\(P=\frac{a+2}{a+3}-\frac{5}{\left(a+3\right)\left(a-2\right)}-\frac{a}{a^2-2a}\)

a) ĐKXĐ : \(\hept{\begin{cases}a\ne0\\a\ne-3\\a\ne2\end{cases}}\)

b)\(=\frac{a+2}{a+3}-\frac{5}{\left(a+3\right)\left(a-2\right)}-\frac{a}{a\left(a-2\right)}\)

\(=\frac{a\left(a-2\right)\left(a+2\right)}{a\left(a+3\right)\left(a-2\right)}-\frac{5a}{a\left(a+3\right)\left(a-2\right)}-\frac{a\left(a+3\right)}{a\left(a+3\right)\left(a-2\right)}\)

\(=\frac{a\left(a^2-4\right)}{a\left(a+3\right)\left(a-2\right)}-\frac{5a}{a\left(a+3\right)\left(a-2\right)}-\frac{a^2+3a}{a\left(a+3\right)\left(a-2\right)}\)

\(=\frac{a^3-4a-5a-a^2-3a}{a\left(a+3\right)\left(a-2\right)}\)

\(=\frac{a^3-a^2-12a}{a\left(a+3\right)\left(a-2\right)}=\frac{a\left(a^2-a-12\right)}{a\left(a+3\right)\left(a-2\right)}\)

\(=\frac{a^2-4a+3a-12}{\left(a+3\right)\left(a-2\right)}=\frac{a\left(a-4\right)+3\left(a-4\right)}{\left(a+3\right)\left(a-2\right)}\)

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

c) \(8a=8a^2\)

⇔ \(8a^2-8a=0\)

⇔ \(8a\left(a-1\right)=0\)

⇔ \(\orbr{\begin{cases}8a=0\\a-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=0\left(ktm\right)\\a=1\left(tm\right)\end{cases}}\)

Với a = 1 =>\(P=\frac{1-4}{1-2}=\frac{-3}{-1}=3\)

x² + 3x + 8

= ( x² + 3x + 9/4 ) + 23/4

= ( x + 3/2 )² + 23/4 ≥ 23/4 ∀ x

Dấu "=" xảy ra khi x = -3/2

=> GTNN của biểu thức = 23/4 <=> x = -3/2

\(P=3x^2+y^2-2xy-3x+2\)

\(=x^2-2xy+y^2+2x^2-3x+2\)

\(=\left(x-y\right)^2+2\left(x-\frac{3}{4}\right)^2+\frac{7}{8}\)

do\(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(x-\frac{3}{4}\right)^2\ge0\end{cases}\Rightarrow P\ge\frac{7}{8}}\)

\(\Rightarrow P_{min}=\frac{7}{8}\)đạt được khi \(x=y=\frac{3}{4}\)

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

Trả lời :

x² - 4xy + 4y²

= x² - 2x . 2y + (2y)²

= (x - 2y)²

Ta có A = x² + 7x + 1

= \(x^2+2.\frac{7}{2}x+\frac{49}{4}-\frac{49}{4}+1\)

= \(\left(x+\frac{7}{2}\right)^2-\frac{45}{4}\ge-\frac{45}{4}\)

Dấu "=" xảy ra <=> \(x+\frac{7}{2}=0\Rightarrow x=-\frac{7}{2}\)

Vậy Min A = -45/4 <=> x = -7/2

A = x² + 7x + 1

= ( x² + 7x + 49/4 ) - 45/4

= ( x + 7/2 )² - 45/4 ≥ -45/4 ∀ x

Dấu "=" xảy ra <=> x + 7/2 = 0 => x = -7/2

=> MinA = -45/4 <=> x = -7/2