a) Ta có: \(M=-x^2-4x+20\)

\(=-\left(x^2+4x-20\right)\)

\(=-\left(x^2+4x+4-24\right)\)

\(=-\left(x+2\right)^2+24\le24\forall x\)

Dấu '=' xảy ra khi x=-2

`(s:48)-(s:50)=20`

`<=>s/48-s/50=20`

`<=>(25s)/1200-(24s)/1200=20`

`<=>s/1200=20`

`<=>s=24000`

Vậy `s=24000`

\(\dfrac{S}{48}-\dfrac{S}{50}=20\)

\(\Leftrightarrow\dfrac{50S-48S}{48\cdot50}=20\)

\(\Leftrightarrow2S=48\cdot50\cdot20=48000\)

\(\Leftrightarrow S=24000\)

\(\dfrac{s}{48}-\dfrac{s}{50}=20\)

\(\Leftrightarrow\dfrac{25s}{1200}-\dfrac{24s}{1200}=\dfrac{24000}{1200}\)

\(\Rightarrow s=24000\)

\(\Rightarrow x^2-5x-36=0\Rightarrow x\left(x-9\right)+4\left(x-9\right)=0\Rightarrow\left(x-9\right)\left(x+4\right)=0\Rightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

\(x+\left(x\cdot2\right)+\left(x\cdot3\right)=20\)

\(\Leftrightarrow x\cdot\left(1+2+3\right)=20\)

\(\Leftrightarrow x\cdot6=20\Leftrightarrow x=\dfrac{10}{3}\)

\(\left[\left(2x-11\right):3+1\right].5=20\\\left(2x-11\right):3+1=20:5\\ \left(2x-11\right):3+1=4\\ \left(2x-11\right):3=4-1\\ \left(2x-11\right):3=3\\ 2x-11=3.3\\ 2x-11=9\\ 2x=11+9\\ 2x=20\\ x=20:2\\ x=10 \)

<=> (2X-11):3+1=20:5
<=> (2X-11):3=4-1
<=> 2X-11=3x3
<=> 2X-11=9
<=> 2X=9+11
<=>X=20:2
<=> X=10

a) y:0,25 +y x 11 = 24

    y x 4 + y x 11 = 24

   y x (4 +11) = 24

   y x 15 = 24

   y = \(\frac{8}{5}\)

a,y:0,25+yx11=24

yx4+yx11       =24

yx(4+11)        =24

yx15              =24

y                    =24:15

y                    =1,6

b,yx8,01-y:100=38

yx8,01-yx0,01 =38

yx(8,01-0,01)  =38

yx8                  =38

y                     =38:8

y                     =4,75

đặt 

\(A=a+b+c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)

\(=>4A=4a+4b+4c+\dfrac{12}{a}+\dfrac{36}{2b}+\dfrac{16}{c}\)

\(=>4A=a+2b+3c+3a+\dfrac{12}{a}+2b+\dfrac{36}{2b}+c+\dfrac{16}{c}\)

áp dụng BDT AM-GM

\(=>\dfrac{12}{a}+3a\ge2\sqrt{12.3}=12\)

\(=>2b+\dfrac{36}{2b}\ge2\sqrt{36}=12\)

\(=>c+\dfrac{16}{c}\ge2\sqrt{16}=8\)

\(=>4A\ge20+12+12+8=52=>A\ge13\)

dấu"=" xảy ra<=>a=2,b=3,c=4