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Xét \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\) (1)
Thay (1) vào P
=> P = \(\dfrac{3k+2.4k+3.5k}{2.5k+3.4k+4.5k}+\dfrac{2.5k+3.4k+4.5k}{3.3k+4.4k+5.5k}\) + \(\dfrac{3.3k+4.4k+5.5k}{4.3k+5.4k+6.5k}\)
=> P = \(\dfrac{26k}{42k}+\dfrac{42k}{50k}\) + \(\dfrac{50k}{62k}\)
=> P = \(\dfrac{13}{21}+\dfrac{21}{25}+\dfrac{25}{31}\approx2,265499232\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Rightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)
\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\\ =\dfrac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{16+9+4}=\dfrac{0}{29}=0\\ \Rightarrow3x=2y;2z=4x;4y=3z\\ \Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12x-8y}{16}=0\\\dfrac{6z-12x}{9}=0\\\dfrac{8y-6z}{4}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}12x-8y=0\\6z-12x=0\\8y-6z=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{z}{4}=\dfrac{x}{2}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
Ta có
\(\dfrac{3x-2y}{4}\)=\(\dfrac{2z-4x}{3}\)=\(\dfrac{4y-3z}{2}\)
=> \(\dfrac{12x-8y}{16}\)=\(\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất DTS bằng nhau
\(\dfrac{12x-8y}{16}\)=\(\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)=\(\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}\)=\(\dfrac{0}{29}\)=0
\(\left\{{}\begin{matrix}12x-8y=0\\6z-12x=0\\8y-6z=0\end{matrix}\right.\)
=>\(\dfrac{x}{2}=\dfrac{y}{3}\),\(\dfrac{y}{3}=\dfrac{z}{4},\dfrac{z}{4}=\dfrac{z}{2}\)
=>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Ta có:
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Rightarrow\left\{{}\begin{matrix}12x-8y=0\\6z-12x=0\\8y-6z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{z}{4}=\dfrac{x}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Vậy \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)(đpcm)
Theo đề ta có:
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
=> \(4.\dfrac{3x-2y}{4}=3.\dfrac{2z-4x}{3}=2.\dfrac{4y-3z}{2}\)
=> \(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
=> \(\dfrac{12x-8y}{16}+\dfrac{6z-12x}{9}+\dfrac{8y-6z}{4}=\dfrac{0}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}12x=8y\\6z=12x\\8y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}12x=8y=6z\\\end{matrix}\right.\)
=> \(\dfrac{12x}{24}=\dfrac{8y}{24}=\dfrac{6z}{24}\)( MSC: 24)
=> \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)(đpcm)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\\ =\dfrac{12x-8x+6x-12x+8y-6z}{16+9+4}\\ =0\\ \Rightarrow3x=2y;2z=4x;4y=3z\\ \Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Rightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)
\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{16+9+4}=0\)\(\Rightarrow\left\{{}\begin{matrix}3x=2y\\2z=4x\\4y=3z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{3}{9}=\dfrac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{1}{3}=\dfrac{2}{3}\\y=3.\dfrac{1}{3}=1\\z=4.\dfrac{1}{3}=\dfrac{4}{3}\end{matrix}\right.\)
Từ \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
\(=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{16+9+4}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-2y}{4}=0\\\dfrac{2z-4x}{3}=0\\\dfrac{4y-3z}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3x=2y\\2z=4x\\4y=3z\end{matrix}\right.\)\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Rightarrow\dfrac{4\left(3x-2y\right)}{3.4}=\dfrac{3\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)
\(\Rightarrow\dfrac{12x-8y}{12}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-8y}{12}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
\(=\dfrac{12x-8y+6z-12x+8y-6z}{12+9+4}\)
\(=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-2y}{4}=0\Rightarrow3x=2y\\\dfrac{2z-4x}{3}=0\Rightarrow2z=4x\\\dfrac{4y-3z}{2}=0\Rightarrow4y=3z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{z}{4}=\dfrac{x}{2}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\rightarrowđpcm\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{16+9+4}=0\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{z}{4}=\dfrac{x}{2}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
a) Ta có: 3x = 2y; 4x = 2z
⇒ \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{x}{2}=\dfrac{z}{4}\)
⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và x + y + z = 27
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{27}{9}=3\)
⇒ \(\dfrac{x}{2}=3\) ⇒ x = 6
\(\dfrac{y}{3}=3\) ⇒ y = 9
\(\dfrac{z}{4}=3\) ⇒ z = 12
Vậy x = 6 ; y = 9 ; z = 12
b) Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
⇒ \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)
⇒ \(\dfrac{2x^2}{8}=\dfrac{3y^2}{27}=\dfrac{5z^2}{80}\)
và 2x2 + 3y2 - 5z2 = -405
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x^2}{8}=\dfrac{3y^2}{27}=\dfrac{5z^2}{80}\)=\(\dfrac{2x^2+3y^2-5z^2}{8+27-80}=\dfrac{-405}{-45}=9\)
+) \(\dfrac{2x^2}{8}=9\) ⇒ 2x2 = 72 ⇒ x2 = 72 : 2
⇒ x2 = 36 ⇒ x = 6 hoặc x = -6
+) \(\dfrac{3y^2}{27}=9\) ⇒ 3y2 = 243 ⇒ y2 = 243 : 3
⇒ y2 = 81 ⇒ y = 9 hoặc y = -9
+) \(\dfrac{5z^2}{80}=9\) ⇒ 5z2 = 720 ⇒ z2 = 720 : 5
⇒ z2 = 144 ⇒ z = 12 hoặc z = -12
Vậy...................................( bạn tự vậy nhé )
c) Giống câu a ( bạn tự chép lại )
d) Mik ko bt lm
CÂU TRẢ LỜI RẤT HAY BẠN NÀO ĐANG CẦN THÌ THAM KHẢO NHÉ!!!!!!!!
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{3xz-2yz}{4z}=\dfrac{2yz-4xy}{3y}=\dfrac{4xy-3xz}{2x}\\ \Rightarrow\dfrac{3xz-2yz}{4z}=\dfrac{2yz-4xy}{3y}=\dfrac{4xy-3xz}{2x}=\dfrac{\left(3xz-3xz\right)+\left(2yz-2yz\right)+\left(4xy-4xy\right)}{4z+3y+2x}=0\\ \Rightarrow3x-2y=2z-4x=4y-3z=0\\ \Rightarrow3x=2y;2z=4x;4y=3z\)
3x=2y => \(\dfrac{x}{2}=\dfrac{y}{3}\)
4x=2z\(\Rightarrow\dfrac{x}{2}=\dfrac{z}{4}\)
\(\dfrac{\Rightarrow x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\\ \Rightarrow x=2k;y=3k;z=4k\)
Thế dô A ; tự tinh !!