Ta có : x - 24 = y

=> x - y = 24

Lại có : \(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{24}{4}=6\)

( theo tính chất của dãy tỉ số bằng nhau )

Nên \(\dfrac{x}{7}=6\) => x = 42

\(\dfrac{y}{3}=6\) => y = 18

Vậy x = 42, y = 18

Ta có :\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}=\dfrac{y-x}{7-5}=\dfrac{48}{2}=24\)

( theo tính chất dãy tỉ số bằng nhau )

Nên \(\dfrac{x}{5}=24\) => x = 120

\(\dfrac{y}{7}=24\) => y = 168

\(\dfrac{z}{2}=24\) => z = 48

Vậy x = 120, y = 168, z = 48

a, Ta có:

\(x-24=y\\ x-y=24\)

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

\(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{24}{4}=6\)

+) \(\dfrac{x}{7}=6\Rightarrow x=6\cdot7=42\)

+) \(\dfrac{y}{3}=6\Rightarrow6\cdot3=18\)

Vậy \(x=42;y=18\)

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

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}=\dfrac{y-z}{7-2}=\dfrac{48}{5}=9,6\)

+) \(\dfrac{x}{5}=9,6\Rightarrow x=9,6\cdot5=48\)

+) \(\dfrac{y}{7}=9,6\Rightarrow y=9,6\cdot7=67,2\)

+) \(\dfrac{z}{2}=9,6\Rightarrow z=9,6\cdot2=19,2\)

Vậy \(x=48;y=67,2;z=19,2\)

mk giải đc bao nhiêu thì bn làm bấy nhiêu nha

a)Vì \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)nên \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{x}{28}\).

Áp dụng t/c dãy tỉ số = nhau, ta có :

\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)

⇒2x = 3.30 = 90 ⇒ x = 45

3y = 3.60 = 180 ⇒ y = 60

z = 3.28 = 84

Ý b) có gì đó sai sai ?

c)Ta có :

\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Áp dụng t/c dãy tỉ số = nhau, ta có :

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)

⇒x = 5.15 = 75

y = 5.10 = 50

z = 5.6 = 30

d)Ta có :

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\left(k\in Z\right)\)

⇒ x = 2k ; y = 3k ; z = 5k

⇒ xyz = 2k.3k.5k = 30k³ = 810

b: 2x^3-1=15

=>2x^3=16

=>x=2

\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)

=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)

=>y-25=32; z+9=50

=>y=57; z=41

d: 3/5x=2/3y

=>9x=10y

=>x/10=y/9=k

=>x=10k; y=9k

x^2-y^2=38

=>100k^2-81k^2=38

=>19k^2=38

=>k^2=2

TH1: k=căn 2

=>\(x=10\sqrt{2};y=9\sqrt{2}\)

TH2: k=-căn 2

=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)

Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x

=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x

=> 2x+3y-1 / 12 = 2x+3y-1 / 6x

=> 12 = 6x => x =2

a, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)

\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)

\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)

Đến đây tự làm tiếp nhé

b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

=> x = 75, y = 50, z = 30

c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)

\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)

\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)

\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)

=> x=... , y=... , z=...

d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)

Ta có: xy = 90 => 2k.5k = 90 => 10k² = 90 => k² = 9 => k = 3 hoặc -3

Với k = 3 => x = 6, y = 15

Với k = -3 => x = -6, y = -15

Vậy...

e, Tương tự câu d

b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)

=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)

     \(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)

      \(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)

1. Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{6}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x+y-z}{6+5-3}=\dfrac{54}{8}=\dfrac{27}{4}\)

+\(\dfrac{x}{6}=\dfrac{27}{4}\Rightarrow x=\dfrac{27.6}{4}=\dfrac{81}{2}\)

+\(\dfrac{y}{5}=\dfrac{27}{4}\Rightarrow y=\dfrac{27.5}{4}=\dfrac{135}{4}\)

+\(\dfrac{z}{3}=\dfrac{27}{4}\Rightarrow z=\dfrac{27.3}{4}=\dfrac{81}{4}\)

Vậy \(x=\dfrac{81}{2};y=\dfrac{135}{4};z=\dfrac{81}{4}\)

2,Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{c}{4}=\dfrac{x+2y-3c}{2+2.3+3.4}=\dfrac{-20}{20}=-1\)

+\(\dfrac{x}{2}=-1\Rightarrow x=-1.2=-2\)

+\(\dfrac{y}{3}=-1\Rightarrow y=-1.3=-3\)

+\(\dfrac{c}{4}=-1\Rightarrow c=-1.4=-4\)

Vậy \(x=-2;y=-3;c=-4\)

Vì \(\dfrac{x}{5}=\dfrac{y}{2};\dfrac{x}{2}=\dfrac{z}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{4}=\dfrac{z}{15}\)

Theo tính chất của dãy tỉ số bằng nhau :

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{4}=\dfrac{z}{15}=\dfrac{x+y+z}{10+4+15}=\dfrac{24}{29}\)

\(\Rightarrow x=\dfrac{240}{29}\)

\(y=\dfrac{64}{29}\)

\(z=\dfrac{360}{29}\)

a: =>x^2+2x-3=x^2-4

=>2x=-1

=>x=-1/2

b: \(\dfrac{12x-15y}{7}=\dfrac{20z-15x}{9}=\dfrac{15y-20z}{11}\)

\(=\dfrac{12x-15y+20z-15x+15y-20z}{7+9+11}=\dfrac{-3x}{27}=\dfrac{-x}{9}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12x-15y}{7}=\dfrac{-x}{9}\\\dfrac{20z-15x}{9}=\dfrac{-x}{9}\\\dfrac{15y-20z}{11}=\dfrac{-x}{9}\\x+y+z=48\end{matrix}\right.\)

\(\Leftrightarrow\begin{matrix}-115x+135y=0\\20z-14x=0\\135y-180z+11x=0\\x+y+z=48\end{matrix}\)

=>\(\left(x,y,z\right)\in\varnothing\)