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a) Ta có: \(6x=4y=3z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{3z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-2}{-4}=\dfrac{1}{2}.\)
Với: \(\dfrac{x}{2}=\dfrac{1}{2}\Rightarrow x=1.\)
\(\dfrac{2y}{6}=\dfrac{y}{3}=\dfrac{1}{2}\Rightarrow y=\dfrac{1}{2}.3=\dfrac{3}{2}.\)
\(\dfrac{3z}{12}=\dfrac{z}{4}=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}.4=\dfrac{4}{2}=2.\)
Vậy: \(x=1;y=\dfrac{3}{2};z=2.\)
a: \(\Leftrightarrow x^3=-216\)
=>x=-6
b: \(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}=\dfrac{3x+5y+7z}{3\cdot2+5\cdot\dfrac{5}{2}+7\cdot\dfrac{7}{4}}=\dfrac{123}{\dfrac{123}{4}}=4\)
=>x=8; y=10; z=7
\(\dfrac{x-2}{4}=\dfrac{y+1}{5}=\dfrac{z+3}{7}\)
\(\Rightarrow\dfrac{2\left(x-2\right)}{8}=\dfrac{y+1}{5}=\dfrac{2\left(z+3\right)}{14}\)
\(\Rightarrow\dfrac{2x-4}{8}=\dfrac{y+1}{5}=\dfrac{2z+6}{14}\)
Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(=\dfrac{2x-4+y+1-2z-6}{8+5-14}\)
\(=\dfrac{2x+y-2z-9}{-1}\)
\(=\dfrac{7-9}{-1}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-2}{4}=2\Rightarrow x-2=8\Rightarrow x=10\\\dfrac{y+1}{5}=2\Rightarrow y+1=10\Rightarrow y=9\\\dfrac{z+3}{7}=2\Rightarrow z+3=14\Rightarrow z=11\end{matrix}\right.\)
Câu 2:
\(\dfrac{x+2000}{x-2000}=\dfrac{y+2001}{y-2001}\)
\(\Leftrightarrow\left(x+2000\right)\left(y-2001\right)=\left(x-2000\right)\left(y+2001\right)\)
\(\Leftrightarrow xy-2001x+2000y-4002000=xy+2001x-2000y-4002000\)
=>-2001x+2000y=2001x-2000y
=>-4002x=-4000y
=>2001x=2000y
hay x/y=2000/2001
a)vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\)=>\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)và 2x+3y+5z=86
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)=\(\dfrac{2x+3y+5z}{6+12+25}\)\(\dfrac{86}{43}\)=2
vì\(\dfrac{2x}{6}\)=2=>2x=2.6=12=>x=12:2=6
\(\dfrac{3y}{12}\)=2=>3y=12.2=24=>y=24:3=8
\(\dfrac{5z}{25}\)=2=>5z=25.2=50=>z=50:5=10
vậy x=6,y=8,z=10
vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)(1)
\(\dfrac{y}{6}\)=\(\dfrac{z}{8}\)=>\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)(2)
từ (1)(2)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)=>\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)và 3x-2y-z=13
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)=\(\dfrac{3x-2y-z}{27-24-16}\)=\(\dfrac{13}{-13}\)=-1
vì\(\dfrac{3x}{27}\)=-1=>3x=-1.27=-27=>x=-27x;3=-9
\(\dfrac{2y}{24}\)=-1=>2y=-1.24=-24=>y=-24:2=-12
\(\dfrac{z}{16}\)=-1=>z=-1.16=-16
vậy...
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\Rightarrow\dfrac{4.\left(3x-2y\right)}{4.4}=\dfrac{3.\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)
=\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{0}{29}\)
\(\Rightarrow\) 12x= 8y
6z=12x
8y=6z
=> 12x=8y=6z
MSC: 24
ta có: \(\dfrac{12x}{24}=\dfrac{8y}{24}=\dfrac{6z}{24}\)= \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)( đpcm)
Đặt \(\dfrac{x}{2}=\dfrac{2y}{3}=\dfrac{3z}{4}=k\Rightarrow x=2k;y=\dfrac{3k}{2};z=\dfrac{4k}{3}\)
Khi đó \(xyz=-108\Leftrightarrow2k\cdot\dfrac{3k}{2}\cdot\dfrac{4k}{3}=-108\)
\(\dfrac{24k^3}{6}=-108\Leftrightarrow4k^3=-108\)
\(\Leftrightarrow k^3=-27\Rightarrow k=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=2\cdot\left(-3\right)=-6\\y=\dfrac{3k}{2}=\dfrac{3\cdot\left(-3\right)}{2}=-\dfrac{9}{2}\\z=\dfrac{4k}{3}=\dfrac{4\cdot\left(-3\right)}{3}=-4\end{matrix}\right.\)