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\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{y-z-25-9}{16-25} \)
\(<=>\frac{x+16}{9}=\frac{2x^3-34}{-9} \)
<=>\(-x-16=2x^3-34\)
<=>\(2x^3+x-18=0\)
=> x=2
=>\(\frac{2+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=2\)
=>y=57
=>z=41
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 \(B=x+y+z=2+57+41=100\)
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 \(B=100\)
Quá đơn giản :
2x3-1 = 15
=> 2x3=16
=> x3 = 8
=> x =2
Thay x vào \(\frac{x+16}{9}\)
=> \(\frac{2+16}{9}=2\)
=> \(2=\frac{y-25}{16}\)
=> y-25 = 32
=> y = 57
=> \(2=\frac{z+9}{25}\)
=> z + 9 = 50
=> ...
Đ/S: ...
Cho \(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}và\dfrac{9-x}{7}+\dfrac{11-x}{9}=2\).Tìm x+y+z
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
\(\dfrac{x+16}{9}=\dfrac{y-25}{-16}=\dfrac{z+49}{25}\) (1)
Ta có: \(4x^3-3=29\)
\(\Rightarrow4x^3=32\Rightarrow x^3=8\)
\(\Rightarrow x=2\)
Thay \(x=2\) vào điều (1) ta có:
\(\dfrac{2+16}{9}=\dfrac{y-25}{-16}=\dfrac{z+49}{25}\)
\(\Rightarrow\dfrac{y-25}{-16}=\dfrac{z+49}{25}=\dfrac{18}{9}\)
\(\Rightarrow\dfrac{y-25}{-16}=\dfrac{z+49}{25}=2\)
\(\Rightarrow\left\{{}\begin{matrix}y-25=2.\left(-16\right)\\z+49=2.25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y-25=-32\\z+49=50\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=-7\\z=1\end{matrix}\right.\)
Vậy giá trị của biểu thức \(A=x+2y+3z\) là:
\(A=2+2.\left(-7\right)+3.1=2-14+3=-9\)
Chúc bạn học tốt!!!
Ta có : \(4x^3-3=29\)
\(\Rightarrow4x^3=32\)
\(\Rightarrow x^3=8\)
\(\Rightarrow x=2\)
Thay x = 2 vào \(\dfrac{x+16}{9}=\dfrac{y-25}{-16}\) ta có :
\(\dfrac{2+16}{9}=\dfrac{y-25}{-16}\)
\(\Rightarrow2=\dfrac{y-25}{-16}\)
\(\Rightarrow y-25=-32\)
\(\Rightarrow y=-7\)
Thay \(y=-7\) vào \(\dfrac{y-25}{-16}=\dfrac{z+49}{25}\) ta có :
\(\dfrac{-7-25}{-16}=\dfrac{z+49}{25}\)
\(\Rightarrow2=\dfrac{z+49}{25}\)
\(\Rightarrow z+49=50\)
\(\Rightarrow z=1\)
Thay x = 2; y = -7; z = 1 vào biểu thức A ta có :
\(A=2+2.\left(-7\right)+3.1\)
\(A=-9\)
Vậy A = -9
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