C=2²⁰²³-(2²⁰²²+2²⁰²¹+...2²+2¹+2⁰)

C=2²⁰²³ - [(2²⁰²³+2²⁰²²+...2²+2¹)-(2²⁰²²+2²⁰²¹+...2²+2¹+2⁰)]

C=2²⁰²³ -(2²⁰²³-1)

C=2²⁰²³ - 2²⁰²³ +1

C = 0+1

C = 1

`2/3 . 6/9+25/54:5/6`

`=2/3 . 2/3+25/54 . 6/5`

`=4/9+5/9=9/9=1`

\(\dfrac{2}{3}\cdot\dfrac{6}{9}+\dfrac{25}{54}:\dfrac{5}{6}=\dfrac{4}{9}+\dfrac{5}{9}=\dfrac{9}{9}=1\)

f) \(\dfrac{-x}{3}=2\dfrac{1}{7}\div\dfrac{5}{7}\)

   \(\dfrac{-x}{3}=\dfrac{15}{7}\div\dfrac{5}{7}\)

   \(\dfrac{-x}{3}=3\)

   \(-x=3\cdot3\)

   \(-x=9\) 

     \(x=-9\)

g) \(\dfrac{5}{4}+\dfrac{1}{4}\div x=1\dfrac{3}{8}\)

    \(\dfrac{5}{4}+\dfrac{1}{4}\div x=\dfrac{11}{8}\)

            \(\dfrac{1}{4}\div x=\dfrac{11}{8}-\dfrac{5}{4}\)

            \(\dfrac{1}{4}\div x=\dfrac{1}{8}\)

                   \(x=\dfrac{1}{4}\div\dfrac{1}{8}\)

                   \(x=2\)

Lời giải:

Gọi $d=ƯCLN(3n+2, 2n+3)$

$\Rightarrow 3n+2\vdots d; 2n+3\vdots d$

$\Rightarrow 3(2n+3)-2(3n+2)\vdots d$

$\Rightarrow 5\vdots d$

$\Rightarrow d=1$ hoặc $d=5$.

Để ps đã cho là tối giản thì $d\neq 5$. Nghĩa là $2n+3\not\vdots 5$

$\Rightarrow 2n-2\not\vdots 5$

$\Rightarrow 2(n-1)\not\vdots 5$

$\Rightarrow n-1\not\vdots 5$

$\Rightarrow n\neq 5k+1$ với $k$ tự nhiên.

NHẦM LỚP R KÌA

\(\dfrac{9}{14}\times7=\dfrac{63}{14}=\dfrac{9}{2}\)

\(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

\(\Rightarrow-7x-189=6x+6\)

\(\Rightarrow-7x-6x=6+189\)

\(\Rightarrow-13x=195\)

\(\Rightarrow x=-15\)

Vay x = -15

\(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

\(\Rightarrow-7\left(x+27\right)=6\left(x+1\right)\)

\(\Rightarrow-7x-189=6x+6\)

\(\Rightarrow-7x-6x=6+189\)

\(\Rightarrow-13x=195\)

\(\Rightarrow x=-15\)

Vậy x=-15

=>3x+1-x+5=6

=>2x+6=6

=>x=0