Đk: x khác 1

a) Ta có: A = \(\frac{x^3-1}{x-1}=\frac{ \left(x-1\right)\left(x^2+x+1\right)}{x-1}=x^2+x+1\)

b) Ta có: \(A=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

Dấu "=" xảy ra <=> \(x+\frac{1}{2}=0\) <=> \(x=-\frac{1}{2}\)

Vậy minA = 3/4 <=> x = -1/2

a,
A=\(\frac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}\)
=\(x^2+x+1\)
b,
Ta có: \(x^2+x+1=x^2+2.\frac{1}{2}.x+\frac{1}{4}-\frac{1}{4}+1 =\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow min_A=\frac{3}{4}\)
Dấu ''='' xảy ra khi :\(x+\frac{1}{2}=0\)
                            \(\Leftrightarrow x=\frac{-1}{2}\)
Vậy \(min_A=\frac{3}{4}\)khi \(x=\frac{-1}{2}\)

Bài làm

a) 3ab² - 12a²b + 9ab = 3ab( b - 4a + 3 )

b) x²( 2x - 3 ) - 5x( 2x - 3 ) = x( 2x - 3 )( x - 5 )

c) x² - 2020x - y² + 2020y 

= ( x² - y² ) - ( 2020x - 2020y )

= ( x - y )( x + y ) - 2020( x - y )

= ( x - y )( x + y - 2020 )

\(\frac{5x+4}{x^2-4x}+\frac{x-2}{x}-\frac{x+2}{x-4}=\frac{5x+4}{x\left(x-4\right)}+\frac{x-2}{x}-\frac{x+2}{x-4}\)

\(=\frac{5x+4}{x\left(x-4\right)}+\frac{\left(x-2\right)\left(x-4\right)}{x\left(x-4\right)}-\frac{x\left(x+2\right)}{x\left(x-4\right)}\)

Khử mẫu : \(5x+4+x^2-4x-2x+8-x^2-2x\)

\(=-x+12\)

Bài làm

Ta có : \(\frac{5x+4}{x^2-4x}+\frac{x-2}{x}-\frac{x+2}{x-4}\)

ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne4\end{cases}}\)

\(=\frac{5x+4}{x\left(x-4\right)}+\frac{x-2}{x}-\frac{x+2}{x-4}\)

\(=\frac{5x+4}{x\left(x-4\right)}+\frac{\left(x-2\right)\left(x-4\right)}{x\left(x-4\right)}-\frac{x\left(x+2\right)}{x\left(x-4\right)}\)

\(=\frac{5x+4}{x\left(x-4\right)}+\frac{x^2-6x+8}{x\left(x-4\right)}-\frac{x^2+2x}{x\left(x-4\right)}\)

\(=\frac{5x+4+x^2-6x+8-x^2-2x}{x\left(x-4\right)}\)

\(=\frac{-3x+12}{x\left(x-4\right)}\)

\(=\frac{-3\left(x-4\right)}{x\left(x-4\right)}=-\frac{3}{x}\)

1)x4+1999x2−1998x+1999.1)x4+1999x2-1998x+1999.

=x4+1999x2+x−1999x+1999=x4+1999x2+x-1999x+1999

=(x4+x)+(1999x2−1999x+1999)=(x4+x)+(1999x2-1999x+1999)

 =x(x3+1)+1999(x2−x+1)=x(x3+1)+1999(x2-x+1)

=x(x+1)(x2−x+1)+1999(x2−x+1)=x(x+1)(x2-x+1)+1999(x2-x+1)

=(x2−x+1)[x(x+1)+1999]=(x2-x+1)[x(x+1)+1999]

=(x2−x+1)(x2+x+1999).=(x2-x+1)(x2+x+1999).

# Chúc bạn học tốt!

Cảm ơn nhìu nhá...:)))

\(\left[\left(\frac{x}{x+3}\right)+\left(\frac{2}{3-x}\right)+\left(\frac{x^2-1}{9-x^2}\right)\right]:\frac{x+1}{x+3}\)

\(=\left[\left(\frac{x}{x+3}\right)+\left(\frac{2}{3-x}\right)+\left(\frac{\left(x-1\right)\left(x+1\right)}{\left(3-x\right)\left(x+3\right)}\right)\right]:\frac{x+1}{x+3}\)

\(=\left[\frac{x\left(3-x\right)}{\left(x+3\right)\left(3-x\right)}+\frac{2\left(x+3\right)}{\left(3-x\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x+1\right)}{\left(3-x\right)\left(x+3\right)}\right]:\frac{x+1}{x+3}\)

\(=\left[\frac{3x-x^2+2x+6+x^2-1}{\left(3-x\right)\left(x+3\right)}\right].\frac{x+3}{x+1}\)

\(=\frac{5x-1}{\left(3-x\right)\left(x+3\right)}.\frac{x+3}{x+1}=\frac{5x-1}{\left(3-x\right)\left(x+1\right)}\)

\(A=\frac{2x^2+2}{x^2+2x+1}=\frac{x^2+1+\left(x^2+1\right)}{x^2+2x+1}\ge\frac{x^2+1+2x}{x^2+2x+1}=1\)

Dấu \(=\)khi \(x=1\)