Tìm x, y, z biết:
\(\frac{1}{x-1}=\frac{2}{y-2}=\frac{3}{z-3}\) và \(x+2y+3z=56\)
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Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)
\(\Rightarrow x=2k+1,y=3k+2,z=4k+3\)
Mà x-2y+3z=-10
Hay 2k+1-2(3k+2)+3(4k+3)=-10
2k+1-6k-4+12k+9=-10
(2k-6k+12k)+(1-4+9)=-10
8k+6=-10
8k=-16
k=-2
\(\Rightarrow x=-2\cdot2+1=-3,y=-2\cdot3+2=-4,z=-2\cdot4+3=-5\)
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=>\frac{x-1}{2}=\frac{2\left(y-2\right)}{6}=\frac{3\left(z-3\right)}{12}=>\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)
Theo t/c dãy tỉ số=nhau:
\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{x-1-\left(2y-4\right)+\left(3z-9\right)}{2-6+12}=\frac{x-1-2y+4+3z-9}{8}\)
\(=\frac{\left(x-2y+3z\right)-\left(1-4+9\right)}{8}=\frac{14-6}{8}=\frac{8}{8}=1\)
Do đó: \(\frac{x-1}{2}=1=>x-1=2=>x=3\)
\(\frac{y-2}{3}=1=>y-2=3=>y=5\)
\(\frac{z-3}{4}=1=>z-3=4=>z=7\)
Vậy x=3;y=5;z=7
=>(x-1)/2=(-2y+4)/-6=(3z-9)/12
=(x-1-2y+4+3z-9)/(2-6+12)
=-16/8=-2
=> (x-1)/2=-2<=>x-1=-4<=>x=-3
=>(y-2)/3=-2<=>y-2=-6<=>y=-4
=>(z-3)/4=-2<=>z-3=-8<=>z=-5
=>(x-1)/2=(-2y+4)/-6=(3z-9)/12
=(x-1-2y+4+3z-9)/(2-6+12)
=-16/8=-2
=> (x-1)/2=-2<=>x-1=-4<=>x=-3
=>(y-2)/3=-2<=>y-2=-6<=>y=-4
=>(z-3)/4=-2<=>z-3=-8<=>z=-5
Vậy x = -3 ; y = -4 ; z = -5
c)\(x:y:z=3:4:5\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)và\(2x^2+2y^2-3z^2=-100\)
đặt\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)
\(\Rightarrow\frac{x}{3}=k\Rightarrow x=3k\)
\(\Rightarrow\frac{y}{4}=k\Rightarrow y=4k\)
\(\Rightarrow\frac{z}{5}=k\Rightarrow z=5k\)
mà\(2x^2+2y^2-3z^2=-100\)
thay\(6k^2+8k^2-15k^2=-100\)
\(k^2\left(6+8-15\right)=-100\)
\(k^2.\left(-1\right)=-100\)
\(k^2=100\)
\(\Rightarrow k=\pm10\)
bạn thế vào nha
Aps dụng tính chất dãy tỉ số bằng nhau Ta có:
\(\frac{1+2+3}{x-1+y-2+z-3}=\frac{1+2+3}{x+y+z-1-2-3}=\frac{1+4+9}{x+2y+3z-\left(-4\right)}=\frac{ }{ }\)
=\(\frac{14}{56+4}=\frac{14}{60}=\frac{7}{30}\)
\(\Rightarrow\)\(\frac{1}{x-1}=\frac{7}{30}\)\(\Rightarrow\)x-1=\(\frac{30}{7}\)\(\Rightarrow\)x=\(\frac{37}{7}\)
\(\Rightarrow\)\(\frac{2}{y-2}=\frac{7}{30}\Rightarrow y-2=\frac{60}{7}\)\(\Rightarrow\)y=\(\frac{74}{7}\)
\(\Rightarrow\)\(\frac{3}{z-3}=\frac{7}{30}\Rightarrow z-3=\frac{90}{7}\)\(\Rightarrow\)x=\(\frac{111}{7}\)
Aps dụng tính chất dãy tỉ số bàng nhau, ta có:
\(\frac{1+2+3}{x-1+2y-2+z-3}=\frac{1+4+9}{x-1+2y-4+3z-9}\)=\(\frac{14}{x+2y+3z-1-2-3}=\frac{14}{56-1-2-3}=\frac{14}{50}=\frac{7}{25}\)
\(\Rightarrow\)\(\frac{1}{x-1}=\frac{7}{25}\Rightarrow x=\frac{32}{7}\)
\(\Rightarrow\)\(\frac{4}{2y-4}=\frac{7}{25}\Rightarrow y=\frac{64}{7}\)
\(\Rightarrow\)\(\frac{9}{3z-9}=\frac{7}{25}\Rightarrow z=\frac{96}{7}\)