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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 \(B=100\)
\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{y-z-25-9}{16-25} \)
\(<=>\frac{x+16}{9}=\frac{2x^3-34}{-9} \)
<=>\(-x-16=2x^3-34\)
<=>\(2x^3+x-18=0\)
=> x=2
=>\(\frac{2+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=2\)
=>y=57
=>z=41
b: 2x^3-1=15
=>2x^3=16
=>x=2
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)
=>y-25=32; z+9=50
=>y=57; z=41
d: 3/5x=2/3y
=>9x=10y
=>x/10=y/9=k
=>x=10k; y=9k
x^2-y^2=38
=>100k^2-81k^2=38
=>19k^2=38
=>k^2=2
TH1: k=căn 2
=>\(x=10\sqrt{2};y=9\sqrt{2}\)
TH2: k=-căn 2
=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)
2. Tham khảo thêm tại đây nha bạn
https://hoc24.vn/hoi-dap/question/417550.html
\(4x^3-3=29\\ \Rightarrow4x^3=32\\ \Rightarrow x^3=8\\ \Rightarrow x=2\)
\(\dfrac{x+16}{9}=\dfrac{2+16}{9}=2\\\Rightarrow\dfrac{y-15}{-16}=\dfrac{z+49}{25}=2\\ \Rightarrow\left\{{}\begin{matrix}y-15=2.\left(-16\right)=-32\\z+49=2.25=50\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-17\\z=1\end{matrix}\right.\)
\(4x^3-3=29\Rightarrow x^3=\dfrac{29+3}{4}=8\Rightarrow x=\sqrt[3]{8}=2\)
Thay số: \(\dfrac{x+16}{9}=\dfrac{2+16}{9}=2\)
Suy ra: \(y=\left(-16\right)\cdot2+25\Leftrightarrow y=-7\) và \(z=25\cdot2-49\Leftrightarrow z=1\)
\(A=x+2y+3z\Leftrightarrow2+\left(-14\right)+3=-9\)
\(4x^3-3=29\Rightarrow x^3=\dfrac{32}{4}=2^3\Rightarrow x=3\)
\(\dfrac{19}{9}=\dfrac{2y-2.25}{-32}=\dfrac{3z+49.3}{75}=\dfrac{2y+3z+49.3-25.2}{75-32}=\dfrac{2y+3z+97}{43}\)
\(\dfrac{\left(2y+3z+3\right)+94}{43}=\dfrac{19}{9}\) \(\Rightarrow\left(x+2y+3z\right)=\dfrac{43.19}{9}-94\)
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