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\(\dfrac{x}{3}=\dfrac{y-5}{7}=\dfrac{z+2}{3}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{2y-10}{14}=\dfrac{5z+10}{15}\)
\(x+2y=5z\Leftrightarrow x+2y-5z=0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{2y-10}{14}=\dfrac{5z+10}{15}=\dfrac{x+2y-10-5z-10}{3+14-15}\)
\(=\dfrac{-20}{2}=-10\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-30\\y=-65\\z=-32\end{matrix}\right.\)
Vậy...
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{2x+y+z}=\frac{1}{(x+y)+(x+z)}\leq \frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\)
\(\Rightarrow \frac{x}{2x+y+z}\leq \frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
Tương tự:
\(\frac{y}{2y+x+z}\leq \frac{1}{4}\left(\frac{y}{y+z}+\frac{y}{y+x}\right)\)
\(\frac{z}{2z+x+y}\leq \frac{1}{4}\left(\frac{z}{z+x}+\frac{z}{z+y}\right)\)
Cộng theo vế:
\(D\leq \frac{1}{4}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{3}{4}\) (dpcm)
Dấu bằng xảy ra khi $x=y=z$
Ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(y+z\right)}\le\dfrac{x}{2\sqrt{\left(x+y\right)\left(y+z\right)}}\)
Tương tự với các phân số khác
\(\Rightarrow VT\le\dfrac{1}{2}\left(\dfrac{x}{\sqrt{\left(x+y\right)\left(z+x\right)}}+\dfrac{y}{\sqrt{\left(y+z\right)\left(x+y\right)}}+\dfrac{z}{\sqrt{\left(z+x\right)\left(x+y\right)}}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x+y}\cdot\sqrt{z+x}}+\dfrac{\sqrt{y}\cdot\sqrt{y}}{\sqrt{y+z}\cdot\sqrt{x+y}}+\dfrac{\sqrt{z}\cdot\sqrt{z}}{\sqrt{z+x}\cdot\sqrt{y+z}}\right)\)
\(\le\dfrac{1}{2}\left(\dfrac{\dfrac{x}{x+y}+\dfrac{x}{z+x}}{2}+\dfrac{\dfrac{y}{y+z}+\dfrac{y}{x+y}}{2}+\dfrac{\dfrac{z}{z+x}+\dfrac{z}{y+z}}{2}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}\right)+\left(\dfrac{y}{y+z}+\dfrac{z}{y+z}\right)+\left(\dfrac{z}{z+x}+\dfrac{x}{z+x}\right)}{2}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{2}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi x = y = z
Áp dụng tính chất dãy tỉ số bằng nhau được:
\(\dfrac{x}{2x+y+z}\)=\(\dfrac{y}{2y+x+z}\)=\(\dfrac{z}{2z+x+y}\)=\(\dfrac{x+y+z}{2x+y+z+2y+x+z+2z+x+y}\)=\(\dfrac{x+y+z}{3x+3y+3z}\)=\(\dfrac{x+y+z}{3.\left(x+y+z\right)}\)=\(\dfrac{1}{3}\)=\(\dfrac{3}{9}\)<\(\dfrac{3}{4}\)(đpcm)
a)Xét \(x=\dfrac{y}{2}=\dfrac{z}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=k\\y=2k\\z=3k\end{matrix}\right.\) (1)
Thay (1) vào 4x - 3y + 2z = 36
\(\Rightarrow4.k-3.2k+2.3k=36\)
\(\Rightarrow4k-6k+6k=36\Rightarrow4k=36\)
\(\Rightarrow k=\dfrac{36}{4}=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=2.4=8\\z=3.4=12\end{matrix}\right.\)
Vậy...............................................................
b) Xét \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{7}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\\z=7k\end{matrix}\right.\) (2)
Thay (2) vào 2x - 3z = 44
\(\Rightarrow2.5k-3.7k=44\)
\(\Rightarrow-11k=44\Rightarrow k=-4\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.\left(-4\right)=-20\\y=4.\left(-4\right)=-16\\z=7.\left(-4\right)=-28\end{matrix}\right.\)
Vậy,................................................
c) Xét \(\dfrac{-x}{7}=\dfrac{y}{11}=\dfrac{-z}{5}=\dfrac{x}{-7}=\dfrac{z}{-5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=-7k\\y=11k\\z=-5k\end{matrix}\right.\) (3)
Thay (3) vào -3z - 2y - x = -88
\(\Rightarrow-3.\left(-5k\right)-2.11k-\left(-7k\right)=-88\)
\(\Rightarrow15k-22k+7k=-88\Rightarrow0k=88\)
\(\Rightarrow k\in\varnothing\)
Suy ra: Không có cặp ( x; y; z) thỏa mãn
Vậy.................................................................
d) Xét \(\dfrac{y}{12}=\dfrac{x}{-5}=\dfrac{z}{11}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5k\\y=12k\\z=11k\end{matrix}\right.\) (4)
Thay (4) vào 5y - 2z = 114
\(\Rightarrow6.12k-2.11k=114\)
\(\Rightarrow50k=114\Rightarrow k=2,28\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5.2,28=-11,4\\y=12.2,28=27,36\\z=25,08\end{matrix}\right.\)
Vậy..............................................
e) Xét \(\dfrac{x}{25}=\dfrac{y}{17}=\dfrac{z}{32}=k\)
\(\left\{{}\begin{matrix}x=25k\\y=17k\\z=32k\end{matrix}\right.\) (5)
Thay (5) vào -2z + 3y - 4x = -452
\(\Rightarrow\left(-2\right).32k+3.17k-4.25k=-452\)
\(\Rightarrow-113k=-452\Rightarrow k=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=25.5=100\\y=17.4=68\\z=32.4=128\end{matrix}\right.\)
Vậy.......................................................
a) Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(x=\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\\ \Rightarrow\dfrac{4x}{4}-\dfrac{3y}{6}+\dfrac{2z}{6}=\dfrac{4x-3y+2z}{4-6+6}=\dfrac{36}{4}=9\)
+) \(\dfrac{x}{1}=9\Rightarrow x=9\)
+) \(\dfrac{y}{2}=9\Rightarrow y=18\)
+) \(\dfrac{z}{3}=9\Rightarrow z=27\)
Vậy x = 9; y = 18; z = 27.
tương tự
\(\dfrac{x-y}{z}=\dfrac{3y}{x-z}=\dfrac{x}{y}=\dfrac{x-y+3y+x}{z+x-z+y}=\dfrac{2x+2y}{x+y}\)
⇒ \(\dfrac{x}{y}=2\) ⇒ x = 2y
Có :\(\dfrac{3y}{x-z}=2\) ⇔ 3y = 2x - 2z
Mà : x = 2y ⇒ 3y = 2. 2y - 2z
⇔ 3y = 4y - 2z
⇔ 2z = y