Ta có :

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

\(=a^2+ab+ac+ba+b^2+bc+ca+cb+c^2-2ab-2bc-2ca\)

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

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

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

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

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

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

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

Chúc bn học tốt

\(\text{(a+b+c).(a+b+c)-2(a.b+b.c+c.a}\)

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

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

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

Vậy.......

Học tốt!

a) Ta có: \(\dfrac{3a^2-10a+3}{2\left(a-3\right)}\)

\(=\dfrac{3a^2-9a-a+3}{2\left(a-3\right)}\)

\(=\dfrac{3a\left(a-3\right)-\left(a-3\right)}{2\left(a-3\right)}\)

\(=\dfrac{\left(a-3\right)\left(3a-1\right)}{2\left(a-3\right)}\)

\(=\dfrac{3a-1}{2}\)

\(=\dfrac{3}{2}a-\dfrac{1}{2}\)(đpcm)

b) Ta có: \(\dfrac{b^2+3b+9}{b^3-27}\)\(=\dfrac{b^2+3b+9}{\left(b-3\right)\left(b^2+3b+9\right)}\)

\(=\dfrac{1}{b-3}\)

\(=\dfrac{b-2}{\left(b-3\right)\left(b-2\right)}\)

\(=\dfrac{b-2}{b^2-5b+6}\)(đpcm)

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

\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=1^3-3.\left(-3\right).1=10\)

Ta có: \(a+b=1\)

\(\Leftrightarrow\left(a+b\right)^2=1\)

\(\Leftrightarrow a^2+b^2+2ab=1\)

\(\Leftrightarrow a^2+b^2-2\cdot3=1\)

\(\Leftrightarrow a^2+b^2=1+6=7\)

Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(=7-\left(-3\right)\)

\(=7+3=10\)

b: \(B=\left(x+2\right)^2-\left(2x-1\right)^2\)

\(=x^2+4x+4-4x^2+4x-1\)

\(=-3x^2+8x+3\)

B

B

Chứng minh đẳng thức sau với a,b,c thuộc Z:

                 a(b-c)-a(b+d)=-a(c+d)

\(ab-ac-ab+ad=-a\left(c+d\right)\)

\(a.\left(b-c-b+d\right)=-a\left(c+d\right)\)

\(-a.\left(c+d\right)\)= VP

\(\Rightarrowđpcm\)

chúc bạn học tốt

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