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

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

=>\(\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

(x-1/2)² =1/2²

x-1/2=1/2

x=1/2 +1/2

x=2/2

x=1

Lời giải:

$4^{2024}-7=(2^2)^{2024}-7=2^{4048}-7$

$=(2^3)^{1349}.2-7=8^{1349}.2-7\equiv (-1)^{1349}.2-7\pmod 9$

$\equiv -2-7\equiv -9\equiv 0\pmod 9$

$\Rightarrow 4^{2024}-7\vdots 9$

a: Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

b: A nằm giữa O và B

=>OA+AB=OB

=>AB+2=6

=>AB=6-2=4(cm)

c: Trên tia Ox, ta có OA<OC

nên A nằm giữa O và C

=>OA+AC=OC

=>AC+2=3

=>AC=1(cm)

Trên tia Ox, ta có: OC<OB

nên C nằm giữa O và B

=>OC+CB=OB

=>CB+3=6

=>CB=3(cm)

=>AC<CB

cảm ơn nhiều nha

2x+3x+17=3x+13

=>2x+17=13

=>2x=-4

=>\(x=-\dfrac{4}{2}=-2\)

\(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+20}\)

\(=1+\dfrac{1}{2\cdot\dfrac{3}{2}}+\dfrac{1}{3\cdot\dfrac{4}{2}}+...+\dfrac{1}{20\cdot\dfrac{21}{2}}\)

\(=\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{20\cdot21}\)

\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{20\cdot21}\right)\)

\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)\)

\(=2\left(1-\dfrac{1}{21}\right)=2\cdot\dfrac{20}{21}=\dfrac{40}{21}\)

?

Đề chọn hsg mà

\(-5^{22}-\left\{-222-\left[-122-\left(100-5^{22}\right)+2022\right]\right\}\)

\(=-5^{22}+222+\left[-122-\left(100-5^{22}\right)+2022\right]\)

\(=-5^{22}+222-122-\left(100-5^{22}\right)+2022\)

\(=-5^{22}+100-100+5^{22}+2022=2022\)

a: \(\left(-\dfrac{5}{3}\right)^3< x< \dfrac{-24}{35}\cdot\dfrac{-5}{6}\)

=>\(\dfrac{-125}{27}< x< \dfrac{120}{210}\)

=>\(-\dfrac{125}{27}< x< \dfrac{4}{7}\)

b: \(\left(12x+11\right)\left(y-3\right)=12\)

mà 12x+11>=11 và 12x+11 chia 12 dư 11 vì x tự nhiên

nên \(\left(x,y\right)\in\varnothing\)

A =4 jv?

Gọi d=ƯCLN(2n+3;4n+4)

=>\(\left\{{}\begin{matrix}2n+3⋮d\\4n+4⋮d\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4n+6⋮d\\4n+4⋮d\end{matrix}\right.\)

=>\(4n+6-4n-4⋮d\)

=>\(2⋮d\)

mà 2n+3 lẻ

nên d=1

=>ƯCLN(2n+3;4n+4)=1

=>\(\dfrac{2n+3}{4n+4}\) là phân số tối giản