Đây là phương trình bậc 3

\(x^3-8-\left(x^2-4x+4\right)=0\Leftrightarrow x^3-8-x^2+4x-4=0\Leftrightarrow x^3-x^2+4x-12=0\Leftrightarrow x=2\)

Vậy phương trình có 1 nghiệm là x=2

x³ - 8 - (x² - 4x + 4) = 0

<=> x³ - x² + 4x - 8 - 4 = 0

<=> x³ - x² + 4x - 12 = 0

<=> (x - 2)(x² + x + 6) = 0

<=> x - 2 = 0 hoặc x² + x + 6 khác 0

<=> x = 2

a) \(M=\frac{x}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\left(x\ne\pm1\right)\)

\(\Leftrightarrow M=\frac{x}{x+1}+\frac{1}{x-1}+\frac{2x}{\left(x-1\right)\left(x+1\right)}\)

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

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

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

Vậy \(M=\frac{x+1}{x-1}\left(x\ne\pm1\right)\)

b) \(M=\frac{x+1}{x-1}\left(x\ne\pm1\right)\)

x-2=1

<=> x=3 (tmđk)

Thay x=3 vào M ta có: \(M=\frac{3+1}{3-1}=\frac{4}{2}=2\)

Vậy M=2 khi x-2=1

c) \(M=\frac{x+1}{x-1}\left(x\ne\pm1\right)\)

M nguyên khi x+1 chia hết cho x-1

=> x-1+2 chia hết cho x-1

 x nguyên => x-1 nguyên => x-1 thuộc Ư (2)={-2;-1;1;2}
Ta có bảng

Vậy x={0;2;3}

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

\(=4x^2-4x+1+x^2+4x+4\)

\(=5x^2+5\)

Ta thấy \(5x^2\ge0\forall x\)

\(\Rightarrow5x^2+5\ge5\)

\(\Rightarrow B\ge5\)

Dấu "=" xảy ra khi \(x=0\)

...

\(B=4x^2-4x+1+x^2+4x+4\)

\(=5x^2+5\ge5\)

Dấu "=" xảy ra <=> x^2 = 0 <=> x = 0

GTNN của B là 5 khi x = 0

\(^{x^2-2x=-1\Leftrightarrow x^2-2x+1=0\Leftrightarrow x=1}\)

x² - 2x = -1

<=> x² - 2x + 1 = 0

<=> x² - 2( x )( 1 ) + 1 = 0

<=> ( x - 1 )² = 0

<=> x - 1 = 0

<=> x = 1