1) Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).3=-3\\y=\left(-1\right).5=-5\\z=\left(-1\right).7=-7\end{matrix}\right.\)

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

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{19}.8=-\dfrac{224}{19}\\y=-\dfrac{28}{19}.12=-\dfrac{336}{19}\\z=-\dfrac{28}{19}.15=-\dfrac{420}{19}\end{matrix}\right.\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{3\cdot2+5\cdot3-7}=\dfrac{-14}{14}=-1\\ \Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)

b, \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{224}{19}\\y=-\dfrac{336}{19}\\z=-\dfrac{420}{19}\end{matrix}\right.\)

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

⇒\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}\)

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

\(\dfrac{2x}{10}=\dfrac{3y}{15}=\dfrac{z}{7}=\dfrac{2x+3y-z}{10+15-7}=\dfrac{-14}{18}=\dfrac{-7}{9}\)

⇒\(\left\{{}\begin{matrix}x=\dfrac{-7}{9}.3=\dfrac{-7}{3}\\y=\dfrac{-7}{9}.5=\dfrac{-35}{9}\\z=\dfrac{-7}{9}.7=\dfrac{-49}{9}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{8}{3}}=\dfrac{x-y+z}{3-\dfrac{5}{2}+\dfrac{8}{3}}=\dfrac{95}{\dfrac{19}{6}}=30\\ \Rightarrow\left\{{}\begin{matrix}x=90\\y=30\cdot\dfrac{5}{2}=75\\z=30\cdot\dfrac{8}{3}=80\end{matrix}\right.\)

\(\dfrac{x}{2}=\dfrac{y}{3}\) ⇒ \(\dfrac{x}{8}=\dfrac{y}{12}\) (1)

\(\dfrac{y}{4}=\dfrac{z}{5}\) ⇒ \(\dfrac{y}{12}=\dfrac{z}{15}\) (2)

Từ (1) và (2) ⇒ \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

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

⇒ \(\left\{{}\begin{matrix}\dfrac{x}{8}=2\\\dfrac{y}{12}=2\\\dfrac{z}{15}=2\end{matrix}\right.\) ⇒\(\left\{{}\begin{matrix}x=16\\y=24\\z=30\end{matrix}\right.\)

Ta có \(\dfrac{x}{2}=\dfrac{y}{3}\) => \(\dfrac{1}{4}\cdot\dfrac{x}{2}=\dfrac{1}{4}\cdot\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\left(1\right)\)

\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{1}{3}\cdot\dfrac{y}{4}=\dfrac{1}{3}\cdot\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\left(2\right)\)

Từ ( 1 ) và ( 2 ) ta có

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\) và x+y-z=10

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

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

\(\Rightarrow\dfrac{x}{8}=2\Rightarrow x=2\cdot8=16\)

\(\dfrac{y}{12}=2\Rightarrow=2\cdot12=24\)

\(\dfrac{z}{15}=2\Rightarrow z=2\cdot15=30\)

vậy x = 16; y = 24; z = 30

Chúc bn học tốt

\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}vàx+y-z=10\)

Ta có :

\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{4}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{12}\\\dfrac{y}{12}=\dfrac{z}{5}\end{matrix}\right.\) => \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\) và \(x+y-z=10\)

Từ : \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}\)

=> \(\dfrac{10}{5}=2\)

Với : \(\dfrac{x}{8}=2\Rightarrow x=16\)

Với : \(\dfrac{y}{12}=2\Rightarrow y=24\)

Với: \(\dfrac{z}{15}=2\Rightarrow z=30\)

Vậy x,y,z là : 16,24,30

a: 2x-3y-4z=24

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)

=>x=-6/7; y=-36/7; z=-18/7

b: 6x=10y=15z

=>x/10=y/6=z/4=k

=>x=10k; y=6k; z=4k

x+y-z=90

=>10k+6k-4k=90

=>12k=90

=>k=7,5

=>x=75; y=45; z=30

d: x/4=y/3

=>x/20=y/15

y/5=z/3

=>y/15=z/9

=>x/20=y/15=z/9

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)

=>x=500; y=375; z=225

e: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x+5}{2}=\dfrac{y-2}{3}=\dfrac{x-y+5+2}{2-3}=\dfrac{10+7}{-1}=-17\)

=>x+5=-34; y-2=-51

=>x=-39; y=-49

g: Áp dụng tính chất của DTSBN, ta được

\(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}=\dfrac{5a-3b-4c-5-9+20}{5\cdot2-3\cdot4-6\cdot4}=\dfrac{-253}{13}\)

=>a-1=-506/13; b+3=-1012/13; c-5=-1518/13

=>a=-493/13; b=-1051/13; c=-1453/13

Lời giải:
e. Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x+5}{2}=\frac{y-2}{3}=\frac{x+5-(y-2)}{2-3}=\frac{(x-y)+5+2}{2-3}=\frac{10+5+2}{-1}=-17$

Suy ra:

$x+5=2(-17)=-34\Rightarrow x=-39$

$y-2=3(-17)=-51\Rightarrow y=-49$

f. Đề thiếu. Bạn xem lại

h. Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}$

$=\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}$

$=\frac{5a-5-(3b+9)-(4c-20)}{10-12-24}$

$=\frac{5a-3b-4c-5-9+20}{-26}=\frac{500-5-9+20}{-26}=\frac{-253}{13}$

Suy ra:
$a-1=2.\frac{-253}{13}\Rightarrow a=\frac{-493}{13}$

$b+3=4.\frac{-253}{13}\Rightarrow b=\frac{-1051}{13}$

$c-5=6.\frac{-253}{13}\Rightarrow c=\frac{-1453}{13}$

Đặt : \(\dfrac{x}{5}=\dfrac{y}{3}=k\)

`=>x=5k,y=3k`

Ta có : \(x^2-y^2=4=>\left(5k\right)^2-\left(3k\right)^2=4\\ =>25k^2-9k^2=4\\ =>16k^2=4\\ =>k^2=\dfrac{1}{4}\\ =>k=\pm\dfrac{1}{2}\)

\(=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

a: \(A=\dfrac{6}{7}x^2y^2\cdot\dfrac{-7}{2}x^2y=-3x^4y^3\)

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

\(\dfrac{x}{-2}=\dfrac{y}{3}=\dfrac{x-y}{-2-3}=\dfrac{5}{-5}=-1\)

Do đó: x=2; y=-3

\(A=-3x^4y^3=-3\cdot2^4\cdot\left(-3\right)^3=3\cdot27\cdot16=81\cdot16=1296\)

\(A=\dfrac{6}{7}x^2y^2.\left(-3\dfrac{1}{2}x^2y\right)\)

\(=\dfrac{6}{7}x^2y^2.\left(-\dfrac{7}{2}\right)x^2y\)

\(=-3x^4y^3\)

b)Có: \(\dfrac{x}{y}=-\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{-y}{3}=\dfrac{x-y}{2+3}=\dfrac{5}{5}=1\)

\(\Rightarrow x=2;y=-3\)

Tại \(x=2;y=-3\) , giá trị của biểu thức là:

 \(-3.2^4.\left(-3\right)^3=-3.16.\left(-27\right)=1296\)

áp dụng dãy tỉ số = nhau ta có \(\dfrac{1+x}{2}=\dfrac{4-2y}{6}=\dfrac{4+z}{5}=\dfrac{x-2y+z+1+4+4}{2+6+5}=\dfrac{11}{13}\)

\(\dfrac{1+x}{2}=\dfrac{11}{13}\Leftrightarrow13\left(1+x\right)=22\Leftrightarrow13x+13=22\Leftrightarrow x=\dfrac{9}{13}\)

\(\dfrac{2-y}{3}=\dfrac{11}{13}\Leftrightarrow13\left(2-y\right)=33\Leftrightarrow-13y+26=33\Leftrightarrow y=-\dfrac{7}{13}\)

\(\dfrac{4+z}{5}=\dfrac{11}{13}\Leftrightarrow13\left(4+z\right)=55\Leftrightarrow13z+52=55\Leftrightarrow z=\dfrac{3}{13}\)

vậy..................

a) \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\) (1)

\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{z}{3}=\dfrac{y}{7}\) (2)

Từ (1) và (2) suy ra: \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

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

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y-z}{9-7-3}=\dfrac{-15}{-1}=15\)

\(\Rightarrow\left\{{}\begin{matrix}x=15.9\\y=15.7\\z=15.3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=135\\y=105\\z=45\end{matrix}\right.\)

Vậy, x = 135, y = 105, z = 45

b, \(\dfrac{x}{-3}=\dfrac{y}{-8}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{64}\)

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

\(\dfrac{x^2}{9}=\dfrac{y^2}{64}=\dfrac{x^2-y^2}{9-64}=-\dfrac{44}{\dfrac{5}{-55}}=-\dfrac{44}{5}:\left(-55\right)=-\dfrac{44}{5}.-\dfrac{1}{55}=\dfrac{44}{275}=0,16\)

+) \(\dfrac{x^2}{9}=0,16\Rightarrow x^2=1,44\Rightarrow x=\pm1,2\)

+) \(\dfrac{y^2}{64}=0,16\Rightarrow y^2=10,24\Rightarrow y=\pm3,2\)

Vậy ...