Cho x,y,z khác 0 thoã mãn 4x+5y-6z=0 và 1/4x+1/5y-1/6z=0 Tính 16x2+25y2+36z2
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Đặt \(\frac{x}{-5}=\frac{y}{6}=\frac{z}{-2}=k\) \(\left(k\ne0\right)\)
\(\Rightarrow x=-5k;y=6k;z=-2k\)
\(\Rightarrow A=\frac{3.k.\left(-5\right)+6.k-2.\left(-2\right).k}{-3.\left(-5\right).k-5.6.k+6.\left(-2\right).k}=\frac{-15k+6k+4k}{15k-30k-12k}=\frac{-5k}{-27k}=\frac{5}{27}\)
Vậy \(A=\frac{5}{27}\).
\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)
\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)
\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)
3z là 3x phải k :v
\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\)
nên :
\(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)
\(\Rightarrow\hept{\begin{cases}\frac{3x}{45}=\frac{4y}{80}=\frac{5z}{120}\\\frac{4x}{60}=\frac{5y}{100}=\frac{6z}{144}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{3x+4y+5z}{45+80+120}=\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\\\frac{4x+5y+6z}{60+100+144}=\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\end{cases}}\)
\(\Rightarrow\frac{3x+4y+5z}{245}=\frac{4x+5y+6z}{304}\)
\(\Rightarrow\frac{3x+4y+5z}{4x+5y+6z}=\frac{245}{304}\)
\(\Rightarrow M=\frac{245}{304}\)
bài này đặt k ez hơn : )
\(\hept{\begin{cases}\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\\\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\end{cases}}\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)
đặt \(k=\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\Rightarrow x=15k,y=20k,z=24k\)
\(\Rightarrow M=\frac{3x+4y+5z}{4x+5y+6z}=\frac{3.15k+4.20k+5.24k}{4.15k+5.20k+6.24k}=\frac{245}{304}\)
khó quá
k nhé tớ k lại cho
hihihiihih ^_^ ~ hihihihihih
Vì \(\left(3x-2y\right)^{100}\ge0\forall x,y\inℤ\)
\(|5y-6z|\ge0\forall y,z\inℤ\Rightarrow|5y-6z|^{153}\ge0\forall y,z\inℤ\)
Nên \(\Rightarrow\hept{\begin{cases}(3x-2y)^{100}=0\\|5y-6z|^{153}=0\end{cases}}\Rightarrow\hept{\begin{cases}3x-2y=0\\5y-6z=0\end{cases}}\Rightarrow\hept{\begin{cases}3x=2y\\5y=6z\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{6}=\frac{z}{5}\end{cases}}}\)
Từ \(\frac{x}{2}=\frac{y}{3};\frac{y}{6}=\frac{z}{5}\)suy ra\(\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)
Ta có
\(\frac{x}{4}=\frac{y}{6}=\frac{z}{5}=\frac{2x}{8}=\frac{5y}{30}=\frac{3z}{15}=\frac{2x-5y+3z}{8-30+15}=\frac{56}{-7}=-8\)
Do đó
\(\frac{x}{4}=-8\Rightarrow x=-32\)
\(\frac{y}{6}=-8\Rightarrow y=-48\)
\(\frac{z}{5}=-8\Rightarrow z=-40\)
Vậy \(x=-32;y=-48;z=-40\)