Question

bai 1 tim x,y,z

a,\(\frac{x}{10}\)=\(\frac{y}{15}\),x=\(\frac{7}{2}\)và x+2y-3z=20

b,2x=3y,49=57 và 4x-3y+5z=7

c,\(\frac{2x}{3}\)=\(\frac{3y}{4}\)=\(\frac{47}{5}\)và x+y+z=49

Answer 1

Đề hình như sai bn ạ

Answer 2

2.

a. \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

\(\Rightarrow7\left(x-3\right)=5\left(x+5\right)\)

\(\Rightarrow7x-21=5x+25\)

\(\Rightarrow7x-21-5x-25=0\)

\(\Rightarrow2x-46=0\)

\(\Rightarrow2x=46\Rightarrow x=23\)

Vậy x = 23

c. \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

\(\Rightarrow\left(x+4\right)\left(x+4\right)=20\cdot5\)

\(\Rightarrow\left(x+4\right)^2=100\)

\(\Rightarrow x+4=10\)

\(\Rightarrow x=6\)

Vậy x = 6

d. x = 4

Answer 3

Answer 4

Answer 5

Answer 6

Bài 5:

Đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)

\(\Rightarrow x=12k;y=9k;z=5k\)

\(\Rightarrow xyz=540k^3=20\Rightarrow k^3=\dfrac{1}{27}\Rightarrow k=\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{1}{3}\cdot12=4;y=\dfrac{1}{3}\cdot9=3;z=\dfrac{1}{3}\cdot5=\dfrac{5}{3}\)

Vậy \(x=4;y=3;z=\dfrac{5}{3}\)

Bài 6:

Đặt \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=k\).

\(\Rightarrow x=5k;y=7k;z=3k\)

\(\Rightarrow x^2+y^2-z^2=\left(5k\right)^2+\left(7k\right)^2+\left(3k\right)^2=585\)

\(\Rightarrow25k^2+49k^2-9k^2=585\)

\(\Rightarrow65k^2=585\Rightarrow k^2=9\Rightarrow k=\pm3\)

\(\Rightarrow\left\{{}\begin{matrix}k=3\Leftrightarrow x=3\cdot5=15;y=3\cdot7=21;z=3\cdot3=9\\k=-3\Leftrightarrow x=-3\cdot5=-15;y=-3\cdot7=-21;z=-3\cdot3=-9\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}k=3\Leftrightarrow x=15;y=21;z=9\\k=-3\Leftrightarrow x=-15;y=-21;z=-9\end{matrix}\right.\)

Answer 7

Bài 1:

b. Ta có:

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

\(4y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{10}=\dfrac{z}{8}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}\)

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

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{4x-3y+5z}{60-30+40}=\dfrac{7}{70}=\dfrac{1}{10}\)

\(\Rightarrow x=\dfrac{1}{10}\cdot15=\dfrac{3}{2};y=\dfrac{1}{10}\cdot10=1;z=\dfrac{1}{10}\cdot8=\dfrac{4}{5}\)

Vậy \(x=\dfrac{3}{2};y=1;z=\dfrac{4}{5}\)

c. Đặt \(\dfrac{2x}{3}=\dfrac{2y}{4}=\dfrac{4z}{5}=a\)

\(\Rightarrow2x=3a;2y=4a;4z=5a\)

\(\Rightarrow x=\dfrac{3a}{2};y=2a;z=\dfrac{5a}{4}\)

\(\Rightarrow x+y+z=\dfrac{3a}{2}+2a+\dfrac{5a}{4}=49\)

\(\Rightarrow\dfrac{6a}{4}+\dfrac{8a}{4}+\dfrac{5a}{4}=49\)

\(\Rightarrow\dfrac{19a}{4}=49\Rightarrow19a=49\cdot4=196\)

\(\Rightarrow a=\dfrac{196}{19}\)

\(\Rightarrow x=\dfrac{\dfrac{196}{19}\cdot3}{2}=\dfrac{294}{19};y=\dfrac{196}{19}\cdot2=\dfrac{392}{19};z=\dfrac{\dfrac{196}{19}\cdot5}{4}=\dfrac{245}{19}\)

Vậy \(x=\dfrac{294}{19};y=\dfrac{392}{19};z=\dfrac{245}{19}\)