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.

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 $(x,y)=(-14,-8)$

giúp mình với mọi người ơi

\(\left(x-2\right)^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\)

\(\Rightarrow24x+25=15\)

\(\Rightarrow24x=-10\)

\(\Rightarrow x=\dfrac{-5}{12}\)

c) 3(x -1 ) + 3(2 – x) = 0
3x-3+6-3x=0
 3=0 (vô lí)
không có giá trị x thỏa mãn 
d) 5x + 3(x – 8) = 8
5x+3x-24=8
8x=32
x=32:8
x=4

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`

Bài 1 

a, -25 . 63 - 25 . 3

= 25 . (-63) - 25 . 3

= 25 . [(-63) - 3]

= 25 . (-66) = -1650

Bài 2

c, (x + 1)² = 16

=> (x + 1)² = 4²

=> x + 1 = 4 

=> x = 3

d, (-38) - (x - 2) = -16

=> (x - 2) = -38 - (-16) 

=> x - 2 = -38 + 16 = -22

=> x = 2 + (-22) 

=> x = -20

c)
\(x^3-3.x^2.6+3.x.6^2-6^3=0\)
\(\left(x-6\right)^3=0\)
x-6=0
x=6
d)
\(x^3-3.x^2.1+3.x.1^2-1-x^3-3x-2=0\)
\(x^3-3x^2+3x-1-x^3-3x^2-2=0\)
\(-6x^2-3=0\)
\(-3\left(2x^2+1\right)=0\)
\(2x^2+1=0\)
2x²=-1
x²=1/2
x=\(\dfrac{\sqrt{2}}{2}\)