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

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

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

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

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

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

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

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 \(x=4;y=3;z=2,5\)

Tk mình đi mọi người mình bị âm nè!

Ai tk mình mình tk lại cho

Ta có : \(3x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{24}=\dfrac{y}{18}\left(1\right)\)

\(5y=6z\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\Rightarrow\dfrac{y}{18}=\dfrac{z}{15}\left(2\right)\)

Từ (1);(2) \(\Rightarrow\dfrac{x}{24}=\dfrac{y}{18}=\dfrac{z}{15}\)

Đặt \(\dfrac{x}{24}=\dfrac{y}{18}=\dfrac{z}{15}=k\Rightarrow x=24k;y=18k;z=15k\)

\(\text{Ta có }:x.y.z=24k.18k.15k=30\\ \Rightarrow k^3.6480=30\\ \Rightarrow k^3=\dfrac{1}{216}\\ \Rightarrow k=\dfrac{1}{6}\\ \Rightarrow x=24.\dfrac{1}{6}=4\\ y=18.\dfrac{1}{6}=3\\ z=15.\dfrac{1}{6}=2.5\)

Vậy x = 4 ; y = 3 ; z = 2,5

a) Tìm các số x, y, z biết rằng

3x=4y ; 5y=6z và xyz = 30

Giải

Ta có: 3x = 4y; 5y = 6z \(\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\); \(\dfrac{y}{6}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\) \(\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)

Có \(xyz=30\) \(\Leftrightarrow\) \(8k.6k.5k=30\)

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

\(\Rightarrow\) \(k^3=8\)

\(\Rightarrow\) \(k=\sqrt[3]{8}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=8.2=16\\y=6.2=12\\z=5.2=10\end{matrix}\right.\)

Ta có: 3x = 4y => \(\frac{x}{4}=\frac{y}{3}\) => \(\frac{x}{8}=\frac{y}{6}\)

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

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

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

       \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=\frac{x+y-z}{8+6-5}=\frac{18}{9}=2\)

=> \(\hept{\begin{cases}\frac{x}{8}=2\\\frac{y}{6}=2\\\frac{z}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.8=16\\y=2.6=12\\z=2.5=10\end{cases}}\)

Vậy ....

ta có :3x=4y,5y=6z

=>\(\dfrac{x}{4}\)=\(\dfrac{y}{3}\); \(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)

=> \(\dfrac{x}{8}\)=\(\dfrac{y}{6}\); \(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)

=> \(\dfrac{x}{8}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)

Đặt \(\dfrac{x}{8}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{5}\)=k

=> x=8k ; y=6k ; z=5k

=> 8k.6k.5k=30

=> 240k³ =30

=>k³ =8

=>k=2

=> x=8.2=16 ; y=6.2=12 ; x =5.2=10

⚡

Câu hỏi của Sương Đặng - Toán lớp 7 | Học trực tuyến - Hoc24.vn

\(3x=4y;5y=6z\)

\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{3};\dfrac{y}{6}=\dfrac{z}{5}\)

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

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

Đặt:

\(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)

\(\Rightarrow8k.6k.5k=30\)

\(\Rightarrow240k^3=30\)

\(k^3=\dfrac{1}{8}\)

\(k=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}.8=4\\y=\dfrac{1}{2}.6=3\\z=\dfrac{1}{2}.5=2,5\end{matrix}\right.\)

\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)

\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)

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

\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)

\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)

mọi người giúp mk câu b, c, d còn lại nha