Question

[Bạch An Nhiên]

1) Tìm GTLN của biểu thức

a) C= -x^2 + 2xy - 4y^2+2x+10y-8

b) D= -2x^2 - 9y^2 + 6xy + 6x + 12y - 2000

2) Tìm x biết

a) x^3 - 5x^2 + 8x - 4 = 0

b) 2x^3 - x^2 + 3x + 6 =0

c) (x^2+x)(x^2 +x + 1 ) = 6

d) (x^2 - 4x)^2 - 8(x^2-4x) +15 = 0

Answer 1

[Nguyễn Nam]

2)

a) \(x^3-5x^2+8x-4=0\)

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

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

\(\Leftrightarrow\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy x=1 ; x=2

b) \(2x^3-x^2+3x+6=0\)

\(\Leftrightarrow2x^3-2x-x^2-x+6x+6=0\)

\(\Leftrightarrow\left(2x^3-2x\right)-\left(x^2+x\right)+\left(6x+6\right)=0\)

\(\Leftrightarrow2x\left(x^2-1\right)-x\left(x+1\right)+6\left(x+1\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)\left(x+1\right)-x\left(x+1\right)+6\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-2x-x+6\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-3x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x^2-3x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x^2-3x=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x^2-3x=-6\left(loai\right)\end{matrix}\right.\)

Vậy x=-1