bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 =>  x-1/3=y-2/4=z-3/5 

áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1

do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự

Bài 1: 

a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )

cũng dễ thôi

dùng dãy tiwr số bằng nhau í dễ thế mà cũng hỏi

dễ thì làm tử đi

đúng rùi đó ! Dễ mà bạn

a) ADTCDTSBN

có: \(\frac{x}{2}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3.\)

=> x/2 = 3 => x = 6

y/3 = 3 => y = 9

z/4 = 3 => z = 12

KL:...

b,c làm tương tự nha

d) ta có: \(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{2x}{10}\)

ADTCDTSBN

có: \(\frac{2x}{10}=\frac{y}{-6}=\frac{z}{7}=\frac{2x+y-z}{10+\left(-6\right)-7}=\frac{49}{-3}\)

=>...

e) ADTCDTSBN

có: \(\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{4}=\frac{x+1+y+2+z+3}{2+3+4}=\frac{\left(x+y+z\right)+\left(1+2+3\right)}{9}\)

\(=\frac{21+6}{9}=\frac{27}{9}=3\)

=>...

g) ta có: \(\frac{x}{4}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=4k\\y=3k\end{cases}}\)

mà xy = 12 => 4k.3k = 12

                          12.k² = 12

                              k² = 1

                        => k = 1 hoặc k = -1

=> x = 4.1 = 4

y = 3.1 = 3

x=4.(-1) = -4 

y=3.(-1) = -3

KL:...

h) ta có: \(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\)

ADTCDTSBN

có: \(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{16}{16}=1\)

=>...

\(\left(x-15\right)\left(y+12\right)\left(z-3\right)=0\)

=>\(\left[{}\begin{matrix}x-15=0\\y+12=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\y=-12\\z=3\end{matrix}\right.\)

TH1: x=15

x+1=y+2=z+3

=>y+2=z+3=15+1=16

=>y=16-2=14;z=16-3=13

TH2: y=-12

x+1=y+2=z+3

=>x+1=z+3=-12+2=-10

=>x=-10-1=-11; z=-10-3=-13

TH3: z=3

x+1=y+2=z+3

=>x+1=y+2=3+3=6

=>x=6-1=5; y=6-2=4

\(\dfrac{4}{x+1}=\dfrac{2}{y-2}=\dfrac{3}{z+2}\)

=>\(\dfrac{x+1}{4}=\dfrac{y-2}{2}=\dfrac{z+2}{3}=k\)

=>x+1=4k; y-2=2k; z+2=3k

=>x=4k-1; y=2k+2; z=3k-2

xyz=12

=>(4k-1)(2k+2)(3k-2)=12

=>(4k-1)(k+1)(3k-2)=6

=>(4k-1)(3k^2-2k+3k-2)=6

=>(3k^2+k-2)(4k-1)=6

=>12k^3-3k^2+4k^2-k-8k+2-6=0

=>12k^3+k^2-9k-7=0

=>

\(\dfrac{4}{x+1}=\dfrac{2}{y-2}=\dfrac{3}{z+2}\)

=>\(\dfrac{x+1}{4}=\dfrac{y-2}{2}=\dfrac{z+2}{3}=k\)

=>x+1=4k; y-2=2k; z+2=3k

=>x=4k-1; y=2k+2; z=3k-2

xyz=12

=>(4k-1)(2k+2)(3k-2)=12

=>(4k-1)(k+1)(3k-2)=6

=>(4k-1)(3k^2-2k+3k-2)=6

=>(3k^2+k-2)(4k-1)=6

=>12k^3-3k^2+4k^2-k-8k+2-6=0

=>12k^3+k^2-9k-4=0

=>k=1

=>x=4k-1=3; y=2k+2=4; z=3k-2=3-2=1