a. A = | x + 2 | + | x - 3 | - 7

=> A = | x + 2 | + | 3 - x | - 7

Áp dụng BĐT | a | + | b |\(\ge\)| a + b | , ta có :

A = | x + 2 | + | 3 - x | - 7 \(\ge\) | x + 2 + 3 - x | - 7 = | 5 | - 7 = 5 - 7 = - 2

Dấu "=" xảy ra <=> \(3\ge x\ge-2\)

Vậy minA = - 2 <=> \(3\ge x\ge-2\)

c. C = - x² + 6x - 4y² - 4y + 5

=> C = - ( x² - 6x + 9 ) - ( 4y² + 4y + 1 ) + 15

=> C = - ( x - 3 )² - 4 ( y - 1/2 )² + 15\(\le\)15\(\forall\)x

Dấu "=" xảy ra <=>\(\orbr{\begin{cases}\left(x-3\right)^2=0\\4\left(y-\frac{1}{2}\right)^2=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=3\\y=\frac{1}{2}\end{cases}}\)

Vậy maxC = 15 <=>\(\orbr{\begin{cases}x=3\\y=\frac{1}{2}\end{cases}}\)

Bài làm

\(\frac{3x-3}{\left(x-1\right)^2}+\frac{2x+2}{1-x^2}\)

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

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

\(=\frac{3}{x-1}-\frac{2}{x-1}=\frac{1}{x-1}\)

Bài làm 

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

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

\(=\frac{1-x}{\left(1-x\right)\left(x-1\right)}=\frac{1}{x-1}\)

Bài làm

a) \(\frac{x^4-2x^2+1}{x^3-3x-2}=\frac{\left(x^2-1\right)^2}{x^3-2x^2+2x^2-4x+x-2}\)

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

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

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

b) \(\frac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^4-1}\)

\(=\frac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)

\(=\frac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)

\(=\frac{\left(x+1\right)\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)

\(=\frac{x^6+x^4+x^2+1}{\left(x-1\right)\left(x^2+1\right)}\)

\(=\frac{\left(x^6+x^4\right)+\left(x^2+1\right)}{\left(x-1\right)\left(x^2+1\right)}\)

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

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

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

\(x\left(2-x\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\2-x=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=2\end{cases}}\)

2x-xx=0

x.(2-x)=0

TH1:2x=0

x=0

TH2:2-x=0

x=2