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\(\frac{x^3}{1000}=\frac{y^3}{3375}=\frac{z^3}{1728}=>\left(\frac{x}{10}\right)^3=\left(\frac{y}{15}\right)^3=\left(\frac{z}{12}\right)^3=>\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=-\frac{49}{7}=-7\)
=>x=-70;y=-105;z=-84
=>x+y+z=-259
tick tớ nhé
\(\frac{x^3}{1000}=\frac{y^3}{3375}=\frac{z^3}{1728}\Rightarrow\left(\frac{x}{10}\right)^3=\left(\frac{y}{15}\right)^3=\left(\frac{z}{12}\right)^3\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}\left(\text{t/c dãy tỉ số = nhau}\right)=\frac{-49}{7}=-7\)
\(\Rightarrow\frac{x}{10}=-7\Rightarrow x=-7.10=-70\)
\(\Rightarrow\frac{y}{15}=-7\Rightarrow y=-7.15=-105\)
\(\Rightarrow\frac{z}{12}=-7\Rightarrow z=-7.12=-84\)
Vậy x+y+z=(-70)+(-105)+(-84)=-259.
\(\frac{x^3}{1000}=\frac{y^3}{3375}=\frac{z^3}{1728}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{-49}{7}=-7\)
\(\Rightarrow\) x = - 70; y = - 105; z = - 84
\(\Rightarrow\) x + y + z = - 259
cho các số z,y,z thỏa mãn :\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)và 2x+3y-z=95. khi đó x+y+z=
Lời giải:
$\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}$
$\Rightarrow (\frac{x}{2})^3=(\frac{y}{4})^3=(\frac{z}{6})^3$
$\Rightarrow \frac{x}{2}=\frac{y}{4}=\frac{z}{6}$
$\Rightarrow \frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}$
Áp dụng TCDTSBN:
$\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}$
$\Rightarrow x^2=1\Rightarrow x=\pm 1$
Nếu $x=1$ thì $\frac{y}{4}=\frac{z}{6}=\frac{1}{2}\Rightarrow y=2; z=3$
$\Rightarrow x+y-z=1+2-3=0$
Nếu $x=-1$ thì $\frac{y}{4}=\frac{z}{6}=\frac{-1}{2}\Rightarrow y=-2; z=-3$
$\Rightarrow x+y-z=(-1)+(-2)-(-3)=0$
Vậy $x+y-z=0$
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}.\)
Áp dụng tc dãy tỉ số bằng nhau ta có :
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{2x+3y-z-5}{9}\)
\(=\frac{50-5}{9}=5\)
\(\left(+\right)\frac{x-1}{2}=5=>x=11\)
\(\left(+\right)\frac{y-2}{3}=5=>y=17\)
\(\left(+\right)\frac{z-3}{4}=5\Rightarrow z=23\)
\(=>x+y+z=11+17+23=51\)
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{\left(2x+3y-z\right)+\left(-2-6+3\right)}{9}\)\(=\frac{50-5}{9}=\frac{45}{9}=5\)
Khi đó:\(\frac{2x-2}{4}=5\Rightarrow2x-2=20\Rightarrow x=11;\frac{3y-6}{9}=5\Rightarrow3y-6=45\Rightarrow y=17;\)
\(\frac{z-3}{4}=5\Rightarrow z-3=20\Rightarrow23\)
Ta có:\(\frac{x-1}{2}=\frac{2.\left(x-1\right)}{2.2}=\frac{2x-2}{4}\)
\(\frac{y-2}{3}=\frac{3.\left(y-2\right)}{3.3}=\frac{3y-6}{9}\)
Theo t/c dãy tỉ số = nhau:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{2x+3y-z-5}{9}=\frac{95-5}{9}=\frac{90}{9}=10\)
=> \(\frac{x-1}{2}=10\Rightarrow x-1=10.2=20\Rightarrow x=20+1=21\)
=> \(\frac{y-2}{3}=10\Rightarrow y-2=10.3=30\Rightarrow y=30+2=32\)
=> \(\frac{z-3}{4}=10\Rightarrow z-3=10.4=40\Rightarrow z=40+3=43\)
Vậy x + y + z = 21 + 32 + 43 = 96.
Ta có:
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2\left(x-1\right)}{4}=\frac{3\left(y-2\right)}{9}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2.1}{4}=\frac{3y-3.2}{9}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}=\frac{2x-2+3y-6-z+3}{9}=\frac{\left(2x+3y-z\right)+\left(-2+-6+3\right)}{9}=\frac{50+\left(-5\right)}{9}=\frac{45}{9}=5\)\(\Rightarrow\frac{x-1}{2}=5\Rightarrow x=5.2+1=11\)
\(\Rightarrow\frac{y-2}{3}=5\Rightarrow y=5.3+2=17\)
\(\Rightarrow\frac{z-3}{4}=5\Rightarrow z=5.4+3=23\)
Vậy \(x+y-z=11+17-23=28-23=5\)
Ta có: \(\frac{x-1}{2}=\frac{2x-2}{4};\frac{y-2}{3}=\frac{3y-6}{9}\)
=> \(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\) và \(2x+3y-z=50\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}\)
\(=\frac{2x+3y-z-\left(2+6-3\right)}{9}=\frac{50-5}{9}=5\)
=> \(x=5.2+1=11\)
\(y=5.3+2=17\)
\(z=5.4+3=23\)