`(x-5)-x(x+3)=6x`

`<=>x-5-x^2-3x-6x=0`

`<=>-x^2-8x-5=0`

`<=>x^2+8x+5=0`

`<=>(x+4)^2-11=0`

`=>` `x+4=\sqrt{11}` hoặc `x+4=-\sqrt{11}`

`=>x=-4+\sqrt{11}` hoặc `x=-4-\sqrt{11}`

( x-5 )2-x ( x+3 )=6 x

<=> (x-5) - x² -3x =6x

<=> x² + 8x + 5= 0

<=> x= -4+ \(\sqrt{11}\);-4- \(\sqrt{11}\)

( x-5 )2-x ( x+3 )=6 x

= (a+1)(a-1)(a+2)(a-2)

= \(\left(a^2-1\right)\left(a^2-4\right)=a^4-4a^2-a^2+4=a^4-5a^2+4\)

(3−x)2−(x+2)2−6(x−1)

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

\(77^2-23^2\)

\(=5929-529\)

\(=5400\)