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\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
\(\Leftrightarrow\frac{y+z-x}{x}+2=\frac{z+x-y}{y}+2=\frac{x+y-z}{z}+2\)
\(\Leftrightarrow\frac{x+y+z}{x}=\frac{x+y+z}{y}=\frac{x+y+z}{z}\)
\(\Rightarrow x=y=z\)
\(\Rightarrow\frac{x}{y}=1;\frac{y}{z}=1;\frac{z}{x}=1\)
\(\Rightarrow B=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2^3=8\)
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:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}\)
\(=\frac{2x+3y-z-\left(2+6-3\right)}{9}=\frac{50-5}{9}=5\)
=> \(x=5.2+1=11\)
\(y=5.3+2=17\)
\(z=5.4+3=23\)
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}.\)
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:\(\frac{2x-2}{4}=5\Rightarrow2x-2=20\Rightarrow x=11;\frac{3y-6}{9}=5\Rightarrow3y-6=45\Rightarrow y=17;\)
\(\frac{z-3}{4}=5\Rightarrow z-3=20\Rightarrow23\)
áp dụng t/c dãy tỉ số = nhau ta có:\
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{\left(x-1\right)-2\left(y-2\right)+3\left(z-3\right)}{2-2.3+3.4}=\frac{\left(x-2y+3z\right)-1+4-9}{8}=\frac{-10-6}{8}=-2\)
=>x-1=-4;y-2=-6;z-3=-8
=>x=-3;y=-4;z=-5
=>x+y+z=-12
\(\frac{x^3}{1000}=\frac{y^3}{3375}=\frac{z^3}{1728}=>\left(\frac{x}{10}\right)^3=\left(\frac{y}{15}\right)^3=\left(\frac{z}{12}\right)^3=>\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=-\frac{49}{7}=-7\)
=>x=-70;y=-105;z=-84
=>x+y+z=-259
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