a^3+3a^2b+3ab^2+b^3\right)\)

\(x^5=\left(a+b\right)^2\left[\left(a^3+2a^2b+ab^2\right)+\left(a^2b+2ab^2+b^3\right)\right]\)

\(x^5=\left(a+b\right)^2\left[a\left(a^2+2ab+b^2\right)+b\left(a^2+2ab+b^2\right)\right]\)

\(x^5=\left(a+b\right)^3\left(a^2+2ab+b^2\right)\)

\(x^5=\left(a+b\right)^5\)

Bạn có thể tự tóm tắt lại

f