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\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x+3y-z-5}{9}=\frac{x+1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) có 2x + 3y - z = 50
\(\Rightarrow\frac{50-5}{9}=5=\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\hept{\begin{cases}x-1=10\\y-2=15\\z-3=20\end{cases}\Rightarrow\hept{\begin{cases}x=11\\y=17\\z=23\end{cases}}}\)
Trả lời:
Ta có:\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}\)\(=\frac{2x+3y-z-5}{9}\)(Tính chất dãy tỉ số bẳng nhau)
Mà\(2x+3y-z=50\)
\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{50-5}{9}=\frac{45}{9}=5\)
\(\Rightarrow\hept{\begin{cases}2x-2=20\\3y-6=45\\z-3=20\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x=22\\3y=51\\z=23\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=11\\y=17\\z=23\end{cases}}\)
Vậy\(\hept{\begin{cases}x=11\\y=17\\z=23\end{cases}}\)
Hok tốt!
Vuong Dong Yet
Ta có :
\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)
\(\Leftrightarrow\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}=\frac{12\left(x+y+z\right)}{18+16+15}=\frac{12\cdot49}{49}=12\) ( do \(x+y+z=49\) )
\(\Rightarrow\hept{\begin{cases}\frac{12x}{18}=12\\\frac{12y}{16}=12\\\frac{12z}{15}=12\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=18\\y=16\\z=15\end{cases}}\) ( thỏa mãn )
Vậy : \(\left(x,y,z\right)=\left(18,16,15\right)\)
\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)
\(\Rightarrow\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}\)
\(\Rightarrow\frac{12x+12y+12z}{18+16+15}=\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)
\(\Rightarrow\frac{12\left(x+y+z\right)}{49}=\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\) có x + y + z = 49
\(\Rightarrow\frac{12\cdot49}{49}=12=\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\)
\(\Rightarrow\hept{\begin{cases}2x=36\\3y=48\\4z=60\end{cases}\Rightarrow\hept{\begin{cases}x=18\\y=16\\z=15\end{cases}}}\)
A) ta có \(\frac{X}{2}=\frac{Y}{3}\)=>\(\frac{X}{8}=\frac{Y}{12}\)(1)
\(\frac{Y}{4}=\frac{Z}{5}\)=>\(\frac{Y}{12}=\frac{Z}{15}\)(2)
Từ (1)và (2)=>\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\) và x-y-z=28
đến đây tự làm
c) \(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)
\(\Rightarrow\left(x-\frac{1}{5}\right)^{2004}=0\) và \(\left(y+0,4\right)^{100}=0\) và \(\left(z-3\right)^{678}=0\)
+) \(\left(x-\frac{1}{5}\right)^{2004}=0\Rightarrow x-\frac{1}{5}=0\Rightarrow x=\frac{1}{5}\)
+) \(\left(y+0,4\right)^{100}=0\Rightarrow y+0,4=0\Rightarrow y=-0,4\)
+) \(\left(z-3\right)^{678}=0\Rightarrow z-3=0\Rightarrow z=3\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\frac{1}{5};-0,4;3\right)\)
Vì \(\left(x-\frac{1}{5}\right)^{2004},\left(y+0,4\right)^{100},\left(z-3\right)^{678}\ge0\forall x,y,z\)
\(\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\frac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{5}\\x=-0,4=-\frac{2}{5}\\z=3\end{matrix}\right.\)
Vậy \(\left(x,y,z\right)\in\left\{\left(\frac{1}{5};-\frac{2}{5};3\right)\right\}\)
(x-\(\frac{1}{5}\))2004+(y+0,4)100+(z-3)678=0
⇒\(\left\{{}\begin{matrix}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{matrix}\right.\)
⇒\(\left\{{}\begin{matrix}x=\frac{1}{5}\\y=-0,4\\z=3\end{matrix}\right.\)
Vậy...
Chúc bạn học tốt!
a) \(\frac{x}{1}=\frac{y}{3}=\frac{4z}{15}=\frac{6x+7y+8z}{1.6+3.7+15.2}=\frac{456}{57}=8\)
x=8
y=24
z=30
\(3x=y\)=> \(\frac{x}{1}=\frac{y}{3}\)
hay \(\frac{x}{4}=\frac{y}{12}\)
\(5y=4z\)=> \(\frac{y}{4}=\frac{z}{5}\)
hay \(\frac{y}{12}=\frac{z}{15}\)
suy ra: \(\frac{x}{4}=\frac{y}{12}=\frac{z}{15}\)
đến đây bạn ADTCDTSBN nhé
Vì \(\left(x-\frac{1}{5}\right)^{2004}\ge0\);\(\left(y+0,4\right)^{100}\ge0\);\(\left(z-3\right)^{678}\ge0\)( Vì mũ chẵn)
Nên để biểu thức bằng 0 \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{5}\\y=-0,4\\z=3\end{cases}}\)
Bài giải
a, Đặt \(\frac{x}{2}=\frac{y}{5}=k\text{ }\Rightarrow\text{ }\hept{\begin{cases}x=2k\\y=5k\end{cases}}\text{ }\Rightarrow\text{ }x\cdot y=2k\cdot5k=10k^2=90\text{ }\Rightarrow\text{ }k^2=9\text{ }\Rightarrow\text{ }k=\pm3\)
\(\Rightarrow\text{ }\hept{\begin{cases}x=2\cdot\left(-3\right)=-6\\y=5\cdot\left(-3\right)=-15\end{cases}}\) hoặc \(\hept{\begin{cases}x=2\cdot3=6\\y=5\cdot3=15\end{cases}}\)
Vậy \(\left(x\text{ ; }y\right)=\left(-3\text{ ; }-15\right)\text{ ; }\left(6\text{ ; }15\right)\)
b, Do \(\hept{\begin{cases}\left(x-\frac{1}{5}\right)^{2004}\ge0\\\left(y+0,4\right)^{100}\ge0\\\left(z-3\right)^{678}\ge0\end{cases}}\text{ mà }\left(x-\frac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-\frac{1}{5}\right)^{2004}\ge0\\\left(y+0,4\right)^{100}\ge0\\\left(z-3\right)^{678}\ge0\end{cases}}\Rightarrow\hept{\begin{cases}\left(x-\frac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{cases}}\Rightarrow\hept{\begin{cases}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{5}\\y=-0,4\\z=3\end{cases}}\)
Vậy \(x=\frac{1}{5}\text{ , }y=-0,4\text{ , }z=3\)
a) ĐẶt \(\frac{x}{2}=\frac{y}{5}=k\)suy ra x=2k, y=5k
Mà x.y=90
suy ra 2k. 5k = 90 suy ra k2=9 suy ra k\(\in\){3;-3}
Với k=3 suy ra x=6, y=15
Với k = -3 suy ra x=-1; y=-15
b) Vì \(\left(x-\frac{1}{5}\right)^{2004}\ge0,\forall x\)
\(\left(y+0,4\right)^{100}\ge0,\forall y\)
\(\left(z-3\right)^{678}\ge0,\forall z\)
Suy ra \(\left(x-\frac{1}{5}\right)^{2004}\)+\(\left(y+0,4\right)^{100}\)+\(\left(z-3\right)^{678}\ge0;\forall x,y,z\)
suy ra \(\left(x-\frac{1}{5}\right)^{2004}=0\)và \(\left(y+0,4\right)^{100}=0\)và \(\left(z-3\right)^{678}=0\)
suy ra x=\(\frac{1}{5}\); y=-0,4 ; z=3