\((x+1)^2=\dfrac{1}{4}\)

\((x+1)^2=(\dfrac{1}{2})^2\) hoặc \((x+1)^2=(-\dfrac{1}{2})^2\)

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

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

\(\left(x+1\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow x+1=\pm\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=\dfrac{-1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

x+1/2=0

= x = 0-1/2

= x = -1/2

TL:

a<b<c
6<a<10->a \(\in\) (7;8;9)
8<c<11->c\(\in\)(9;10)
1)=>a=7
b=8
c=9
2) hoặc =>a=7
b=8
c=10
3) hoặc=>a=7
b=9
c=10
4) hoặc a=8;b=9;c=10.

ta gọi số đó là \(x\left(1000< x< 6000\right)\)

theo đề bài ta có:

\(x< 6000\) và: \(\left\{{}\begin{matrix}x+15⋮40\\x+1560\\x+15⋮126\end{matrix}\right.\)

\(\Rightarrow x+15\in BC\left(60,40,126\right)=\left\{0;2520;5040;7560;....\right\}\)

mà 1000<x<6000\(\Rightarrow x+15\in\left\{2520;5040\right\}\)

bạn tự trừ nốt nhé

\(=\dfrac{2^3.3^2.2.3^3}{2^2.3^3}=2^2.3^2=36\)

qiup toi (:(

`8-(7-x+3)=-11`

`7-x+3=8-(-11)=8+11=19`

`7-x=19-3=16`

`x=7-16=-9`

Vậy `x=-9`

Lời giải:

$3x-15=2+4+6+8+...+100=(100+2).50:2=2550$

$3x=2550+15$

$3x=2565$

$x=855$

Bài 3:
$8\not\in A$

$5\in A$

$\left\{3;7\right\}\subset A$

$\left\{5\right\}\subset A$

$\\left\{3;5;7\right\}\subset A$

$\left\{7\right\}\subset A$

$\varnothing \subset A$

$7\in A$