`(x+1)(x+3)-x.(x+5)=2022`

`x^2 + 4x +3-x^2 -5x = 2022`

`-x+3=2022`

`-x=2022-3`

`-x=-2019`

`=> x = 2019`

_____________________
`(2x-1)(12x+3)-(3x+1).(8x-1)=2022`

`24x^2 -6x-3-4x^2 -5x+1=2022`

`-11x-2=2022`

`-11x=2022+2`

`-11x=2024`

`x=2024:(-11)`

`x=-184`

a) \(\left(x+1\right)\left(x+3\right)-x\left(x+5\right)=2022\)

\(\Leftrightarrow x^2+3x+x+3-x^2-5x=2022\)

\(\Leftrightarrow-x=2019\)

\(\Leftrightarrow x=-2019\)

b) \(\left(2x-1\right)\left(12x+3\right)-\left(3x+1\right)\left(8x-1\right)=2022\)

\(\Leftrightarrow24x^2+6x-12x-3-24x^2+3x-8x+1=2022\)

\(\Leftrightarrow-11x=2024\)

\(\Leftrightarrow x=-184\)

Ai giúp mình với ạ!

`A = 2(x^2 + 3/2 x - 1)`

`= 2(x^2 + 2 . 3/4 x + 9/16 - 25/16)`

`= 2(x+3/4)^2 - 25/8 >= 0 - 25/8 = -25/8`

Dấu bằng xảy ra `<=> x = -3/4`.

Bạn làm tt với `b, c` nhé.

`A = x^2 - 1 - x^6 + 1`

`= -x^6 + x^2`

`B = x^6 + 27 - 27 - 27x^2 - 9x^4 - x^6`

`= -9x^2(3+x^2)`

ncl

Ta có: 

\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(y+z\right)+\left(y+z\right)^2\)

\(=\left[\left(x+y+z\right)-\left(y+z\right)\right]^2\)

\(=x^2\)

\(=x.x\)

\(VT=x^2-8x+16+y^2+4y+4+7=\)

\(=\left(x-4\right)^2+\left(y+2\right)^2+7\ge7\forall x;y\)

\(\Rightarrow VT\ge0\forall x;y\)