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a) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{-4}=\frac{x-y-z}{2-3+4}=\frac{27}{3}=9\)
=> \(\hept{\begin{cases}\frac{x}{2}=9\\\frac{y}{4}=9\\\frac{z}{-4}=9\end{cases}}\) => \(\hept{\begin{cases}x=9.2=18\\y=9.3=27\\z=9.\left(-4\right)=-36\end{cases}}\)
Vậy ...
a, ÁP DỤNG DÃY TỈ SỐ BĂNG NHAU TA CÓ
\(\frac{x}{2}=\frac{y}{3}=\frac{x}{-4}=\frac{x-y-z}{2-3+4}=\frac{27}{3}=9\)
\(\Rightarrow\hept{\begin{cases}x=9.2=18\\y=9.3=27\\z=9.\left(-4\right)=-36\end{cases}}\)
a) Thiếu đề
b) Áp dụng t/c của dãy tỉ số bằng nhau, ta có :
\(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) => \(\frac{4x}{4}=\frac{3y}{6}=\frac{2z}{6}=\frac{4x+3y+2z}{4+6+6}=\frac{14}{16}=\frac{7}{8}\)
=> \(\hept{\begin{cases}\frac{x}{1}=\frac{7}{8}\\\frac{y}{2}=\frac{7}{8}\\\frac{z}{3}=\frac{7}{8}\end{cases}}\) => \(\hept{\begin{cases}x=\frac{7}{8}.1=\frac{7}{8}\\y=\frac{7}{8}.2=\frac{7}{4}\\z=\frac{7}{8}.3=\frac{21}{8}\end{cases}}\)
Vậy ...
Sửa lại xíu :
\(a)\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)và \(x-2y+3z=14\)
\(b)\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)và \(4x+3y+2z=36\)
Ta có : \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\)
=>\(\frac{4\left(3x-2y\right)}{16}=\frac{3\left(2z-4y\right)}{9}=\frac{2\left(4y-3z\right)}{4}\)
Hay \(\frac{12x-8y}{16}=\frac{6z-12y}{9}=\frac{8y-6z}{4}\)= \(\frac{12x-8y+6z-12y+8y-6z}{16+9+4}=0\)
+, \(\frac{12x-8y}{16}=0\)=>\(12x-8y=0\)=>\(12x=8y\Rightarrow3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\left(1\right)\)
+, \(\frac{6z-12x}{9}=0\Rightarrow6z-12x=0\Rightarrow6z=12x\Rightarrow z=2x\Rightarrow\frac{z}{4}=\frac{x}{2}\left(2\right)\)
+, \(\frac{8y-6z}{4}=0\Rightarrow8y-6z=0\Rightarrow8y=6z\Rightarrow4y=3z\Rightarrow\frac{y}{3}=\frac{z}{4}\left(3\right)\)
Từ (1) , (2) và (3) ta suy ra : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)(đpcm)
đặt \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=a\)
\(\Rightarrow z=\frac{4y-2a}{3}\Rightarrow\frac{z}{4}=\frac{y-2a}{3}\)
\(x=\frac{4a+2y}{3}\Rightarrow\frac{x}{2}=\frac{2a+y}{3}\)
\(\left\{\begin{matrix}6x-4y=16y-12z\\4z-8x=12y-9z\\9x-6y=8z-16x\end{matrix}\right.\)\(\Leftrightarrow\) \(\left\{\begin{matrix}6x-20y+12z=0\\-8x-12y+13z=0\end{matrix}\right.\)
\(\left\{\begin{matrix}48x-160y+96z=0\\-48x-72y+78z=0\end{matrix}\right.\)
\(-232y+174z=0\Rightarrow174z=232y\)
\(\Leftrightarrow\frac{174z}{174.4}=\frac{232y}{174.4}\Leftrightarrow\frac{z}{4}=\frac{y}{3}\left(1\right)\)
\(\left\{\begin{matrix}9x-6y=8z-16x\\12y-9z=4z-8x\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}25x-6y-8z=0\\8x+12y-13z=0\end{matrix}\right.\)
\(\left\{\begin{matrix}50x-12y-16z=0\\8x+12y-13z=0\end{matrix}\right.\)
\(58x-29z=0\Leftrightarrow58x=29z\Leftrightarrow\frac{58x}{58.2}=\frac{29z}{58.2}\)
\(\Leftrightarrow\frac{x}{2}=\frac{z}{4}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\left(đpcm\right)\)
\(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{3xz-2yz}{4z}=\frac{2zy-4xy}{3y}=\frac{4yx-3zx}{2x}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{3xz-2yz}{4z}=\frac{2zy-4xy}{3y}=\frac{4yx-3zx}{2x}=\frac{3zx-2yz+2zy-4xy+4xy-3xz}{4z+3y+2z}=0\)
\(\frac{3x-2y}{4}=0\Rightarrow3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}\left(1\right)\)
\(\frac{2z-4x}{3}=0\Rightarrow2z=4x\Rightarrow\frac{x}{2}=\frac{z}{4}\left(2\right)\)
\(\frac{4y-3z}{2}=0\Rightarrow4y=3z\Rightarrow\frac{y}{3}=\frac{z}{4}\left(3\right)\)
\(\text{T}ừ\left(1\right),\left(2\right),\left(3\right)\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\left(\text{đpcm}\right)\)
\(\frac{3x-2y}{4}=\frac{2z-4y}{3}=\frac{4y-3z}{2}\)
=>\(\frac{4\left(3x-2y\right)}{4.4}=\frac{3\left(2z-4x\right)}{3.3}=\frac{2\left(4y-3z\right)}{2.2}\)
=\(\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}\)
=\(\frac{\left(12x-8y\right)+\left(6z-12y\right)+\left(8y-6z\right)}{16+9+4}\)
=\(\frac{12x-8y+6z-12x+8y-6z}{29}\)
=\(\frac{\left(12x-12x\right)+\left(8y-8y\right)+\left(6z-6z\right)}{29}\)
=\(\frac{0}{29}=0\)
Ta có: \(\frac{3x-2y}{4}=0\)=> 3x = 2y => \(\frac{x}{2}=\frac{y}{3}\) (1)
\(\frac{2z-4x}{3}=0\)=> 2z = 4x => \(\frac{x}{2}=\frac{z}{4}\) (2)
Từ (1) và (2) => \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
Chúc bạn học tốt!
\(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=0\)
\(\Rightarrow\hept{\begin{cases}3x-2y=0\\2z-4x=0\end{cases}\Rightarrow\hept{\begin{cases}3x=2y\\2z=4x\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{x}{2}=\frac{z}{4}\end{cases}\Rightarrow}\frac{x}{2}=\frac{y}{3}=\frac{z}{4}}\)