a) Ta có : 2x = 3y = 5z

=> \(\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\)

=> \(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x-y-z}{15-10-6}=\frac{-33}{-1}=33\)(dãy tỉ số bằng nhau)

=> \(\hept{\begin{cases}x=33.15=495\\y=33.10=330\\z=33.6=198\end{cases}}\)

b) Ta có 10x = 15y = 6z

=> \(\frac{10x}{30}=\frac{15y}{30}=\frac{6z}{30}\)

=> \(\frac{x}{3}=\frac{y}{2}=\frac{z}{5}\)

=> \(\frac{10x}{30}=\frac{5y}{10}=\frac{z}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có 

\(\frac{x}{3}=\frac{y}{2}=\frac{z}{5}=\frac{10x}{30}=\frac{5y}{10}=\frac{z}{5}=\frac{10x-5y+z}{30-10+5}=\frac{-33}{25}=-1.32\)

=> \(\hept{\begin{cases}x=-3,96\\y=-2,64\\z=-6,6\end{cases}}\)

c) Ta có \(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}\)

=> \(\frac{x^2}{25}=\frac{y^2}{49}=\frac{z^2}{9}=\frac{x^2+y^2-z^2}{25+49-9}=\frac{585}{65}=9\)

=> \(\hept{\begin{cases}x=\pm15\\y=\pm21\\z=\pm9\end{cases}}\)

Vì  \(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}\)=> x ; y ; z cùng dấu

=> Các cặp x;y;z thỏa mãn là (15;21;9) ; (-15;-21;-9)

a) \(\hept{\begin{cases}2x=3y=5z\\x-y-z=23\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}\\x-y-z=23\end{cases}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x-y-z}{\frac{1}{2}-\frac{1}{3}-\frac{1}{5}}=\frac{23}{-\frac{1}{30}}=-690\)

\(\hept{\begin{cases}\frac{x}{\frac{1}{2}}=-690\\\frac{y}{\frac{1}{3}}=-690\\\frac{z}{\frac{1}{5}}=-690\end{cases}}\Rightarrow\hept{\begin{cases}x=-345\\y=-230\\z=-138\end{cases}}\)

b) \(\hept{\begin{cases}10x=15y=6z\\10x-5y+z=-33\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{\frac{1}{10}}=\frac{y}{\frac{1}{15}}=\frac{z}{\frac{1}{6}}\\10x-5y+z=-33\end{cases}}\Rightarrow\hept{\begin{cases}\frac{10x}{1}=\frac{5y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}\\10x-5y+z=-33\end{cases}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{10x}{1}=\frac{5y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}=\frac{10x-5y+z}{1-\frac{1}{3}+\frac{1}{6}}=\frac{-33}{\frac{5}{6}}=-\frac{198}{5}\)

\(\hept{\begin{cases}\frac{10x}{1}=-\frac{198}{5}\\\frac{5y}{\frac{1}{3}}=-\frac{198}{5}\\\frac{z}{\frac{1}{6}}=-\frac{198}{5}\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{99}{25}\\y=-\frac{66}{25}\\z=-\frac{33}{5}\end{cases}}\)

c) \(\hept{\begin{cases}\frac{x}{5}=\frac{y}{7}=\frac{z}{3}\\x^2+y^2-z^2=585\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x^2}{5^2}=\frac{y^2}{7^2}=\frac{z^2}{3^2}\\x^2+y^2-z^2=585\end{cases}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x^2}{5^2}=\frac{y^2}{7^2}=\frac{z^2}{3^2}=\frac{x^2+y^2-z^2}{5^2+7^2-3^2}=\frac{585}{65}=9\)

\(\hept{\begin{cases}\frac{x^2}{5^2}=9\\\frac{y^2}{7^2}=9\\\frac{z^2}{3^2}=9\end{cases}}\Rightarrow\hept{\begin{cases}x=\pm15\\y=\pm21\\z=\pm9\end{cases}}\)

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}\)cùng dấu

=> ( x ; y ; z ) = ( 15 ; 21 ; 9 ) hoặc ( x ; y ; z ) = ( -15 ; -21 ; -9 )

a) Ta có: 2x=3y=5z

\(\frac{x}{\frac{1}{2}}\)=\(\frac{y}{\frac{1}{3}}\)=\(\frac{z}{\frac{1}{5}}\)

Áp dụng dãy tỉ số bằng nhau, ta có:

\(\frac{x}{\frac{1}{2}}\)=\(\frac{y}{\frac{1}{3}}\)=\(\frac{z}{\frac{1}{5}}\)=\(\frac{x-y-z}{\frac{1}{2}-\frac{1}{3}-\frac{1}{5}}\)=\(\frac{23}{\frac{-1}{30}}\)=-690

Lại có:

\(\frac{x}{\frac{1}{2}}\)=-690 \(=>\) x=-690.\(\frac{1}{2}\)=-345

\(\frac{y}{\frac{1}{3}}\)=-690 \(=>\) y=-690.\(\frac{1}{3}\)=-230

\(\frac{z}{\frac{1}{5}}\)=-690 \(=>\) z=-690.\(\frac{1}{5}\)=-138

Vậy............................................................................................

1.Tính giá trị của biểu thức: A=\(\frac{5x^2+3y^2}{10x^2-3y^2}\left(1\right)biết\frac{x}{3}=\frac{y}{5}suyra:5x=3y;suyra:x=\frac{3y}{5};thayvào\left(1\right)taco:\frac{5\left(\frac{3y}{5}\right)^2+3y^2}{10\left(\frac{3y}{5}\right)^2-3y^2}=\frac{\frac{9y^2}{5}+3y^2}{\frac{18y^2}{5}-3y^2}=\frac{24y^2}{5}\cdot\frac{5}{3y^2}=8\)

2.\(\frac{x}{y}=\frac{7}{10}suyra;\frac{x}{7}=\frac{y}{10}\left(1\right)và\frac{y}{z}=\frac{5}{8}suyra;\frac{y}{5}=\frac{z}{8}suyra;\frac{y}{5}\cdot\frac{1}{2}=\frac{z}{8}\cdot\frac{1}{2}suyra;\frac{y}{10}=\frac{z}{16}\left(2\right)Tù\left(1\right)và\left(2\right)suyra\frac{x}{7}=\frac{y}{10}=\frac{z}{16}và2x+5y-2z=9;suyra:\frac{2x}{14}=\frac{5y}{50}=\frac{2z}{32}ápdụngtínhchấtcủadãytỉsốbằngnhautacó\frac{2x}{14}=\frac{5y}{50}=\frac{2z}{32}=\frac{2x+5y-2z}{14+50-32}=\frac{9}{32}suyra;x=\frac{63}{32};y=\frac{45}{16};z=\frac{9}{2}\)

1. -2x=5y =>\(\frac{x}{y}=\frac{-5}{2}=>y=\frac{-2x}{5}\)

Thế y=\(\frac{-2x}{5}\) ta được:

x+\(\frac{-2x}{5}\)=30     \(\Rightarrow\frac{5x-2x}{5}=30\)

\(\Rightarrow3x=150\)\(\Rightarrow x=50\)

=>y=30-x=30-50=-20.

Vậy x=50; y=-20.

Những bài khác tương tự bạn nhé!

bạn kia làm đúng rồi

k tui nha 

thank

3(x-2)-4(2x+1)-5(2x+3)=50

<=>(3x-6)-(8x+4)-(10x+15)=50

<=>3x-6-8x-4-10x-15=50

<=>(3x-8x-10x)+(-6-4-15)=50

<=>-15x-25=50

<=>-15x=75

<=>x=-5

\(3\frac{1}{2}:\left(4-\frac{1}{3}\left|2x+1\right|\right)=\frac{21}{22}\)

<=>\(4-\frac{1}{3}\left|2x+1\right|=\frac{7}{2}:\frac{21}{22}=\frac{11}{3}\)

<=>\(\frac{1}{3}\left|2x+1\right|=4-\frac{11}{3}=\frac{1}{3}\)

<=>\(\left|2x+1\right|=1\)

<=>2x+1=1 hoặc 2x+1=-1

<=>2x=0 hoặc 2x=-2

<=>x=0 hoặc x=-2

Vậy......................

\(A=\frac{2x-y}{3x-y}+\frac{5y-x}{3x+y}\)

\(=\frac{\left(2x-y\right)\left(3x+y\right)+\left(5y-x\right)\left(3x-y\right)}{\left(3x-y\right)\left(3x+y\right)}\)

\(=\frac{3x^2+15xy-6y^2}{9x^2-y^2}\)

\(=\frac{3\left(x^2+5xy-2y^2\right)}{9x^2-y^2}\)

\(=\frac{3\left(10x^2+5xy-3y^2-9x^2+y^2\right)}{9x^2-y^2}\)

\(=-\frac{3\left(9x^2-y^2\right)}{9x^2-y^2}\)

= - 3 (đpcm)

~~~

\(A=\frac{1}{x}+\frac{1}{x+2}+\frac{x-2}{x^2+2x}\)

\(=\frac{x+2+x+x-2}{x^2+2x}\)

\(=\frac{3x}{x\left(x+2\right)}\)

\(=\frac{3}{x+2}\)

\(A\in Z\)

\(\Leftrightarrow3⋮x+2\)

\(\Leftrightarrow x+2\in\text{Ư}\left(3\right)=\left\{-3:-1;1;3\right\}\)

\(\Leftrightarrow x\in\left\{-5;-3;-1;1\right\}\)

cực dễ , có máy tính là ok