x + 50 = 60

x = 60 -50 

x = 10

\(P=\sqrt{x-5}+\sqrt{13-x}\)

\(P^2=\left(\sqrt{x-5}+\sqrt{13-x}\right)^2\)

= x-5 +13 - x + 2\(\sqrt{\left(x-5\right)\left(13-x\right)}\)

=8+2\(\sqrt{\left(x-5\right)\left(13-x\right)}\)

theo BDT cosi ta co 

2\(\sqrt{\left(x-5\right)\left(13-x\right)}\) \(\le x-5+13-x\)=8

8+2\(\sqrt{\left(x-5\right)\left(13-x\right)}\le8+8=16\)

\(P^2\le16\Leftrightarrow P\le4\Rightarrow maxP=4\)

dau = xay ra <=> x-5=13 -x <=> x=9

1/6

đáp số:1/6

\(\text{Ta có​: }\)​:     \(\frac{2006}{2008}\text{ x }\frac{2001}{2004}\text{ x }\frac{2008}{2002}\text{ x }\frac{2004}{2006}\text{ x }\frac{1001}{2001}\)

\(=\frac{2006\text{ x }\text{ }2001\text{ x }2008\text{ x }2004\text{ x }1001}{2008\text{ x }2004\text{ x }2002\text{ x }2006\text{ x }2001}\text{ }\)

Rút gọn các số ( 2006 ; 2001 ; 2008 ; 2004) ở cả tử và mẫu ta có:

\(=\frac{1001}{2002}=\frac{1}{2}\)

Theo đầu bài ta có:
\(\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{6}\)
\(\Rightarrow\frac{2\cdot\left(x+1\right)}{2\cdot2}=\frac{3\cdot\left(y+3\right)}{3\cdot4}=\frac{4\cdot\left(z+5\right)}{4\cdot6}\)
\(\Rightarrow\frac{2x+2}{4}=\frac{3y+9}{12}=\frac{4z+20}{24}\)
\(=\frac{\left(2x+2\right)+\left(3y+9\right)+\left(4z+20\right)}{4+12+24}\)
\(=\frac{\left(2x+3y+4z\right)+\left(2+9+20\right)}{4+12+24}\)
\(=\frac{9+31}{40}=1\)
\(\Rightarrow\hept{\begin{cases}x=1\cdot2-1=1\\y=1\cdot4-3=1\\z=1\cdot6-5=1\end{cases}}\)

Là bàn chân

câu này quá dễ

theo mình nghĩ thì là vô số điểm