1)

a) 3x = 4y \(\Rightarrow\frac{x}{4}=\frac{y}{3}\)\(\Rightarrow\frac{x}{8}=\frac{y}{6}\)( 1 )

5y = 6z \(\Rightarrow\frac{y}{6}=\frac{z}{5}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y+z}{8+6+5}=\frac{1}{19}\)

\(\Rightarrow x=\frac{8}{19};y=\frac{6}{19};z=\frac{5}{19}\)

b) \(\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-3}{5}\Rightarrow\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}\)

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

\(\frac{3x-3}{9}=\frac{4y-8}{16}=\frac{5z-15}{25}=\frac{\left(3x-3\right)+\left(4y-8\right)+\left(5z-15\right)}{9+16+25}=\frac{-25}{50}=\frac{-1}{2}\)

\(\Rightarrow x=\frac{-1}{2};y=0;z=\frac{1}{2}\)

a) \(\frac{2x}{3}=\frac{3y}{4}\Leftrightarrow8x=9y\Rightarrow x=\frac{9y}{8}\left(1\right)\)

     \(\frac{3y}{4}=\frac{4z}{5}\Leftrightarrow15y=16z\Rightarrow z=\frac{15y}{16}\left(2\right)\)

THay (1) và (2) vào biểu thức \(x+y+z=41\);ta được : \(\frac{9y}{8}+y+\frac{15y}{16}=41\)

\(\Rightarrow18y+16y+15y=656\Rightarrow y=\frac{656}{49}\)

Do đó : \(x=\frac{\frac{9.656}{49}}{8}=\frac{738}{49}\)

             \(z=\frac{\frac{15.656}{49}}{16}=\frac{615}{49}\)

KL : \(x=\frac{738}{49};y=\frac{656}{49};z=\frac{615}{49}\)

b) Ta có : \(4x=3y\Rightarrow x=\frac{3y}{4}\)(1)  

                \(5y=6z\Rightarrow z=\frac{5y}{6}\)(2)

Thay (1) và (2) vào biểu thức \(x^2+y^2+z^2=500\);ta được :

\(\left(\frac{3y}{4}\right)^2+y^2+\left(\frac{5y}{6}\right)^2=500\)

\(\Rightarrow\frac{9y^2}{16}+y^2+\frac{25y^2}{36}=500\Rightarrow324y^2+576y^2+400y^2=288000\)

\(\Rightarrow1300y^2=288000\Rightarrow y^2=\frac{2880}{13}\Rightarrow\orbr{\begin{cases}y=\frac{24\sqrt{65}}{13}\\y=-\frac{24\sqrt{65}}{13}\end{cases}}\)

Với \(y=\frac{24\sqrt{65}}{13}\Rightarrow x=\frac{3\cdot\frac{24\sqrt{65}}{13}}{4}=\frac{18\sqrt{65}}{13};z=\frac{5\cdot\frac{24\sqrt{65}}{13}}{6}\)

     \(y=-\frac{24\sqrt{65}}{13}\Rightarrow x=-\frac{18\sqrt{65}}{13};z=\frac{5\cdot-\frac{24\sqrt{65}}{13}}{6}\)

Đặt \(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}=k\)=>\(\hept{\begin{cases}x=5k+1\\y=3k+2\\z=2k+2\end{cases}}\)

Có \(3x^2-5y^2-6z^2=43\)<=>\(3\left(5k+1\right)^2-5\left(3k+2\right)^2-6\left(2k+2\right)^2=43\)

\(\Leftrightarrow3\left(25k^2+10k+1\right)-5\left(9k^2+12k+4\right)-6\left(4k^2+8k+4\right)=43\)

\(\Leftrightarrow75k^2+30k+3-45k^2-60k-20-24k^2-48k-24=43\)

\(\Leftrightarrow6k^2-78k-41=43\)\(\Leftrightarrow6k^2-78-84=0\)\(\Leftrightarrow6\left(k-14\right)\left(k+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}k-14=0\\k+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}k=14\\k=-1\end{cases}}\)

+) Với k=14 thì: x=14.5+1=71;y=14.3+2=44;z=14.2+2=30

+) Với k=-1 thì: x=(-1).5+1=-4;y=(-1).3+2=-1;z=(-1).2+2=0

Vậy .....................

Đặt \(\frac{x}{-5}=\frac{y}{6}=\frac{z}{-2}=k\)  \(\left(k\ne0\right)\)

\(\Rightarrow x=-5k;y=6k;z=-2k\)

\(\Rightarrow A=\frac{3.k.\left(-5\right)+6.k-2.\left(-2\right).k}{-3.\left(-5\right).k-5.6.k+6.\left(-2\right).k}=\frac{-15k+6k+4k}{15k-30k-12k}=\frac{-5k}{-27k}=\frac{5}{27}\)

Vậy \(A=\frac{5}{27}\).

\(\frac{x-1}{5}=\frac{y-2}{3}=\frac{z-2}{2}\)\(\Leftrightarrow\frac{3\left(x-1\right)}{15}=\frac{5\left(y-2\right)}{15}=\frac{6\left(z-2\right)}{12}\)

\(\Leftrightarrow\frac{3x-3}{15}=\frac{5y-10}{15}=\frac{6z-12}{12}\).Áp dụng tc dãy tỉ số "=" nhau ta có:

\(\frac{3x-3}{15}=\frac{5y-10}{15}=\frac{6z-12}{12}=\frac{\left(3x-3\right)-\left(5y-10\right)+\left(6z-12\right)}{15-15+12}=\frac{9-5}{12}=\frac{1}{3}\)

\(\Rightarrow\hept{\begin{cases}\frac{3x-3}{15}=\frac{1}{3}\Rightarrow x=\frac{8}{3}\\\frac{5y-10}{15}=\frac{1}{3}\Rightarrow y=3\\\frac{6z-12}{12}=\frac{1}{3}\Rightarrow z=\frac{8}{3}\end{cases}}\)

a) Ta có : \(\frac{x-1}{2}=\frac{y+3}{4}\Leftrightarrow\left(x-1\right).4=\left(y+3\right).2\Leftrightarrow4x-4=2y+6\Leftrightarrow4x-2y=10\Leftrightarrow x=\frac{10+2y}{4}\left(1\right)\)

 \(\frac{y+3}{4}=\frac{z-5}{6}\Leftrightarrow\left(y+3\right).6=\left(z-5\right).4\Leftrightarrow6y+18=4z-20\Leftrightarrow6y-4z=-38\Rightarrow z=\frac{6y+38}{4}\left(2\right)\)Thay (1) và (2) vào biểu thức \(5x-3y-4z=20\); ta được : 

\(\frac{5.\left(10+2y\right)}{4}-3y-\frac{4.\left(6y+38\right)}{4}=20\)

\(\Leftrightarrow50+10y-12y-24y-152=80\)

\(\Leftrightarrow-26y=182\Rightarrow y=-7\)

Với \(y=-7\Rightarrow x=\frac{10+2.-7}{4}=-1;z=\frac{6.-7+38}{4}=-1\)

Vậy .... 

mk ko bt 

bạn cute quá ; 

tặng bạn , tk mk nhé ; 

@Phạm thu phương sao lại cute ạ ><

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............................................................................................

\(\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+\frac{3}{5}\right|\)

\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+0,6\right|\)

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

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

\(\Rightarrow\left|x-\frac{1}{2}\right|=1-\frac{3}{4}\)

\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{1}{4}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{4}\\x-\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{4}\end{cases}}}\)

Vậy ...

\(1)\) Ta có : 

\(3x=4y\)\(\Leftrightarrow\)\(\frac{x}{4}=\frac{y}{3}\)\(\Leftrightarrow\)\(\frac{x}{8}=\frac{y}{6}\)

\(5y=6z\)\(\Leftrightarrow\)\(\frac{y}{6}=\frac{z}{5}\)

\(\Rightarrow\)\(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}\)

Đặt \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=k\)\(\Rightarrow\)\(\hept{\begin{cases}x=8k\\y=6k\\z=5k\end{cases}}\) \(\left(1\right)\)

Thay \(\left(1\right)\) vào \(xyz=30\) ta được : 

\(8k.6k.5k=30\)

\(\Leftrightarrow\)\(240k^3=30\)

\(\Leftrightarrow\)\(k^3=\frac{30}{240}\)

\(\Leftrightarrow\)\(k^3=\frac{1}{8}\)

\(\Leftrightarrow\)\(k^3=\left(\frac{1}{2}\right)^3\)

\(\Leftrightarrow\)\(k=\frac{1}{2}\)

Suy ra : 

\(x=8k=8.\frac{1}{2}=\frac{8}{2}=4\)

\(y=6k=6.\frac{1}{2}=\frac{6}{2}=3\)

\(z=5k=5.\frac{1}{2}=\frac{5}{2}\)

Vậy \(x=4\)\(;\)\(y=3\) và \(z=\frac{5}{2}\)

Chúc bạn học tốt ~ 

Ta có : 

\(3x=4y\Rightarrow\frac{x}{4}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{6}\left(1\right)\)

\(5y=6z\Rightarrow\frac{y}{6}=\frac{z}{5}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\)

\(\Rightarrow\frac{x}{8}=\frac{y}{6}=\frac{z}{5}\)

Đặt \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=k\left(k\ne0\right)\)

\(\Rightarrow x=8k;y=6k;z=5k\)

\(\Rightarrow xyz=8k.6k.5k\)

\(\Rightarrow xyz=240k^3\)

Do \(xyz=30\)

\(\Rightarrow240k^3=30\)

\(\Rightarrow k^3=\frac{1}{8}\)

\(\Rightarrow k=\frac{1}{2}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{2}.8=4\\y=\frac{1}{2}.6=3\\z=\frac{1}{2}.5=2,5\end{cases}}\)

Vậy \(x=4;y=3;z=2,5\)

Áp dụng tính chất dãy tỉ số bằng nhau: 

\(\frac{x+3}{5}=\frac{y-2}{3}=\frac{z-1}{7}=\frac{3x+9}{15}=\frac{5y-10}{15}=\frac{7z-7}{49}=\frac{3x+9-5y+10+7z-7}{15-15+49}=\frac{86+12}{49}=\frac{98}{49}=2\)

=>x=2.5-3=7;y=2.3+2=8;z=2.7+1=15

cais đầu nhân cả tử và mẫu với 3, cái thứ 2 nhân vs 5 ,cái thứ 3 nhân vs 7 sd DTSBN là xong

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