d) (x - 2)^2 = 1 

= x = 2 + 1 = 3  

c) (x^2 + 1). (x + 2011) = 0

Tim x:

a) x^2 + 2x = 0 

= \(x^2+2x=0\)

= \(x^2=0:2=0\)

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

Rút gọn thừa số chung : 

\(2x^2-5x-3=0\)

x = \(\frac{-1}{2}\)x = 3

=\(x^2=0\)

=> x = 0 

Bạn Nguyễn Phương Trung làm đúng rồi, các bn tk cho bn ấy nha !

\(f\left(x\right)=\left(x-1\right).g\left(x\right)\)

\(\Rightarrow3x^3-2x^2+x+5=\left(x-1\right)\left(3x^2+ax+b\right)\)

\(\Rightarrow3x^3-2x^2+x+5=3x^3+ax^2+bx-3x^2-ax-b\)

\(\Rightarrow-2x^2+x+5=x^2\left(a-3\right)+x\left(b-a\right)-b\)

-Bạn kiểm tra lại đề.

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Ta có: \(B=\frac{x}{\sqrt{x}}-\frac{2\sqrt{x}}{\sqrt{x}}+\frac{4}{\sqrt{x}}=\sqrt{x}-2+\frac{4}{\sqrt{x}}=\left(\sqrt[4]{x}\right)^2-2.\sqrt[4]{x}.\frac{2}{\sqrt[4]{x}}+\left(\frac{2}{\sqrt[4]{x}}\right)^2+2\)

\(=\left(\sqrt[4]{x}-\frac{2}{\sqrt[4]{x}}\right)^2+2\ge2\)

Vậy Min B = 2  khi x = 4.

Chúc em học tốt :)

Bài 1:

\(x^3-x^2-x+1=0\)

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

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

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

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

Vậy x = 1 hoặc x = -1

Bài 2:
\(2x-2x^2-1=-2\left(x^2-x+\dfrac{1}{2}\right)\)

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

\(=-2\left(x^2-\dfrac{1}{2}\right)^2-\dfrac{1}{2}< 0\)

\(\Rightarrowđpcm\)

a: =>|2x-4|=-2x+4

=>2x-4<=0

=>x<=2

b: =>|x+5|=x+5

=>x+5>=0

hay x>=-5

c: =>|x-2|=x-2

=>x-2>=0

hay x>=2

Ta có \(A=\frac{x^2-2x+2011}{x^2}\)

\(=\frac{x^2}{x^2}-\frac{2x}{x^2}+\frac{2011}{x^2}\)

\(=1-\frac{2}{x}+\frac{2011}{x^2}\)

Đặt \(\frac{1}{x}=y\)ta có:

\(A=1-2y+2011y^2\)

\(A=2011y^2-2y+1\)

\(A=2011\left(y^2-\frac{2}{2011}y+\frac{2}{2011}\right)\)

\(=2011\left(y^2-2\times y\times\frac{1}{2011}+\frac{1}{2011^2}-\frac{1}{2011^2}+\frac{1}{2011}\right)\)

\(=2011\left(\left(y-\frac{1}{2011}\right)^2\right)+\frac{2010}{2011^2}\)

\(=2011\left(y-\frac{1}{2011}\right)^2+\frac{2010}{2011}\)

Vì (y-\(\frac{1}{2011}\))\(^2\)>=0

\(\Rightarrow2011\left(y-\frac{1}{2011}\right)^2+\frac{2010}{2011}\)

Hay \(A>=\frac{2010}{2011}\)

quyet