Em nên gõ công thức trực quan để được hỗ trợ tốt nhất nhé

Em nên gõ công thức trực quan để được hỗ trợ tốt nhất nhé

\(x^2+y^2+z^2=xy+3x+2z-4\)

\(\Leftrightarrow x^2+y^2+z^2-xy-3x-2z+4=0\)

\(\Leftrightarrow\left(\dfrac{1}{4}x^2-xy+y^2\right)+\left(z^2-2z+1\right)+3.\left(\dfrac{1}{4}x^2-x+1\right)=0\)

\(\Leftrightarrow\left(\dfrac{1}{2}x-y\right)^2+\left(z-1\right)^2+3.\left(\dfrac{1}{2}x-1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=y\\z=1\\\dfrac{1}{2}x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\z=1\\x=2\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(2;1;1\right)\)

=>x^2+y^2+z^2-xy-3y-2z+4=0
=>x^2-xy+1/4y^2+3/4y^2-3y+3+z^2-2z+1=0
=>(x-1/2y)^2+3/4(y-2)^2+(z-1)^2=0
=>(x-1/2y)^2=0 (y-2)^2=0 (z-1)^2=0 
=>x=1/2y y=2 z=1
=>x=1,y=2,z=1

     (x+1)² + (x-2)(x+3) -4x

=    x² + 2x + 1 + x²+3x - 2x   -6 - 4x

=2 x² -x  - 5

\(\left(1-3x^2\right)^2=\left(5x+2\right)^2\\ =>\left[{}\begin{matrix}1-3x^2=5x+2\\1-3x^2=-\left(5x+2\right)\end{matrix}\right.< =>\left[{}\begin{matrix}3x^2+5x+1=0\\3x^2-5x-3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{-5\pm\sqrt{13}}{6}\\x=\dfrac{5\pm\sqrt{61}}{6}\end{matrix}\right.\\ =>S=\left\{\dfrac{-5+\sqrt{13}}{6};\dfrac{-5-\sqrt{13}}{6};\dfrac{5+\sqrt{61}}{6};\dfrac{5-\sqrt{61}}{6}\right\}\)

= (3x)⁴ + 4(3x)³(4y) + 6(3x)²(4y)² + 4(3x)(4y)³ + (4y)³

= 81x⁴ + 432x³y + 864x²y² + 768xy³ + 64y³