\(a+b=3\)

\(\Rightarrow\left(a+b\right)^2=9\)

\(\Rightarrow a^2+2ab+b^2=9\)

\(\Rightarrow7+2ab=9\)

\(\Rightarrow2ab=2\Rightarrow ab=1\)

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=3.\left(7-1\right)=18\)

1 sai

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

2 đúng

3 đúng

4 sai

(x-3)^2=-(3-x)^2

5 sai

(x-3)^3=-(3-x)^3

Ta có:\(x+y=a\)

=>\(x^2+2xy+y^2=a^2\)

=>\(x^2+y^2=a^2-2xy=a^2-2b\left(đpcm\right)\)

Ta lại có:\(x^3+3x^2y+3xy^2+y^3=a^3\)

=>\(x^3+y^3+3xy\left(x+y\right)=a^3\)

=>\(x^3+y^3=a^3-3xy\left(x+y\right)=a^3-3ab\left(đpcm\right)\)

b)\(a+b+c=0\) =>\(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3a^2c+6abc=0\) =>\(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\) =>\(a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)=0\) =>\(a^3+b^3+c^3=3abc\left(đpcm\right)\)

a) Nếu a = 2 và b = 1 thì a – b = 2 – 1 = 1.

b) Nếu m = 6 và n = 3 thì: m + n = 6 + 3 = 9.

m – n = 6 -3 = 3.

m × n = 6× 3 = 18.

m : n = 6 : 3 = 2.

\(\dfrac{a+2}{a-2}=\dfrac{b+3}{b-3}\\ \Leftrightarrow\dfrac{a+2}{b+3}=\dfrac{a-2}{b-3}=\dfrac{a+2+a-2}{b+3+b-3}=\dfrac{2a}{2b}=\dfrac{a}{b}\\ \dfrac{a+2}{b+3}=\dfrac{a-2}{b-3}=\dfrac{4}{6}=\dfrac{2}{3}\\ \Rightarrow\dfrac{a}{b}=\dfrac{2}{3}\)

Ta có:

\(\dfrac{a+2}{a-2}=\dfrac{b+3}{b-3}.\)

\(\Leftrightarrow\dfrac{a+2}{b+3}=\dfrac{a-2}{b-3}=\dfrac{a+2+a-2}{b+3+b-3}=\dfrac{2a}{2b}=\dfrac{a}{b}.\)

\(\Rightarrow\dfrac{a+2}{b+3}=\dfrac{a-2}{b-3}=\dfrac{a}{b}=\dfrac{4}{6}=\dfrac{2}{3}.\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{2}{3}.\)

\(\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}\left(đpcm\right).\)

a) Nếu a : 3 và b : 3 thì tổng a + b chia hết cho  3

b) Nếu a : 2 và b : 4 thì tổng a + b chia hết cho  2 

c) Nếu a : 6 và b : 9 thì tông a + b chia hết cho  3 

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Gạch dưới số mà bạn chọn :

a) Nếu a : 3 và b : 3 thì tổng a + b chia hết cho 3

b) Nếu a : 2 và b : 4 thì tổng a + b chia hết cho 2

c) Nếu a : 6 và b : 9 thì tông a + b chia hết cho  3 

Nếu a = 8, b = 5, c =2 thì: a + b + c = 8 + 5 + 2 = 15.

a – b – c = 8 – 5 -2 = 1.

a × b × c = 8 × 5 × 2 = 80.

Ta có:

\(N=\left(a-2\right)\left(a+3\right)-\left(a-3\right)\left(a+2\right)\)

\(N=a^2+3a-2a-6-\left(a^2+2a-3a-6\right)\)

\(N=a^2+a-6-a^2+a-6\)

\(N=2a\)

Mà: \(2a\) luôn chẵn với mọi a

\(\Rightarrow N\) chẵn với mọi a

N=(a+3)(a-2)-(a-3)(a+2)

=a^2-2a+3a-6-(a^2+2a-3a-6)

=a^2+a-6-a^2+a+6

=2a là số chẵn