2x^3-1=15

=>2x^3=16

=>x=2

(x+16)/9=(y-5)/16=(z+9)/25

=>(y-5)/16=(z+9)/25=2

=>y-5=32 và z+9=50

=>y=37 và z=41

B=x+y+z=2+37+41=80

Ta có: \(2x^3-1=15\Leftrightarrow x^3=8\Rightarrow x=2\)

\(\Rightarrow\dfrac{18}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\Rightarrow\left\{{}\begin{matrix}\dfrac{y-25}{16}=2\Rightarrow y=57\\\dfrac{z+9}{25}=2\Rightarrow z=41\end{matrix}\right.\)

Vậy \(B=x+y+z=2+57+41=100\)

`2x^3-1=15=>2x^3=16=>x^3=8=>x=2`

Có:`[x+16]/9=[y-25]/16`

`=>[2+16]/9=[y-25]/16=>y=57`

Có:`[x+16]/9=[z+9]/25`

`=>[2+16]/9=[z+9]/25=>z=41`

Ta có:`B=x+y+z=2+57+41=100`

theo bài ra ta có:

\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{x+16+y-25+z+9}{9+16+25}=\dfrac{x+y+z}{50}\\ \Rightarrow\dfrac{x+16}{9}=\dfrac{x+y+z}{50}\left(1\right)\)ta lại có:

\(\dfrac{9-x}{7}+\dfrac{11-x}{9}=2\\ \Rightarrow\dfrac{7+2-x}{7}+\dfrac{9+2-x}{9}=2\\ \Rightarrow\left(1+\dfrac{2-x}{7}\right)+\left(1+\dfrac{2-x}{9}\right)=2\\ \Rightarrow\left(1+1\right)+\left(\dfrac{2-x}{7}+\dfrac{2-x}{9}\right)=2\\ \Rightarrow2+\left(2-x\right)\left(\dfrac{1}{7}+\dfrac{1}{9}\right)=2\\ \Rightarrow\left(2-x\right)\left(\dfrac{1}{7}+\dfrac{1}{9}\right)=0\\ \Rightarrow2-x=0\\ \Rightarrow x=2\)

thay x = 2 vào 1 ta có:

\(\Rightarrow\dfrac{2+16}{9}=\dfrac{x+y+z}{50}\\ \Rightarrow\dfrac{18}{9}=\dfrac{x+y+z}{50}\\ \Rightarrow2=\dfrac{x+y+z}{50}\\ \Rightarrow x+y+z=2.50\\ \Rightarrow x+y+z=100\)

vậy x + y + z = 100

Ta có :

\(2x^3-1=15\)

\(\Leftrightarrow2x^3=16\)

\(\Leftrightarrow x^3=8\)

\(\Leftrightarrow x=2\)

Thay \(x=2\) zô : \(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

\(\Leftrightarrow\dfrac{2+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

\(\Leftrightarrow\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)

+) \(\dfrac{y-25}{16}=2\)

\(\Leftrightarrow y-25=32\)

\(\Leftrightarrow y=57\)

+) \(\dfrac{z+9}{25}=2\)

\(\Leftrightarrow z+9=50\)

\(\Leftrightarrow z=41\)

Ta có :

\(\left\{{}\begin{matrix}x=2\\y=57\\z=41\end{matrix}\right.\) \(\Leftrightarrow x+y+z=2+57+41=100\)

\(2x^3-1=15\)

\(\Rightarrow2x^3=16\)

\(\Rightarrow x^3=8\)

\(\Rightarrow x=2\)

Thay x vào \(\dfrac{x+16}{9}=\dfrac{y-25}{16}+\dfrac{z+9}{25}\) thì tìm được y và z

Tính nốt x + y + z

\(2x^3-1=15\)

\(2x^3=16\)

\(x^3=8\)

\(\Rightarrow x=2\)

\(\dfrac{x+16}{9}=\dfrac{y+25}{16}=\dfrac{z+9}{25}\)

\(\Leftrightarrow\dfrac{2+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

\(\Leftrightarrow\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)

\(\Rightarrow\dfrac{y-25}{16}=2\)

\(\Rightarrow y-25=32\)

\(\Rightarrow y=57\)

\(\Rightarrow\dfrac{z+9}{25}=2\)

\(\Rightarrow z+9=50\)

\(\Rightarrow z=41\)

\(\Rightarrow\)\(x=2\) , \(y=57\) , \(z=41.\)

\(B=x+y+z\)

\(B=2+57+41\)

\(B=100\)

Vậy \(B=100\)

mình k ghi lại đề nữa ta có

\(1\ge\dfrac{4^2}{x+24}+\dfrac{5^2}{y+16}+\dfrac{3^2}{z+4}\ge\dfrac{\left(4+5+3\right)^2}{x+y+z+24+16+4}=\dfrac{12^2}{x+y+z+44}\)

=>x+y+z+44>=12^2=144=> x+y+z=100

đặt x+y+z=a(a>=100)

\(x+y+z+\dfrac{1}{x+y+z}=a+\dfrac{1}{a}=\dfrac{a}{10000}+\dfrac{1}{a}+\dfrac{9999a}{10000}\ge\dfrac{2}{100}+\dfrac{9999a}{10000}\)

do a>=100 nên

\(a+\dfrac{1}{a}\ge\dfrac{2}{100}+\dfrac{9999}{100}=\dfrac{10001}{100}\) khi a= 100 hay x+y+z=100