a) Ta có : \(\frac{x}{y}=\frac{2}{3}\) => \(\frac{x}{2}=\frac{y}{3}\) => \(\frac{2x}{4}=\frac{3y}{9}\)

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

  \(\frac{2x}{4}=\frac{3y}{9}=\frac{2x+3y}{4+9}=\frac{208}{13}=16\)

=> \(\hept{\begin{cases}\frac{x}{2}=16\\\frac{y}{3}=16\end{cases}}\) => \(\hept{\begin{cases}x=16.2=32\\y=16.3=48\end{cases}}\)

Vậy ...

b) \(\frac{3}{x}=\frac{4}{y}\) => \(\frac{x}{3}=\frac{y}{4}\)=> \(\frac{-3x}{-9}=\frac{5y}{20}\)

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

        \(\frac{-3x}{-9}=\frac{5y}{20}=\frac{-3x+5y}{-9+20}=\frac{33}{11}=3\)

=> \(\hept{\begin{cases}\frac{x}{3}=3\\\frac{y}{4}=3\end{cases}}\) => \(\hept{\begin{cases}x=3.3=9\\y=3.4=12\end{cases}}\)

Vậy ...

a) \(\text{Ta có : }\frac{x}{y}=\frac{2}{3}\Leftrightarrow\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{2x}{4}=\frac{3y}{9}\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có :}\frac{2x}{4}=\frac{3y}{9}=\frac{2x+3y}{4+9}=\frac{208}{13}=16\)

\(\Rightarrow\frac{2x}{4}=16\Rightarrow2x=64\Rightarrow x=32\)

\(\Rightarrow\frac{3y}{9}=16\Rightarrow3y=144\Rightarrow y=48\)

\(\text{Vậy }x=32;y=48\)

b) \(\text{Ta có : }\frac{3}{x}=\frac{4}{y}\Leftrightarrow\frac{y}{4}=\frac{x}{3}\Leftrightarrow\frac{5x}{20}=-\frac{3x}{-9}\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có : }\frac{5x}{20}=\frac{-3x}{-9}=\frac{5y+\left(-3x\right)}{20+\left(-9\right)}=\frac{33}{11}=3\)

\(\text{Nếu }\frac{-3x}{-9}=3\Rightarrow-3x=-27\Rightarrow x=9\)

\(\text{Nếu}\frac{5y}{20}=3\Rightarrow5y=60\Rightarrow y=12\)

\(\text{Vậy}x=9;y=12\)

c) \(\text{Ta có : }8x=5y\Rightarrow\frac{x}{5}=\frac{y}{8}\Leftrightarrow\frac{2x}{10}=\frac{y}{8}\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có :}\frac{2x}{10}=\frac{y}{8}=\frac{y-2x}{10-8}=\frac{-10}{2}=-5\)

\(\text{Nếu }\frac{2x}{10}=-5\Rightarrow2x=-50\Rightarrow x=-25\)

\(\text{Nếu }\frac{y}{8}=-5\Rightarrow y=-40\)

\(\text{Vậy}x=-25;y=-40\)

mk cung hoc lop 7 nhung cai bai do ma ko lam dc thi chet di

Ta có:2x+y=z−38⇒2x+y−z=−382x+y=z−38⇒2x+y−z=−38

Vì 3x=4y=5x−3x−4y3x=4y=5x−3x−4y nên 3x=5z−3x−3x3x=5z−3x−3x

⇒3x−5z−6x⇒3x−5z−6x

⇒9x=5z⇒9x=5z

⇒x5=z9⇒x20=z36⇒x5=z9⇒x20=z36(1)

Vì 3x=4y⇒x4=y3⇒x20=z153x=4y⇒x4=y3⇒x20=z15 (2)

Từ (1) và (2)⇒x20=y15=z36⇒x20=y15=z36

Áp dụng tính chất dãy tỉ số bằng nhau:

x20=y15=z36=2x+y−z2.20+15−36=−3819=−2x20=y15=z36=2x+y−z2.20+15−36=−3819=−2

x20=−2⇒x=20.(−2)=−40x20=−2⇒x=20.(−2)=−40

y15=−2⇒y=15.(−2)=−30y15=−2⇒y=15.(−2)=−30

z36=−2⇒z=36.(−2)=−72z36=−2⇒z=36.(−2)=−72

Vậy x=−40;y=−30;z=−72

a )  

Ta có : 

\(\hept{\begin{cases}\frac{x}{5}=\frac{y}{6}\\\frac{y}{8}=\frac{z}{7}\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{20}=\frac{y}{24}\\\frac{y}{24}=\frac{z}{21}\end{cases}}}\)

và \(x+y-z=69\)

ADTCDTSBN , ta có : 

\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{20}=3\\\frac{y}{24}=3\\\frac{z}{21}=3\end{cases}\Rightarrow\hept{\begin{cases}x=3.20=60\\y=3.24=72\\z=3.21=63\end{cases}}}\)

Vậy ...

b )  

Ta có : 

\(5y=72\Rightarrow y=\frac{72}{5}=14,4\)

\(\Rightarrow x=14,4.3:2=21,6\)

và \(3x+5y-7z=30\)

Thay vào làm tiếp : 

c ) 

\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\)

\(=\frac{3\left(x-1\right)}{6}=\frac{4\left(y+3\right)}{16}=\frac{5\left(z-5\right)}{30}\)

\(=\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}\)

\(=\frac{5z-25-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\)( ADTCDTSBN ) 

\(=\frac{5z-25-3x+3-4y-12}{8}=\frac{5z-3x-4y-34}{8}\)

\(=\frac{50-34}{8}=\frac{16}{8}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x-1}{2}=2\\\frac{y+3}{4}=2\\\frac{z-5}{6}=2\end{cases}\Rightarrow\hept{\begin{cases}x-1=2.2=4\\y+3=2.4=8\\z-5=2.6=12\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\y=5\\z=17\end{cases}}}\)

Vậy ...

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405

Ta có: 2x + 3y + 5z - 119 = 0

=>  2x + 3y + 5z = 119

 \(\frac{x+2}{3}=\frac{y+3}{5}=\frac{z-4}{7}\Leftrightarrow\frac{2x+4}{6}=\frac{3y+9}{15}=\frac{5z-20}{35}\)

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

\(\frac{2x+4}{6}=\frac{3y+9}{15}=\frac{5z-20}{35}=\frac{2x+4+3y+9+5z-20}{6+15+35}=\frac{119+4+9-20}{56}=\frac{112}{56}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x+2}{3}=2\\\frac{y+3}{5}=2\\\frac{z-4}{7}=2\end{cases}\Rightarrow}\hept{\begin{cases}x+2=6\\y+3=10\\z-4=14\end{cases}}\Rightarrow\hept{\begin{cases}x=4\\y=7\\z=18\end{cases}}\)

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:

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

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