Bạn nên viết đề bằng công thức tán để được hỗ trợ tốt hơn. 

2021 + \(\dfrac{3}{(3+\dfrac{3}{3-x})}\)     =    2023

            \(\dfrac{3}{(3+\dfrac{3}{3-x})}\)    = 2023 - 2021 

           \(\dfrac{3}{(3+\dfrac{3}{3-x})}\) = 2

              3 + \(\dfrac{3}{3-x}\) = \(\dfrac{3}{2}\) 

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

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

                       3 - \(x\) = 3 : ( -\(\dfrac{3}{2}\))

                     3 - \(x\) = -2

                          \(x\) = 3  - ( -2) 

                           \(x\) = 5

Vậy \(x\) = 5

ai biết ko cứu tôi với

( \(\dfrac{2}{123}\) + \(\dfrac{2023}{2022}\) )( \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) - \(\dfrac{2}{15}\))

=( \(\dfrac{2}{123}\) + \(\dfrac{2023}{2022}\) )( \(\dfrac{5}{15}\) - \(\dfrac{3}{15}\) - \(\dfrac{2}{15}\))

=( \(\dfrac{2}{123}\) + \(\dfrac{2023}{2022}\))( \(\dfrac{5-3-2}{15}\))

=( \(\dfrac{1}{123}\) + \(\dfrac{2023}{2022}\)). \(\dfrac{0}{15}\)

= ( \(\dfrac{1}{123}+\dfrac{2023}{2022}\)).0

= 0

\(x+xy+y=1\)

\(2x+2xy+2y=2\)

\(2x\left(1+y\right)+2y=2\)

\(2x\left(y+1\right)+2y+2=4\)

\(2x\left(y+1\right)+2\left(y+1\right)=4\)

\(\left(2x+2\right)\left(y+1\right)=4\)

\(2\left(x+1\right)\left(y+1\right)=4\)

\(\left(x+1\right)\left(y+1\right)=2\)

\(TH1:\left\{{}\begin{matrix}x+1=1\\y+1=2\end{matrix}\right.\)

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

\(TH2:\left\{{}\begin{matrix}x+1=2\\y+1=1\end{matrix}\right.\)

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

\(TH3:\left\{{}\begin{matrix}x+1=-1\\y+1=-2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

\(TH4:\left\{{}\begin{matrix}x+1=-2\\y+1=-1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

\(Vậy...\)

x+xy+y=1⇔x(y+1)+y+1=2⇔(x+1)(y+1)=2

⇒(x+1;y+1)=(-1;-2),(-2;-1),(1;2),(2;1)

sau tự tính nhé :3

Lời giải:

$2^x+2^{x+1}+2^{x+2}+...+2^{x+2019}=2^{x+2023}-8$

$2^x(1+2+2^2+...+2^{2019})=2^{x+2023}-8$

Xét:

$A=1+2+2^2+...+2^{2019}$

$2A=2+2^2+2^3+...+2^{2020}$

$\Rightarrow A=2A-A=2^{2020}-1$

Khi đó:

$2^x.A=2^{x+2023}-8$

$2^x(2^{2020}-1)=2^{x+2023}-2^3$

$2^x(2^{2023}-2^{2020}+1)-2^3=0$

$2^x(2^{2020}.7+1)=2^3$

$x$ ra số sẽ khá xấu. Bạn coi lại.

\(2ab-a+b=3\)

\(4ab-2a+2b=6\)

\(2a\left(2b-1\right)+2b=6\)

\(2a\left(2b-1\right)+\left(2b-1\right)=7\)

\(\left(2a+1\right)\left(2b-1\right)=7\)

\(TH1:\left\{{}\begin{matrix}2a+1=1\\2b-1=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=4\end{matrix}\right.\)

\(TH2:\left\{{}\begin{matrix}2a+1=7\\2b-1=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=\dfrac{1}{2}\end{matrix}\right.\)

\(TH3:\left\{{}\begin{matrix}2a+1=-1\\2b-1=-7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=-3\end{matrix}\right.\)

\(TH4:\left\{{}\begin{matrix}2a+1=-7\\2b-1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-4\\b=0\end{matrix}\right.\)

Rồi em tự tính ab ra nha

1/5 = 20%

54 cây chiếm:

\(100\%-\left(20\%+35\%\right)=45\%\left(tổng.số.cây\right)\)

Số cây các bạn khối 6 đã trồng là:

\(54:45\%=120\left(cây\right)\)

Đ/s: 120 cây

Đổi 35%=35/100=7/20

Số cây ngày thứ 3 trồng chiếm:

   1-(1/5+7/20)=9/20( số cây khối 6)

Số cây khối 6 trồng là :

     54:9/20=120(cây)

         Đáp số:120 cây

\(\left(3\dfrac{3}{4}\times x+75\%\right):\dfrac{2}{3}=-1\)
\(\Rightarrow\dfrac{15}{4}\times x+\dfrac{3}{4}=-\dfrac{2}{3}\)
\(\dfrac{15}{4}\times x=-\dfrac{2}{3}-\dfrac{3}{4}\)
\(\dfrac{15}{4}\times x=-\dfrac{17}{12}\)
\(x=-\dfrac{17}{12}:\dfrac{15}{4}\)
\(x=-\dfrac{17}{45}\)