2/\(ĐKXĐ:x\ne-1\)

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

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

Đặt \(\frac{2}{x+1}=t\)

\(\Rightarrow Q=t^2-2t+2=\left(t-1\right)^2+1\ge1\forall t\)

\(\Rightarrow minQ=1\Leftrightarrow t=1\)

                           \(\Leftrightarrow\frac{2}{x+1}=1\)

                         \(\Leftrightarrow x=1\left(tmđkxđ\right)\)             

Ta có: \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}=\frac{2^2}{2}=2\)

=> \(A\le\frac{2019}{2.2+2016}=\frac{2019}{2020}\)

Dấu "=" xảy ra <=> a = b = 1

x(x²+6x+9) - 3x= x³+6x²+12x+8+1

\(\Leftrightarrow\)x³+6x²+9x-3x=x³+6x²+12x+9

\(\Leftrightarrow\)6x=12x+9

\(\Leftrightarrow\)6x=-9

\(\Leftrightarrow\)x=-3/2

Vậy phương trình có 1 nghiệm duy nhất x=-3/2

x(x + 3)^2  - 3x = (x + 2)^3 + 1

<=> x(x^2 + 6x + 9) = x^3 + 6x^2 + 12x + 8 + 1

<=> x^3 + 6x^2 + 9x = x^3 + 6x^2 + 12x + 9

<=> 3x + 9 = 0

<=> 3x = -9

<=> x = -3

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

\(\Leftrightarrow x^3-6x^2+9x-3x^2+18x-27-2x+2=x^3-4x^2+4x-5x^2\)

\(\Leftrightarrow x^3-9x^2+25x-25=x^3-9x^2+4x-5x^2\)

\(\Leftrightarrow x^3-9x^2+25x-25=x^3-9x^2+4x\)

\(\Leftrightarrow-9x^2+25x-25=-9x^2+4x\)

\(\Leftrightarrow25x-25=4x\)

\(\Leftrightarrow-25=4x-25x\)

\(\Leftrightarrow-25=-21x\)

\(\Leftrightarrow x=\frac{21}{25}\)

\(Pt\Leftrightarrow x^3-1-3x^2+3x-2x+2-x^3+4x^2-4x+5x^2=0\)

\(\Leftrightarrow6x^2-5x+1=0\)

\(\Leftrightarrow x=\frac{3\pm\sqrt{3}}{6}\)

5²-5=25-5=20 không chia hết cho 6 nhé

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

\(\Leftrightarrow4x^2-4x+1+5=4x^2-9-x\)

\(\Leftrightarrow4x^2-4x^2-4x+x=-9-5-1\)

\(\Leftrightarrow-3x=-15\)

\(\Leftrightarrow x=5\)

Vậy x=5