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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!!!!!!!!
\(\dfrac{7x-3z}{5}=\dfrac{3y-5x}{7}=\dfrac{5z-7y}{3}\)
\(\Rightarrow\dfrac{35x-15z}{25}=\dfrac{21y-35x}{49}=\dfrac{15z-21y}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{35x-15z}{25}=\dfrac{21y-35x}{49}=\dfrac{15z-21y}{9}\)
\(=\dfrac{35x-15z+21y-35x+15z-21y}{25+49+9}\)
\(=\dfrac{0}{25+49+9}=0\)
\(\Rightarrow\left\{{}\begin{matrix}7x=3z\Rightarrow\dfrac{x}{3}=\dfrac{z}{7}\\3y=5x\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\\5z=7y\Rightarrow\dfrac{z}{7}=\dfrac{y}{5}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{3+5+7}=\dfrac{30}{15}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.5=10\\z=2.7=14\end{matrix}\right.\)
Tương tự
a) \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
Từ \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Rightarrow\dfrac{x^3}{2^3}=\dfrac{y^3}{4^3}=\dfrac{z^3}{6^3}\)
\(\Leftrightarrow\dfrac{x^2}{2^2}=\dfrac{y^2}{4^2}=\dfrac{z^2}{6^2}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{1}{4}\Rightarrow x^2=\dfrac{1}{4}\cdot4\Rightarrow x^2=1\Rightarrow x=1\)
\(\dfrac{y^2}{16}=\dfrac{1}{4}\Rightarrow y^2=\dfrac{1}{4}\cdot16\Rightarrow y^2=4\Rightarrow y=2\)
\(\dfrac{z^2}{36}=\dfrac{1}{4}\Rightarrow z^2=\dfrac{1}{4}\cdot36\Rightarrow z^2=9\Rightarrow z^2=3\)
Xin lỗi mình chỉ làm được câu a)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\Rightarrow\dfrac{4.\left(3x-2y\right)}{4.4}=\dfrac{3.\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)
=\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{0}{29}\)
\(\Rightarrow\) 12x= 8y
6z=12x
8y=6z
=> 12x=8y=6z
MSC: 24
ta có: \(\dfrac{12x}{24}=\dfrac{8y}{24}=\dfrac{6z}{24}\)= \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)( đpcm)
Từ giả thiết \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{4}\)
\(\Rightarrow\dfrac{3xz-2yz}{4z}=\dfrac{2yz-4xy}{3y}=\dfrac{4xy-3xz}{4x}\)
Áp dung tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{3xz-2yz}{4z}=\dfrac{2yz-4xy}{3y}=\dfrac{4xy-3xz}{4x}=\dfrac{3xz-2yz+2yz-4xy+4xy-3xz}{4z+3y+4x}=\dfrac{0}{4z+3y+4x}=0\)
\(\Rightarrow3xz=2yz=4xy\)
\(\Rightarrow\dfrac{3xyz}{y}=\dfrac{2xyz}{x}=\dfrac{4xyz}{z}\)
\(\Rightarrow\dfrac{3}{y}=\dfrac{2}{x}=\dfrac{4}{z}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)(đpcm)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Rightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)
\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\\ =\dfrac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{16+9+4}=\dfrac{0}{29}=0\\ \Rightarrow3x=2y;2z=4x;4y=3z\\ \Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Giải
Ta có:
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)\(=\dfrac{12x-8y+6z-12+8y-6z}{16+9+4}\)\(=\)\(\dfrac{0}{29}=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x-2y=0\Rightarrow3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\\2z-4x=0\Rightarrow2z=4x\Rightarrow\dfrac{x}{2}=\dfrac{z}{4}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.2=4\\y=2.3=6\\z=2.4=8\end{matrix}\right.\)
Vậy \(x=4;y=6;z=8\)
Bạn học tốt!
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
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)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\\ =\dfrac{12x-8x+6x-12x+8y-6z}{16+9+4}\\ =0\\ \Rightarrow3x=2y;2z=4x;4y=3z\\ \Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)