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

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

\(\Rightarrowđpcm\)

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

\(VP=a^4+b^4+2a^2b^2-2a^2b^2=a^4+b^4=VT\)\(\Rightarrowđpcm\)

c) ​\(a^6+b^6=\left(a^2+b^2\right)\left[\left(a^2+b^2\right)^2-3a^2b^2\right]\)

\(VP=\left(a^2+b^2\right)\left(a^4-a^2b^2+b^4\right)=a^6+b^6\)

\(VP=VT\Rightarrowđpcm\)

d)\(a^6-b^6=\left(a^2-b^2\right)[\left(a^2+b^2\right)^2-a^2b^2]\)

\(VP=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)=a^6-b^6=VT\)

\(VP=VT\Rightarrowđpcm\)

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

b. (a+b)^3= (a+b)(a+b)(a+b) = (a^2 + 2ab + b^2)(a + b) = a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3 = a^3 + 3a^2b + 3b^2a + b^3

c. (a-b)^3= (a - b)(a-b)(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

e. (a-b) ( a^2 + ab +b^2) = a^3 + a^2b + b^2a - ba^2 - ab^2 - b^3 = a^3 - b^3

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

Chọn B

B

a) \(\left(2x-3y\right)^2=4x^2-12xy+9y^2\)

b) \(\left(5p-q\right)^2=25p^2-10pq+q^2\)

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

d) \(\left(1+3s\right)^2=1+6s+9s^2\)

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

f) \(\left(3u-v\right)^3=27u^3-27u^2v+9uv^2-v^3\)

a,\(\left(2x-3y\right)=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2\)

=\(4x^2-12xy+6y^2\)

b,\(\left(5p-q\right)^2=\left(5p\right)^2-2.5p.q+q^2\)

=\(25p^2-10pq+q^2\)

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

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

d,\(\left(1+3s\right)^2=1+6s+9s^2\)

e,(a\(^2b+2b)^2=(a^2b)^2+2.a^2b.2b^2+\left(2b\right)^2\)

=\(a^4b^2+4a^2b^2+4b^2\)

f,\(\left(3u-v\right)^3=27u^3-27u^2v+9uv^2-v^3\)

a) Ta có: \(a+b+c=0\)

\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)

\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(b+a+c\right)\right]\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow a^4+b^4+c^4=4\left(a^2b^2+b^2c^2+c^2a^2\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

b) Ta có: \(a+b+c=0\)

\(\Rightarrow2abc\left(a+b+c\right)=0\)

\(\Rightarrow2a^2bc+2ab^2c+2abc^2=0\)

Ta lại có:

\(a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)^2\)(chứng minh câu a)

\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+4a^2bc+4ab^2c+4abc^2\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)

\(\Rightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)