\((2x-1)(x-2)=2x^2-4x-x+2=2x^2-5x+2\)

Bài làm

   | a | . a

=   a   . a

= a² 

# Chúc bạn học tốt #

đáp án

|a|.a

=a.a

=a²

~~~/a/=GTTĐ a nhé

+) a=0=>/a/.a=0

+) a<0=>/a/.a=-a.a=-a²

+) a>0=>/a/.a=a.a=a²

(a+b).(a+b)=(a+b)²

(a+b)(a+b)=(a+b)^2

\(\left(a+b\right)\left(a+b\right)\)

\(=\left(a+b\right)^2\)

\(=a^2+2ab+b^2\)

I a I - a

= a - a

=0

TH1:|a|=a

=> |a|-a=a-a=0

TH2:|a|=-a

=> |a|-a=(-a)-a=-2a

  | a | - a

= a - a = 0

Lời giải:
$(2x+3)^2+(2x+5)^2=4x^2+12x+9+4x^2+20x+25$

$=8x^2+32x+34$

đk: x>=0; x khác 3

a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)

b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)

ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)