cách giải chi tiết nè bạn j đó ơi
ta có: x/2=y/3;y/4=z/5 và x+y-z=10
x/2=y/3=>x/8=y/12 1
y/4=z/5=>y/12=z/15 2
Từ 1, 2=> x/8=y/12=z/15
Áp dụng tính chất dãy tỉ số bằng nhau, ta có
x/8=y/12=z/15=x+y-z/8+12-15=10/5=2
Ta có:
x/8=2=>x=2.8=16
y/12=2=.=>y=2.12=24
z/15=2=>z=2.15=30
Vậy x=16;y=24;z=30
(Bài này mình chắc đúng luôn)

Ta có : \(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}vax+y-z=10\)0

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\left[1\right]\)

\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\left[2\right]\)

\(Tu1va2\Rightarrow:\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

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

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

\(\Leftrightarrow\frac{x}{8}=2\Rightarrow x=16;\frac{y}{12}=2\Rightarrow y=24\)

\(\frac{z}{15}=2\Rightarrow z=30.Vayx=16;y=24;z=30\)

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

\(\frac{y-2}{3}=\frac{3\left(y-2\right)}{3}=\frac{3y-6}{3}\)

\(\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-\left(z-3\right)}{4+9-4}\)

\(=\frac{2x+3y-z+3-2-6}{9}=\frac{50-5}{9}=\frac{45}{9}=5\)

=>x-1=5.2=10

=>x=11

y-2=5.3=15

=>y=17

z-3=5.4=20

=>z=23

vậy (x;y;z)=(11;17;23)

đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)

=>\(x=2x+1;y=3k+2;z=4k+3\)

thay \(x=2k+1;y=3k+2;z=4k+3\)vào 2x+3y-z=50 ta được:

2(2k+1)+3(3k+2)-(4k+3)=50

<=>4k+2+9k+6-4k-3=50

<=>9k+5=50

<=>9k=45

<=>k=5

=>x=2.5+1=11

y=3.5+2=17

z=4.5+3=23

x=3;y=5;z=7

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\Rightarrow x=2k+1;y=3k+2;z=4k+3\)

ta có: 2(2k+1)+3(3k+2)-(4k+3)=50

4k+2+9k+6-4k-3=50

9k+5=50

9k=50-5=45

k=45:9=5

x=5x2+1=11

y=5x3+2=17

z=5x4+3=23

a, 3x-(2x+1)\(=2\)

\(\Leftrightarrow3x-2x-1=2\)

\(\Leftrightarrow3x-2x=2+1\)

\(\Leftrightarrow x=3\)

a) \(3x-\left(2x+1\right)=2\)

\(3x-2x-1=2\)

\(x-1=2\)

\(x=3\)

vay \(x=3\)

Với các bài khá nâng cao như vậy bạn đăng tách ra nhé!

Answer:

a) Ta có: \(x:y:z=3:4:5\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

Ta đặt: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(k\inℕ^∗\right)\)

\(\Rightarrow\hept{\begin{cases}x=3k\\y=4k\\z=5k\end{cases}}\)

Ta có: \(5z^2-3x^2-2y^2=594\)

\(\Rightarrow5.\left(5k\right)^2-3.\left(3k\right)^2-2.\left(4k\right)^2=594\)

\(\Rightarrow5.5^2k^2-3.3^2k^2-2.4^2k^2=594\)

\(\Rightarrow5.25k^2-3.9k^2-2.16.k^2=594\)

\(\Rightarrow125k^2-27k^2-32k^2=594\)

\(\Rightarrow k^2.\left(125-27-32\right)=594\)

\(\Rightarrow k^2.66=594\)

\(\Rightarrow k^2=9\)

\(\Rightarrow k=\pm3\)

Với \(k=3\Rightarrow\hept{\begin{cases}x=3.3=9\\y=3.4=12\\z=3.5=15\end{cases}}\)

Với \(k=-3\Rightarrow\hept{\begin{cases}x=\left(-3\right).3=-9\\y=\left(-4\right).3=-12\\z=\left(-5\right).3=-15\end{cases}}\)

Answer:

b) \(3.\left(x-1\right)=2.\left(y-2\right)\Rightarrow6.\left(x-1\right)=4.\left(y-2\right)\)

Mà: \(4.\left(y-2\right)=3.\left(z-3\right)\)

\(\Rightarrow6.\left(x-1\right)=4.\left(y-2\right)=3.\left(z-3\right)\)

\(\Rightarrow\frac{6.\left(x-1\right)}{12}=\frac{4.\left(y-2\right)}{12}=\frac{3.\left(z-3\right)}{12}\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}==\frac{\left(2x-2\right)+\left(3y-6\right)-z}{4+9-4}=\frac{2x-2+3y-6-z}{9}=\frac{\left(2x+3y-z\right)-\left(2+6\right)}{9}=\frac{50-8}{9}=\frac{14}{3}\)

\(\Rightarrow\hept{\begin{cases}x-1=2.\frac{14}{3}=\frac{28}{3}\\y-2=3.\frac{14}{3}=14\\z-3=4.\frac{14}{3}=\frac{56}{3}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{31}{3}\\y=16\\z=\frac{68}{3}\end{cases}}\)

c) \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}\)

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y-z}{18+16-15}=\frac{38}{19}=2\)

\(\Rightarrow\frac{x}{18}=2\Rightarrow x=18.2=36\)

\(\Rightarrow\frac{y}{16}=2\Rightarrow y=16.2=32\)

\(\Rightarrow\frac{z}{15}=2\Rightarrow z=15.2=30\)