a) \(4x-2=x\)

\(4x-x=2\)

\(3x=2\)

\(x=\dfrac{2}{3}\)

b) Thay \(x=1,y=3\) ta có \(3=a.1\Rightarrow a=3\)

Vậy hàm số cần tìm là \(y=3x\)

c) Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{x+y+z}{1+2+3}=\dfrac{180}{6}=30\)

\(\Rightarrow\left\{{}\begin{matrix}x=30\times1=30\\y=30\times2=60\\z=30\times3=90\end{matrix}\right.\)

x/2 = y/3 = z/6 = 180

=> x = 180 : 2 = 90

     y = 180 : 3 = 60

     z = 180 : 6 = 30

~~~ HT ~~~

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 đề bài, ta có:

\(\frac{x}{1}=\frac{y}{4};3z=4y\) (-) \(\frac{z}{4}=\frac{y}{3}\)

Vậy ta có: \(\frac{x}{3}=\frac{y}{12}=\frac{z}{16}\)

(=) \(\frac{4x}{12}=\frac{y}{12}=\frac{z}{16}\)

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

\(\frac{4x}{12}+\frac{y}{12}+\frac{z}{16}=\frac{4x+y+z}{12+12+16}=\frac{8}{40}=0,2\)

=> x=0,2.3= 0,6

     y=0,2.12= 2,4

    z= 0,2.16= 3,2

Vậy x=0,6

       y=2,4

       z=3,2

<333

Cái dấu <33 này là tym thôi bạn ;v đừng để ý nó

Ta có \(\frac{x}{1}=\frac{y}{4}\)(1) ; \(\frac{y}{z}=\frac{3}{4}\)(2).

(2) => \(\frac{y}{3}=\frac{z}{4}\)(3)

Nhân cả hai vế của (3) cho \(\frac{1}{4}\), ta có: \(\frac{y}{12}=\frac{z}{16}\)

Nhân cả hai vế của (1) cho \(\frac{1}{3}\), ta có \(\frac{x}{3}=\frac{y}{12}\)

=> \(\frac{x}{3}=\frac{y}{12}=\frac{z}{16}\)=> \(\frac{4x}{12}=\frac{y}{12}=\frac{z}{16}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{4x}{12}=\frac{y}{12}=\frac{z}{16}=\frac{4x+y-z}{12+12-16}=\frac{8}{8}=1\)

=> \(\hept{\begin{cases}\frac{4x}{12}=1\\\frac{y}{12}=1\\\frac{z}{16}=1\end{cases}\Rightarrow\hept{\begin{cases}4x=12\\y=12\\z=16\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=12\\z=16\end{cases}}}\)

Vậy x = 3; y = 12; z = 16.

bài 2 :

ta có x:y:z=3:5:(-2)

=>x/3=y/5=z/-2

=>5x/15=y/5=3z/-6

áp dụng tc dãy ... ta có :

5x/15=y/5=3z/-6=5x-y+3z/15-5+(-6)=-16/4=-4

=>x/3=-=>x=-12

=>y/5=-4=>y=-20

=>z/-2=-4=>z=8

Theo đề được:

 \(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{4}}\) và x.y.z=180

=> \(\left(\frac{x}{\frac{1}{2}}\right)^3=\left(\frac{y}{\frac{1}{3}}\right)^3=\left(\frac{z}{\frac{1}{4}}\right)^3=\frac{x.y.z}{\frac{1}{2}.\frac{1}{3}.\frac{1}{4}}=\frac{180}{\frac{1}{24}}=4320\)

Vậy \(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{4}}=\sqrt[3]{4320}\)

=> Không tìm được x,y,z