e, Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)

\(\Rightarrow\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)

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

\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{\left(x-2y+3z\right)+\left(-1+4-9\right)}{8}=\frac{14-6}{8}=1\)

Do đó: \(\frac{x-1}{2}=1\Rightarrow x=2.1+1=3\)

\(\frac{2y-4}{6}=1\Rightarrow y=\frac{6.1+4}{2}=5\)

\(\frac{3z-9}{12}=1\Rightarrow z=\frac{12.1+9}{3}=7\)

Vậy x=3; y=5; z=7

h, Ta có: \(\frac{x}{2}=\frac{y}{3}=\left(\frac{x}{2}\right)^2=\left(\frac{y}{3}\right)^2=\frac{x^2}{4}=\frac{y^2}{9}=\frac{x.y}{2.3}=\frac{54}{6}=9\)

Do đó: \(\frac{x^2}{4}=9\Rightarrow x^2=4.9=36\Rightarrow x=6;x=-6\)

\(\frac{y^2}{9}=9\Rightarrow y^2=9.9=81\Rightarrow y=9;y=-9\)

\(\frac{x}{12}=\frac{y}{9}=\frac{z}{8}\) => \(\frac{x-y}{12-9}=\frac{15}{3}=5\)=> x = 60 ; y= 45 ; z=40 => x-y-z = 60-45-40 =-25

\(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\Leftrightarrow\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}\)

Áp dụng t/c dãy tỉ số=nhau:

\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x-y}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)

=>x/2=30=>x=60

y/3/2=30=>y=30.3/2=45

z/4/3=30=>z=30.4/3=40

Vậy x-y-z=60-45-40=-25

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

\(\Rightarrow x=2k+1,y=3k+2,z=4k+3\)

Mà x-2y+3z=-10

Hay 2k+1-2(3k+2)+3(4k+3)=-10

2k+1-6k-4+12k+9=-10

(2k-6k+12k)+(1-4+9)=-10

8k+6=-10

8k=-16

k=-2

\(\Rightarrow x=-2\cdot2+1=-3,y=-2\cdot3+2=-4,z=-2\cdot4+3=-5\)

Mấy bài còn lại tương tự nhé cậu

 =>(x-1)/2=(-2y+4)/-6=(3z-9)/12 
=(x-1-2y+4+3z-9)/(2-6+12) 
=-16/8=-2 
=> (x-1)/2=-2<=>x-1=-4<=>x=-3 
=>(y-2)/3=-2<=>y-2=-6<=>y=-4 
=>(z-3)/4=-2<=>z-3=-8<=>z=-5

Vậy x = -3 ; y = -4 ; z = -5

a)

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

\(!x+2y!=5\Rightarrow\orbr{\begin{cases}x+2y=5\\x+2y=-5\end{cases}\Rightarrow\orbr{\begin{cases}x+2.\frac{2}{3}x=5\Rightarrow x=\frac{15}{7}\\x+2.\frac{2}{3}x=-5\Rightarrow x=-\frac{15}{7}\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}y=\frac{10}{7}\\y=\frac{-10}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}z=\frac{6}{7}\\z=\frac{6}{7}\end{cases}}\)

(x,y,z)=(15/7,10/7,6/7)

(x,y,z)=(-15/7,-10/7,-6/7)

Ta có: \(\frac{x-1+1}{2+1}=\frac{y-2+2}{3+2}=\frac{z-3+3}{4+3}=\frac{x}{3}=\frac{y}{5}=\frac{z}{7}=\frac{x}{3}=\frac{2y}{10}=\frac{3z}{21}=\frac{-10}{14}=\frac{-5}{7}\)

\(\Rightarrow\frac{x}{3}=\frac{-5}{7}\Rightarrow x=\frac{-15}{7};\frac{y}{5}=\frac{-5}{7}\Rightarrow y=\frac{-25}{7};\frac{z}{7}=\frac{-5}{7}\Rightarrow z=-5\)

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

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

\(\frac{x-1}{2}\)= \(\frac{2y-4}{6}\)= \(\frac{3z-9}{12}\)= \(\frac{x-1-2y-4+3z-9}{2-6+12}\)=\(\frac{\left(-10\right)-6}{8}\)=\(\frac{-16}{8}\)= -2

-> \(\frac{x-1}{2}\)= - 2 => x = -3 (1)

-> \(\frac{y-2}{3}\)= - 2 => y = -7 (2)

-> \(\frac{z-3}{4}\)= - 2 => z = -5 (3)

Từ (1), (2) và (3) suy ra: x + y + z = (-3) + (-7) + (-5) = - 15

-19

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