Ta có :

2x=3y = \(\dfrac{x}{3}\) = \(\dfrac{y}{2}\) = \(\dfrac{x}{15}\) = \(\dfrac{y}{10}\)

3y=5z = \(\dfrac{y}{5}\) = \(\dfrac{z}{3}\) = \(\dfrac{y}{10}\) = \(\dfrac{z}{6}\)

=> \(\dfrac{x}{15}\) = \(\dfrac{y}{10}\) = \(\dfrac{z}{6}\) = \(\dfrac{x.y.z}{15.10.6}\) = \(\dfrac{36}{900}\)= \(\dfrac{1}{25}\)

=> x= \(\dfrac{1}{25}\) . 15 =\(\dfrac{3}{5}\)

y=\(\dfrac{1}{25}\) . 10 = \(\dfrac{2}{5}\)

z=\(\dfrac{1}{25}\).6 = \(\dfrac{6}{25}\)

Vậy ...

1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)

2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)

3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)

Áp dụng t/c dtsbn:

\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)

Theo bài ra ta cs 

\(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\)

\(4y=5z\Rightarrow\frac{y}{5}=\frac{z}{4}\)

T lại cs 

\(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{15}=\frac{y}{10}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{x}{10}=\frac{z}{8}\left(2\right)\)

Từ (1);(2) \(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)

ADTC dãy tỉ số bằng nhau ta cs 

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}=\frac{2x+3y-4z}{2.15+3.10-4.8}=\frac{56}{28}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{15}=2\\\frac{y}{10}=2\\\frac{z}{8}=2\end{cases}\Rightarrow\hept{\begin{cases}x=30\\y=20\\z=16\end{cases}}}\)

\(2x=3y;4y=5z\) => \(8x=12y;12y=15z\)

=>  \(\frac{8x}{120}=\frac{12y}{120}=\frac{15z}{120}\)=> \(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)

=>   \(\frac{2x}{30}=\frac{3y}{30}=\frac{4z}{32}\)

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

\(\frac{2x}{30}=\frac{3y}{30}=\frac{4z}{32}=\frac{2x+3y-4z}{30+30-32}=\frac{56}{28}\)

=> \(\frac{2x}{30}=2=>2x=60=>x=30\)

\(\frac{3y}{30}=2=>3y=60=>y=20\)

\(\frac{4z}{32}=2=>4z=64=>z=16\)

2x = 3y => \(\frac{x}{3}=\frac{y}{2}\) =>\(\frac{x}{15}=\frac{y}{10}\)

4y = 5z => \(\frac{y}{5}=\frac{z}{4}\) => \(\frac{y}{10}=\frac{z}{8}\)

=> x/15 = y/10 = z/8

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{8}\)=> \(\frac{2x}{30}=\frac{3y}{30}=\frac{4z}{32}=\frac{2x+3y-4z}{30+30-32}=\frac{56}{28}=2\)

=> \(\hept{\begin{cases}\frac{x}{15}=2\\\frac{y}{10}=2\\\frac{z}{8}=2\end{cases}}\) => \(\hept{\begin{cases}x=30\\y=20\\z=16\end{cases}}\)

Vậy ...

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}\Rightarrow\dfrac{2x+4}{6}=\dfrac{3y-15}{-12}=\dfrac{z+1}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x+4}{6}=\dfrac{3y-15}{-12}=\dfrac{z+1}{5}=\dfrac{2x+4-3y+15+z+1}{6-\left(-12\right)+5}=\dfrac{\left(2x-3y+z\right)+\left(4+15+1\right)}{23}=\dfrac{72+20}{23}=\dfrac{92}{23}=4\)

\(\dfrac{x+2}{3}=4\Rightarrow x+2=12\Rightarrow x=10\\ \dfrac{y-5}{-4}=4\Rightarrow y-5=-16\Rightarrow y=-11\\ \dfrac{z+1}{5}=4\Rightarrow z+1=20\Rightarrow z=19\)

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

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}=\dfrac{2x-3y+z+4+15+1}{2\cdot3-3\cdot\left(-4\right)+5}=\dfrac{92}{23}=4\)

Do đó: x=10; y=-11; z=4

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}\text{ và }2x-3y+z=72\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau:}\)

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}=\dfrac{2\left(x+2\right)-3\left(y-5\right)+z+1}{2.3-3.\left(-4\right)+5}=\dfrac{92}{23}=4\)

\(\Rightarrow\dfrac{x+2}{3}=4\Rightarrow x+2=12\Rightarrow x=10\)

\(\dfrac{y-5}{-4}=4\Rightarrow y-5=-16\Rightarrow y=-11\)

\(\dfrac{z+1}{5}=4\Rightarrow z+1=20\Rightarrow z=19\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{2x+3y-5z}{10+12-15}=\dfrac{2x-3y+5z}{10-12+15}\\ \Rightarrow A=\dfrac{10+12-15}{10-12+15}=\dfrac{7}{13}\)

A=7/13

Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{-4k-3.\left(-7k\right)-6.3k}=\dfrac{16k}{-1k}=-16\)

a, \(A+B=x^2-2x-y^2+3y-1+\left(-2x^2+3y^2-5z+3\right)\)

\(=x^2-2x-y^2+3y-1-2x^2+3y^2-5z+3\)

\(=-x^2-2x+2y^2+3y-5z+2\)

b, \(A-B=x^2-2x-y^2+3y-1-\left(-2x^2+3y^2-5z+3\right)\)

\(=x^2-2x-y^2+3y-1+2x^2-3y^2+5z-3\)

\(=3x^2-2x-4y^2+3y+5z-4\)

c, Thay x=-2,y=1 vào biểu thức A-B ta được:

\(A-B=3.\left(-2\right)^2-2.\left(-2\right)-4.1^2+3.1+5z-4=12+4-4+3+5z-4=11+5z\)

\(A=x^2-2x-y^2+3y-1\)

\(B=-2x^2+3y^2-5z+3\)

a)  A+B =

\(\left(x^2-2x-y^2+3y-1\right)+\left(-2x^2+3y^2-5z+3\right)\)

\(=\left(x^2-2x^2\right)-\left(y^2+3y^2\right)-2x+3y-5z-1+3\)

\(=-x^2-4y^2-2x+3y-5z-1+3\)

\(=\left(-1-4-2+3-5-1+3\right).\left(x^2-x\right).y^2.z\)

\(=-7xy^2z\)

b ) Tính A-B ( tương tự A+B )

C)  Thay x=-2 và y=1 vào biểu thức ta có :

\(-7xy^2z\)

\(=-7.-2.1.z\)

\(=14z\)

a) A+B = x² - 2x - y² + 3y - 1 - 2x² + 3y² - 5z + 3

           = -x² + 2y² - 2x + 3y - 5x + 2

b) A-B = x^2 - 2x - y^2 + 3y - 1 - ( -2x² + 3y² - 5z + 3)

          = x² - 2x - y^2 + 3y - 1 + 2x² - 3y² +5z - 3

          = 3x² - 4y² -2x + 3y + 5x - 3

c) Tự thay vào kết quả trên