`3x^2+4x=2x`

`<=>3x^2+2x=0`

`<=>x(3x+2)=0`

`<=>x=0,3x+2=0`

`<=>x=0,x=(-2)/3`

Vậy `x=0,x=(-2)/3`

help me

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

=\(\left(x+y+3\right)\left(x+y-3\right)\)

xin tiick

`(x^2-2xy+y^2)-9`

`=(x-y)^2-3^2`

`=(x-y-3)(x-y+3)`

\((3x-2).(3x+2)-4x.(2x+3)-(2x-1)^2\)

\(=(3x)^2-2^2-8x^2-12x-[(2x)^2-2.2x.1+1^2]\)

\(=9x^2-4-8x^2-12x-4x^2+4x-1\)

\(=(9x^2-8x^2-4x^2)+(4x-12x)-(4+1)\)

\(=-3x^2-8x-5\)

\((x+2).(x^2+3x+1)-(x+1).(x^2-x+1)+(x-2)^3\)

\(=x^3+3x^2+x+2x^2+6x+2-(x^3-1^3)+(x^3-3x^2.2+3x.2^2-2^3)\)

\(=x^3+3x^2+x+2x^2+6x+2-x^3-1+x^3-6x^2+12x-8\)

\(=(x^3-x^3+x^3)+(3x^2+2x^2-6x^2)+(x+6x+12x)+(2-1-8)\)

\(=x^3-x^2+19x-7\)

ta có: 

ta có :

Mình cần gấp lắm rồi

Từ a⁴ + b⁴ \(\ge\)2a²b² cộng a² + b² vào 2 vế

\(a^4+b^4\ge\frac{1}{2}\left(a^2+b^2\right)^2\)

Tương tự\(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\)

Từ đó suy ra \(a^4+b^4\ge\frac{1}{8}\left(a+b\right)^2\)

Cái cuối là \(a^4+b^4\ge\frac{1}{8}\left(a+b\right)^4\)nha mình nhầm

\(S=2x+4y+6z\le2\sqrt{\left[x^2+\left(2y\right)^2+\left(3z\right)^2\right]\left(1^2+1^2+1^2\right)}=2\sqrt{3.3}=6\)

Dấu \(=\)khi \(\hept{\begin{cases}x^2+4y^2+9z^2=3\\\frac{x}{1}=\frac{2y}{1}=\frac{3z}{1}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\\z=\frac{1}{3}\end{cases}}\).

\(4=x^2+y^2-xy=\frac{1}{2}\left(x^2+y^2\right)+\frac{1}{2}\left(x-y\right)^2\ge\frac{1}{2}\left(x^2+y^2\right)\)

\(\Leftrightarrow x^2+y^2\le8\)

Dấu \(=\)khi \(x=y=\pm2\).

\(4=x^2+y^2-xy=\frac{3}{2}\left(x^2+y^2\right)-\frac{1}{2}\left(x+y\right)^2\le\frac{3}{2}\left(x^2+y^2\right)\)

\(\Leftrightarrow x^2+y^2\ge\frac{8}{3}\)

Dấu \(=\)khi \(x=-y=\pm\frac{2}{\sqrt{3}}\).