x(x-1)=1-x²

x²-x=1-x²

2x²-x-1=0

\(\hept{\begin{cases}x=1\\x=\frac{-1}{2}\end{cases}}\)

x(x-1)= (1-x)(1+x)

x(x-1)+(x-1)(x+1)=0

(x-1) (x + x+1)= 0

(x-1)(2x+1)=0

x=1 hoặc x = -1/2

\(x^2-5\)

\(=x^2-\left(\sqrt{5}\right)^2\)

\(=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

Với đề bài và đã có x ta chỉ cần thay x vào là được : 

\(101^3-3.101^2+3.101-1=\)

\(97^3+9.97^2+27.97+27=\)

Dùng hằng đẳng thức đi bạn :)

a)\(x^3-3x^2+3x-1=\left(x-1\right)^3=\left(101-1\right)^3=100^3=1000000\)

b)\(x^3+9x^2+27x+27=\left(x+3\right)^3=\left(97+3\right)^3=100^3=1000000\)

2/

a, \(A=2x^2+6x-5=2\left(x^2+3x-\frac{5}{2}\right)=2\left(x^2+2x\cdot\frac{3}{2}+\frac{9}{4}-\frac{19}{4}\right)=2\left[\left(x+\frac{3}{2}\right)^2-\frac{19}{4}\right]=2\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0\Rightarrow A=\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\ge-\frac{19}{2}\)

Dấu "=" xảy ra khi x=-3/2

Vậy Amin=-19/2 khi x=-3/2

b,bài này phải tìm min 

 \(B=\left(2x-x\right)\left(x+4\right)=x\left(x+4\right)=x^2+4x=x^2+4x+4-4=\left(x+2\right)^2-4\)

Vì \(\left(x-2\right)^2\ge0\Rightarrow B=\left(x-2\right)^2+4\ge4\)

Dấu "=" xảy ra khi x = 2

Vậy Bmin=4 khi x=2

Bài 2)Ta có:

\(2x^2+6x-5\)

\(=2x^2+6x+\frac{9}{2}-\frac{19}{2}\)

\(=2\left(x^2+3x+\frac{9}{4}\right)-\frac{19}{2}\)

\(=2\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\ge-\frac{19}{2}\)

A= 2006 X 2008 - 2007²

A = 2006 . 2008 - 2007 . 2007

A = 2006 . ( 2007 + 1 ) - 2007 . ( 2006 + 1 )

A = 2006 . 2007 + 2006 - 2007 . 2006 + 2007

A = -1

B= 2016 X 2018 - 2017²

B= 2016 . 2018 - 2017 . 2017

B = 2016 . ( 2017 + 1 ) - 2017 . ( 2016 + 1 )

B = 2016 . 2017 + 2016 - 2017 . 2016 + 2017

B = -1

cảm ơn bạn nhé....

\(2x^2+6x-8=0\)

<=> \(2x^2-2x+8x-8=0\)

<=> \(2x\left(x-1\right)+8\left(x-1\right)=0\)

<=> \(\left(2x+8\right)\left(x-1\right)=0\)

<=> \(\hept{\begin{cases}2x+8=0\\x-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=-4\\x=1\end{cases}}\)

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

<=> \(2x^2-2x+x-1=0\)

<=> \(2x\left(x-1\right)+\left(x-1\right)=0\)

<=> \(\left(2x+1\right)\left(x-1\right)=0\)

<=> \(\hept{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

\(4x^2-5x-9=0\)

<=> \(4x^2+4x-9x-9=0\)

<=> \(4x\left(x+1\right)-9\left(x+1\right)=0\)

<=> \(\left(4x-9\right)\left(x+1\right)=0\)

<=> \(\hept{\begin{cases}4x-9=0\\x+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{3}{2}\\x=-1\end{cases}}\)

học tốt

\(2x^2+6x-8=0\)

\(< =>2x^2-2x+8x-8=0\)

\(\Leftrightarrow2x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x+8\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x+8\right)\left(x-1\right)=0\)

\(\Leftrightarrow2x+8=0\)hoặc \(x-1=0\)

\(\Leftrightarrow x=-4\)hoặc \(x=1\)