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Đặt: \(k=\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
\(\Rightarrow k^3=\frac{xyz}{3.4.5}=\frac{1620}{60}=27\)
=> k = 3
Nên \(\frac{x}{3}=3\Rightarrow x=9\)
\(\frac{y}{4}=3\Rightarrow y=12\)
\(\frac{z}{5}=3\Rightarrow z=15\)
Vậy x = 9 , y = 12 , z = 15
a)
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\Leftrightarrow x=3k;y=4k;z=5k\)và \(xyz=1620\)
\(\Rightarrow3k.4k.5k=1620\Leftrightarrow60k^3=1620\)
\(\Rightarrow k=\sqrt[3]{1620:60}=3\)
\(\hept{\begin{cases}\frac{x}{3}=3\Rightarrow x=3.3=9\\\frac{y}{4}=3\Rightarrow y=3.4=12\\\frac{z}{5}=3\Rightarrow z=3.5=15\end{cases}}\)
Vậy \(x=9;y=12;z=15\)
b)
Ta có:
\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{6}\Leftrightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{18}\) và \(x+y+z=334\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{18}=\frac{x+y+z}{10+15+18}=\frac{334}{43}\)
\(\hept{\begin{cases}\frac{x}{10}=\frac{334}{43}\Rightarrow x=\frac{334}{43}.10=\frac{3340}{43}\\\frac{y}{15}=\frac{334}{43}\Rightarrow y=\frac{334}{43}.15=\frac{5010}{43}\\\frac{z}{18}=\frac{334}{43}\Rightarrow z=\frac{334}{43}.18=\frac{6012}{43}\end{cases}}\)
Vậy \(x=\frac{3340}{43};y=\frac{5010}{43};z=\frac{6012}{43}\)
\(\frac{x}{3}=\frac{y}{4};\) \(\frac{y}{4}=\frac{z}{5}\)
suy ra: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)
\(\Rightarrow\)\(x=3k;\)\(y=4k;\)\(z=5k\)
Ta có: \(x.y.z=1620\)
\(\Rightarrow\)\(3k.4k.5k=1620\)
\(\Leftrightarrow\)\(60k^3=1620\)
\(\Leftrightarrow\)\(k^3=27\)
\(\Leftrightarrow\)\(k=3\)
suy ra: \(x=9;\)\(y=12;\)\(z=15\)
\(\left(5x-1\right)^2+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
\(=\left(5x-1\right)^2-2\left(5x-1\right)\left(5x+4\right)+\left(5x+4\right)^2\)
\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2\)
\(=\left(5x-1-5x-4\right)^2\)
\(=\left(-5\right)^2\)
\(=25\)
(5x-1)^2+2(5x-1)(5x+1)+5(2x-1)
=25x^2-10x+1+2(25x^2-1)+10x-5
=25x^2-4+50x^2-2
=75x^2-6
\(\left(5x-1\right)^2+2\left(5x-1\right)\left(5x+1\right)+5\left(2x-1\right)\)
\(=25x^2-10x+1+2\left(25x^2-1\right)+10x-5\)
\(=25x^2-10x+1+50x^2-2+10x-5\)
\(=\left(25x^2+50x^2\right)+\left(10x-10x\right)+\left(1-2-5\right)\)
\(=75x^2-6\)
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
\(\Rightarrow5^x\left(1+5+5^2\right)=3875\\ \Rightarrow5^x=\dfrac{3875}{31}=125=5^3\\ \Rightarrow x=3\)
\(6x-5x^2=2-5x^2\Rightarrow6x=2\Rightarrow x=\frac{1}{3}\)
6x-5x2=2-5x2
<=>6x=2-5x2+5x2
<=>6x=2
<=>x=1/3
k mik nhé
Thanks
\(5x^2=1620\)
\(\Rightarrow x^2=1620:5\)
\(\Rightarrow x^2=324\)
\(\Rightarrow x^2=18^2\)
\(\Rightarrow\left[{}\begin{matrix}x=18\\x=-18\end{matrix}\right.\)
Vậy: x=18 hoặc x=-18