Question

démontrer que les égalité suivantes sont vraies pour n'importe quelles valeurs de a et b

4ab=(a+b)²-(a-b)²

(a+b)(a-b)+b²=ab+a(a-b)

Answer 1

Answer:

(a+b)^2 = (a+b) (a+b) = A^2+B^2+AB+AB

(a-b)^2= (a-b) (a-b) = A^2+B^2-AB-AB

Equate the two equations:

(A^2+B^2+2AB) - (A^2+B^2-2AB)

The minus sign means that both the A^2 and B^2 cancel:

2AB - - 2AB= 4AB; Both minus signs next to each other means that they are replaced with a + sign.

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Step-by-step explanation:

Answer 2

Step-by-step explanation:

Hi friend, Here is the required answer:-

LHS-. (a-b)²= a²+b²-2ab

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

Since LHS = RHS

So this equation is verified.

I hope you understood..

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