5x = 8y = 20z = \(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{8}\)

Dựa vào tính chất dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-y-z}{8-5-8}=\dfrac{3}{-5}\)

=> 5x = 8y = 20z = \(\dfrac{3}{-5}\)

=> \(\left\{{}\begin{matrix}x=-\dfrac{3}{25}\\y=-\dfrac{3}{40}\\z=-\dfrac{3}{100}\end{matrix}\right.\)

8y = 20z = \(\dfrac{y}{20}=\dfrac{z}{8}\)

Mà 5x = 8y = \(\dfrac{x}{8}=\dfrac{y}{5}\)

Nên mik cho \(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{8}\) luôn.

c: -62+x=-14+46

=>\(x-62=46-14=32\)

=>\(x=32+62=94\)

d: \(25+\left(x-5\right)=-123-\left(15-123\right)\)

=>\(x-5+25=-123-15+123\)

=>\(x+20=-15\)

=>\(x=-15-20=-35\)

`c) -62 + x = -14 + 46`

`-62 + x = 32`

`x = 32 - (-62)`

`x = 94`

Vậy `x = 94`

`d) 25 + ( x - 5 ) = -123 - ( 15 - 123 )`

`25 + ( x - 5 ) = -123 - 15 + 123`

`25 + ( x - 5 ) = - 15`

`x - 5 = -15 - 25`

`x - 5 = -40`

`x = -40 + 5`

`x = -35`

Vậy `x = -35`

\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)

b, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)

c, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)

d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)

\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)

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

  \(\frac{x}{3}=\frac{y}{5}=\frac{z}{6}=\frac{x+y+z}{3+5+6}=\frac{48}{14}=\frac{24}{7}\)

suy ra:  \(\frac{x}{3}=\frac{24}{7}\)=>   \(x=\frac{72}{7}\)

             \(\frac{y}{5}=\frac{24}{7}\) =>   \(y=\frac{120}{7}\)

             \(\frac{z}{6}=\frac{24}{7}\) =>  \(z=\frac{144}{7}\)

Vậy...

b) c)  bạn làm tương tự

d) Đặt:    \(\frac{x}{3}=\frac{y}{5}=k\)  =>    \(x=3k;\)  \(y=5k\)

Ta có:  \(x.y=60\)

<=>  \(3k.5k=60\)

<=>  \(k^2=4\)

<=>  \(k=\pm2\)

a) 

x/5-2/3+3/4=17/60

<=> x/5=17/60-3/4+2/3

<=> x/5=1/5

<=> x=1

b)

18−75−2x=−22312018−75−2x=−223120  

<=> −5140−2x=−223120−5140−2x=−223120

<=>2x=−5140+2231202x=−5140+223120

<=>2x=7122x=712 <=> 7x=2.127x=2.12

                             <=> x=247x=247

vậyx=247

Ta có : x:y:z = 3:4:5

 \(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{2y}{8}=\frac{3z}{15}=\frac{z-2y+3z}{3-8+15}=\frac{60}{10}=6\)

Nên : \(\frac{x}{3}=6\Rightarrow x=18\)

         \(\frac{y}{4}=6\Rightarrow y=24\)

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

Vậy x = 18 ; y = 24 ; z = 30

x:y:z=3:4:5 => x/3 = y/4 = z/5 => x/3 = 2y/8 = 3z/15

Theo TCDTSBN ta có:

x/3 = 2y/8 = 3z/15 = x-2y+3z/3-8+15 = 60/10 = 6

=> x = 18, y = 24, z = 30

a) Đặt \(\frac{x}{3}=\frac{y}{5}=k\) => x = 3k ; y = 5k

Do đó x . y = 3k . 5k = 15k² = 60

=> k² = 4 => k = + 2

- Với k = 2 thì x = 6 ; y = 10

- Với k = - 2 thì x = -6 ; y = -10

b) Tương tự     

Ta có: 4x = 7y

<=> \(\dfrac{4}{y}=\dfrac{7}{x}\)

<=> \(\dfrac{16}{y^2}=\dfrac{49}{x^2}\)

=> \(\dfrac{16+49}{x^2+y^2}=\dfrac{65}{260}=\dfrac{1}{4}\)

=> \(\left\{{}\begin{matrix}x=28\\y=16\end{matrix}\right.\)

Lời giải:

$4x=7y\Rightarrow \frac{x}{7}=\frac{y}{4}=k$

$\Rightarrow x=7k; y=4k$

Ta có: $x^2+y^2=260$

$\Leftrightarrow (7k)^2+(4k)^2=260$

$\Leftrightarrow  65k^2=260$

$\Leftrightarrow k^2=4$

$\Rightarrow k=\pm 2$
Nếu $k=2$ thì $x=7k=14; y=4k=8$. Ta có $(x,y)=(14,8)$

Nếu $k=-2$ thì $x=7k=-14; y=4k=-8$. Ta có $(x,y)=(-14,-8)$