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

ta có

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

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

= a\(^2\) + b\(^2\) ( đpcm)

1) \(\left(a+b\right)^2\)

\(=\left(a+b\right)\left(a+b\right)\)

\(=a^2+ab+ab+b^2\)

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

2) \(\left(a-b\right)^3\)

\(=\left(a-b\right)\left(a-b\right)\left(a-b\right)\)

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

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

\(=a^3-a^2b-2a^2+2ab^2+ab^2-b^3\)

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

`a)` 

`(a+b)^2`

`=(a+b)(a+b)`

`=a^2+ab+ab+b^2`

`=a^2+2ab+b^2`

`->` ĐPCM

`b)` `(a-b)^3`

`=(a-b)(a-b)(a-b)`

`=(a^2-2ab+b^2)(a-b)`

`=a^3-3a^2b+3ab^2-b^3`

`->` ĐPCM

(a+b)²

= (a+b).(a+b)

= a.a+a.b+b.a+b.b

= a²+ab+ab+b²

= a²+2ab+b²

=> đpcm

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

(a-b)^2=(a-b)(a-b)=a^2-ab-ab+b^2=a^2-2ba+b^2

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

(a+3)^3=(a+b)^2*(a+b)

=(a^2+2ab+b^2)(a+b)

=a^3+a^2b+2a^2b+2ab^2+b^2a+b^3

=a^3+3a^2b+3ab^2+b^3

b) \(\left(a-b\right)\left(a-b\right)\)

\(=a\left(a+b\right)-b\left(a-b\right)\)

\(=a^2-ab-ba+b^2\)

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

a) \(\left(a+b\right)\left(a+b\right)\)

\(=a\left(a+b\right)+b\left(a+b\right)\)

\(=a^2+ab+ba+b^2\)

\(=a^2+2ab+b^2\)

c) \(\left(a-b\right)\left(a+b\right)\)

\(=a\left(a+b\right)-b\left(a+b\right)\)

\(=a^2+ab-ba-b^2\)

\(=a^2-b^2\)

3 số thực dương nhé.

Áp dụng bất đẳng thức Cauchy Schwarz dạng Engel có :

\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\ge\frac{\left(1+1+1\right)^2}{\left(a^2+2bc\right)+\left(b^2+2ca\right)+\left(c^2+2ab\right)}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1^2}=9\)

Dấu bằng xảy ra \(\Leftrightarrow\frac{1}{a^2+2bc}=\frac{1}{b^2+2ca}=\frac{1}{c^2+2ab}\)và \(a+b+c=1\)

\(\Leftrightarrow a^2+2bc=b^2+2ca=c^2+2ab\)

Mong có ai giúp mình từ đẳng thức trên giải ra a=b=c.

a=b=c ket hop voi a+b+c=<1 =>a=b=c=1/3 nhe

\(a^2+2bc=b^2+2ca=c^2+2ab\)

Ta có: \(\hept{\begin{cases}a^2+2bc=b^2+2ca\\b^2+2ca=c^2+2ab\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)\left(a+b\right)-2c\left(a-b\right)=0\\\left(b-c\right)\left(b+c\right)-2a\left(b-c\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)\left(a+b-2c\right)=0\\\left(b-c\right)\left(b+c-2a\right)=0\end{cases}}\)

Tới đây thì suy được ra là \(a=b=c\) rồi nhé Trần Thùy Dung - Trang của Trần Thùy Dung - Học toán với OnlineMath

Ta dùng hằng đẳng thức

(a^2 + b^2) - 2ab = a^2 - 2ab + b^2 = (a-b)^2.

Mà (a-b)^2 >= 0 nên ta suy ra a^2+b^2 >= 2ab.

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Bài này mà là toán lớp 3 á ??????????