d: \(\frac{3}{x}+\frac{y}{3}=\frac56\)

=>\(\frac{3}{x}=\frac56-\frac{y}{3}=\frac{5-2y}{6}\)

=>x(5-2y)=18

=>x(2y-5)=-18

mà 2y-5 lẻ và x>0

nên (x;2y-5)∈{(18;1);(2;9);(6;3)}

=>(x;2y)∈{(18;6);(2;14);(6;8)}

=>(x;y)∈{(18;3);(2;7);(6;4)}

c: \(\frac{x}{9}-\frac{3}{y}=\frac{1}{18}\)

=>\(\frac{x}{9}-\frac{1}{18}=\frac{3}{y}\)

=>\(\frac{2x-1}{18}=\frac{3}{y}\)

=>y(2x-1)=54

mà 2x-1 lẻ

nên (2x-1;y)∈{(1;54);(3;18);(9;6);(27;2)}

=>(2x;y)∈{(2;54);(4;18);(10;6);(28;2)}

=>(x;y)∈{(1;54);(2;18);(5;6);(14;2)}

b: \(\frac{x}{6}-\frac{2}{y}=\frac{1}{30}\)

=>\(\frac{x}{6}-\frac{1}{30}=\frac{2}{y}\)

=>\(\frac{5x-1}{30}=\frac{2}{y}\)

=>y(5x-1)=60

mà 5x-1 chia 5 dư 4

nên (5x-1;y)∈{(4;15)}

=>(5x;y)∈{(5;15)}

=>(x;y)∈{(1;15)}

a: \(\frac{x}{2}-\frac{4}{y}=\frac15\)

=>\(\frac{x}{2}-\frac15=\frac{4}{y}\)

=>\(\frac{5x-2}{10}=\frac{4}{y}\)

=>y(5x-2)=40

mà 5x-2 chia 5 dư 3

nên (5x-2;y)∈{(8;5)}

=>(5x;y)∈{(10;5)}

=>(x;y)∈{(2;5)}

Bài 1.

a.\(\left(x-8\right)\left(x^3+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

b.\(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)

\(\Leftrightarrow4x-3-x-5=30-3x\)

\(\Leftrightarrow4x-x+3x=30+5+3\)

\(\Leftrightarrow6x=38\)

\(\Leftrightarrow x=\dfrac{19}{3}\)

Bài 1:

a. $(x-8)(x^3+8)=0$

$\Rightarrow x-8=0$ hoặc $x^3+8=0$

$\Rightarrow x=8$ hoặc $x^3=-8=(-2)^3$

$\Rightarrow x=8$ hoặc $x=-2$

b.

$(4x-3)-(x+5)=3(10-x)$

$4x-3-x-5=30-3x$

$3x-8=30-3x$

$6x=38$
$x=\frac{19}{3}$

b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

b: \(\Leftrightarrow x-15-27-x+x-13=-1\)

\(\Leftrightarrow x-55=-1\)

hay x=54

\(a,\left(x-8\right)\left(x^3+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

\(b,\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\\ \Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow3x-8-30+3x=0\\ \Leftrightarrow6x-38=0\\ \Leftrightarrow x=\dfrac{19}{3}\)

TK

`a.(x-8)(x+8)=0`

`⇔³{x−8=0x³+8=2 `

`⇔³³{x=8x³=−2³ `

`⇔{x=8x=−2`

Vậy ` x = 8;-2`

`b. ( 4 x − 3 ) − ( x + 5 ) = 3 . ( 10 − x )`

`⇔ 4 x − 3 − x − 5 = 30 − 3 x`

`⇔ 3 x − 8 = 30 − 3 x`

`⇔ 3 x + 3 x = 30 + 8`

`⇔ 6 x = 38`

`⇔ x = 19/ 3`

Vậy ` x = 19/ 3`

-210 = (-1) + (-2) + (-3) + ... + (-x - 1) + (-x)

=> -210 = -(1 + 2 + 3 + ... + x + 1 + x)

=> 210 = 1 + 2 + 3 + ... + x + x + 1

=> 210 = \(\frac{\left(x+1+1\right).\left(x+1\right)}{2}\)

=> 420 = (x + 2) . (x + 1)

=> 21 . 20 = (x + 2) . (x + 1)

=> (19 + 2) . (19 + 1) = (x + 2) + (x + 1)

=> x = 19.

\(y=\frac{1}{x^2+\sqrt{x}}\)

(-x)= -20

=> x =20

1+2+3+...+x=\(\frac{x\left(x+1\right)}{2}\)

-\(\frac{x\left(x+1\right)}{2}\)=-210

\(\frac{x\left(x+1\right)}{2}\)=210

x(x+1)=420

=>x=20