Ta có \(\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)

\(\Leftrightarrow1+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=-\dfrac{1}{2}\)

\(\Leftrightarrow\left(ab+bc+ac\right)^2=\dfrac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}\)

-> \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=1\)

\(\Leftrightarrow a^4+b^4+c^4=1-\dfrac{2.1}{4}=\dfrac{1}{2}\)

\(a+b+c=0\)

\(\left(a+b+c\right)^2=0\)

\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(2\left(ab+bc+ca\right)=-1\left(a^2+b^2+c^2=1\right)\)

\(ab+bc+ca=-\dfrac{1}{2}\)

\(\left(ab+bc+ca\right)=\dfrac{1}{4}\)

\(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)

\(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\dfrac{1}{4}\left(a+b+c=0\right)\)

+) \(a^2+b^2+c^2=1\)

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

\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)

\(a^4+b^4+c^4=1-\dfrac{1}{2}=\dfrac{1}{2}\left(a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}\right)\)

\(\dfrac{x}{18}\) = \(\dfrac{y}{15}\) = \(\dfrac{x-y}{18-15}\) = \(\dfrac{-6}{3}\) = -2

x = 18. (-2) = -36

y = 15 x (-2) = -30

x/18 = y/15 và x-y = -6

=> x/18 = y/15 => x-y/ 18-15 => -6 / 3 = -2

=> x = -2 . 18 = -36

=> y = -2 . 15 = -30

a) A\(=x^2+y^2+2x+15-6y\)

\(=x^2+y^2+2x+9+5+1-6y\)

\(=\left(x^2+2x+1\right)+\left(y^2-6y+9\right)+5\)

\(=\left(x+1\right)^2+\left(y-3\right)^2+5\)

Vì \(\left(x+1\right)^2\) luôn dương với mọi x

     \(\left(y-3\right)^2\) luôn dương với mọi y

\(\Rightarrow\left(x+1\right)^2+\left(y-3\right)^2+5\) \(\ge5>0\)

Vậy biểu thức A luôn dương với mọi x, y

Câu b) bạn làm tương tự nha

77^2-23^2

=5929-529

=5400

\(77^2-23^2\)

\(=5929-529\)

\(=5400\)

À đề là j ạ

\(17^2+2.17.33+33^2\)

\(=289+1122+1089\)

\(=2500\)

`x^3-6x.(x+2)`

`=x^3-6x^2 -12x`

\(1^2-2^2+3^2-4^2+...+999^2-1000^2\)

\(=\left(1-2\right)\left(1+2\right)+\left(1-2\right)\left(3+4\right)+...+\left(999-1000\right)\left(999+1000\right)\)

\(=-\left(1+2+3+4+...+999+1000\right)\)

\(=-\dfrac{\left(\dfrac{1000-1}{1}+1\right).\left(1000+1\right)}{2}=-500500\)

Không làm :)

đk a ≠ 1  ; a ≥ 0

\(=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{1-\sqrt{a}}\right)\\ =\left(1+\sqrt{a}\right)\left(1+\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right)\\ =\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)