Áp dụng dãy tỉ số bằng nhau:

b.

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=-5.\left(-1\right)=5\end{matrix}\right.\)

d.

\(\dfrac{4}{x}=\dfrac{7}{y}\Rightarrow\dfrac{y}{7}=\dfrac{x}{4}=\dfrac{y-x}{7-4}=\dfrac{-12}{3}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.\left(-4\right)=-16\\y=7.\left(-4\right)=-28\end{matrix}\right.\)

a) Ta có: \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)

nên \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{2x}{3}=12\\\dfrac{3y}{4}=12\\\dfrac{4z}{5}=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=36\\3y=48\\4z=60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=20\end{matrix}\right.\)

Vậy: (x,y,z)=(18;16;20)

b) Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=3k\end{matrix}\right.\)

Ta có: \(x^2-y^2=4\)

\(\Leftrightarrow\left(5k\right)^2-\left(3k\right)^2=4\)

\(\Leftrightarrow16k^2=4\)

\(\Leftrightarrow k\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

Trường hợp 1: \(k=\dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5k=5\cdot\dfrac{1}{2}=\dfrac{5}{2}\\y=3k=3\cdot\dfrac{1}{2}=\dfrac{3}{2}\end{matrix}\right.\)

Trường hợp 2: \(k=-\dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5k=5\cdot\dfrac{-1}{2}=\dfrac{-5}{2}\\y=3k=3\cdot\dfrac{-1}{2}=\dfrac{-3}{2}\end{matrix}\right.\)

Vậy: \(\left(x,y\right)\in\left\{\left(\dfrac{5}{2};\dfrac{3}{2}\right);\left(-\dfrac{5}{2};-\dfrac{3}{2}\right)\right\}\)

a)

Theo tính chất của dãy tỉ số bằng nhau, ta có : 

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Suy ra : 

\(x=\dfrac{12.3}{2}=18\\ y=\dfrac{12.4}{3}=16\\ z=\dfrac{12.5}{4}=15\)

b)

\(x=\dfrac{y}{3}.5=\dfrac{5y}{3}\\ x^2-y^2=4\\ \Leftrightarrow\left(\dfrac{5y}{3}\right)^2-y^2=4\\ \Leftrightarrow\dfrac{16y^2}{9}=4\Leftrightarrow y=\pm\dfrac{3}{2} \)

Với $y = \dfrac{3}{2}$ thì $x = \dfrac{5}{2}$

Với $y = \dfrac{-3}{2}$ thì $x = \dfrac{-5}{2}$

c)

\(\dfrac{x}{y+z+1}=\dfrac{y}{z+x+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2x+2y+2z}=\dfrac{1}{2}\)

Suy ra : 

\(2x=y+z+1\Leftrightarrow y+z=2x-1\)

Mặt khác : 

\(x+y+z=\dfrac{1}{2}\Leftrightarrow x+2x-1=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}\)

\(2y=x+z+1=z+\dfrac{3}{2}\)

Mà \(y+z=0\Leftrightarrow z=-y\)

nên suy ra:  \(y=\dfrac{1}{2};z=-\dfrac{1}{2}\)

Bài 2: 

Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)

Ta có: xy=12

\(\Leftrightarrow12k^2=12\)

\(\Leftrightarrow k^2=1\)

Trường hợp 1: k=1

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=3\\y=4k=4\end{matrix}\right.\)

Trường hợp 2: k=-1

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=-3\\y=4k=-4\end{matrix}\right.\)

a)\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{21}\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{98}{48}=\frac{49}{23}\)

suy ra :

\(\frac{x}{10}=\frac{49}{23}\Rightarrow x=\frac{490}{23}\)

\(\frac{y}{15}=\frac{49}{23}\Rightarrow y=\frac{735}{23}\)

\(\frac{z}{21}=\frac{49}{23}\Rightarrow z=\frac{1029}{23}\)

bạn xem lại đề ra số hơi xấu

khiếp,ít ít thôi, t giải phụ chứ nhìn lóa mắt quá

a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)

b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)

\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)

d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)

\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)

a) �2=�5=�7;�+�+�=562x​=5y​=7z​;x+y+z=56

�2=�5=�7=�+�+�2+5+7=5614=42x​=5y​=7z​=2+5+7x+y+z​=1456​=4

⇒{�=4.2=8�=4.5=20�=4.7=28⇒⎩⎨⎧​x=4.2=8y=4.5=20z=4.7=28​

b) �1,1=�1,3=�1,4(1);2�−�=5,51,1x​=1,3y​=1,4z​(1);2x−y=5,5

(1)⇒2�−�1,1.2−1,3=5,50,9(1)⇒1,1.2−1,32x−y​=0,95,5​
    

⇒⎩⎨⎧​x=1,1.0,95,5​=0,96,05​y=1,3.0,95,5​=0,97,15​z=1,11,4​.x=1,11,4​.0,96,05​=0,998,47​​

d) �2=�3=�5;���=−302x​=3x​=5z​;xyz=−30

�2=�3=�5=���2.3.5=−3030=−12x​=3x​=5z​=2.3.5xyz​=30−30​=−1

⇒{�=2.(−1)=−2�=3.(−1)=−3�=5.(−1)=−5⇒⎩⎨⎧​x=2.(−1)=−2y=3.(−1)=−3z=5.(−1)=−5​
 

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Bài a:

\(Theo.tính.chất.dãy.tỷ.số.bằng.nhau.ta.có:\\ \dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{6}=\dfrac{x-y-z}{3-2-6}=\dfrac{30}{-5}=-6\\ Vậy:x=-6.3=-18;y=-6.2=-12;z=-6.6=-36\)

Bài b:

Theo t/c dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{4}=\dfrac{b}{5}=\dfrac{c}{6}=\dfrac{a+b-c}{4+5-6}=\dfrac{15}{3}=5\\ \Rightarrow a=5.4=20;b=5.5=25;c=5.6=30\\ Vậy:a=20;b=25;c=30\)