\(b,\left(x+5\right)^2=100\)

TH1:\(\left(x+5\right)^2=10^2\)

        \(\Rightarrow x+5=10\\ \Rightarrow x=10-5\\ \Rightarrow x=5\)

TH2:\(\left(x+5\right)^2=\left(-10\right)^2\)

         \(\Rightarrow x+5=-10\\ \Rightarrow x=-10-5\\ \Rightarrow x=-15\)

\(c,\left(2x-4\right)^2=0\\ \Rightarrow2x-4=0\\ \Rightarrow2x=0+4\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)

\(d,\left(x-1\right)^3=-27\\ \Rightarrow\left(x-1\right)^3=\left(-3\right)^3\\ \Rightarrow x-1=-3\\ \Rightarrow x=-3+1\\ \Rightarrow x=-2\)

b) (x+5)²=100
    (x+5)²=10²
     x+5=10
     x=10-5
     x=5
c) (2x-4)²=0
    2x-4=0
    2x=4
      x=4:2
      x=\(\pm\)2
d)(x-1)³=-27
   (x-1)³=-3³
    x-1=-3
    x=-3+1
    x=-2

Ko vì -5/2 khác -3

a: f(0)=1/2

f(1/2)=3/2

\(P=A.B=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}.\dfrac{\sqrt{x}+6}{\sqrt{x}-1}=\dfrac{\sqrt{x}+6}{\sqrt{x}-3}\)

\(=1+\dfrac{9}{\sqrt{x}-3}\le1+\dfrac{9}{0-3}=1-3=-2\)

\(maxP=-2\Leftrightarrow x=0\)

\(1,x=16\Leftrightarrow A=\dfrac{4-1}{4-3}=\dfrac{3}{1}=3\\ 2,B=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\\ 3,P=AB=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}+6}{\sqrt{x}-1}=\dfrac{\sqrt{x}+6}{\sqrt{x}-3}\\ P=1+\dfrac{9}{\sqrt{x}-3}\\ Vì.\sqrt{x}-3\ge-3\Leftrightarrow\dfrac{9}{\sqrt{x}-3}\le-3\\ \Leftrightarrow P=1+\dfrac{9}{\sqrt{x}-3}\le1-3=-2\\ P_{max}=-2\Leftrightarrow x=0\)

Bài 2: 

Ta có: \(3n^3+10n^2-5⋮3n+1\)

\(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow3n\in\left\{0;-3;3\right\}\)

hay \(n\in\left\{0;-1;1\right\}\)

Phần BA là phần nào á b mình zoom ảnh thấy phần C