yynbjkgyg

x= 2002/3000

ko bt đúng ko mong bn nhắc nhở

thank you

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2022}\)

\(\Rightarrow\dfrac{yz+zx+xy}{xyz}=\dfrac{1}{x+y+z}\)

\(\Rightarrow\left(yz+zx+xy\right)\left(x+y+z\right)=xyz\)

\(\Rightarrow xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+3xyz-xyz=0\)

\(\Rightarrow xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+2xyz=0\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Rightarrow x=-y\) hoặc \(y=-z\) hoặc \(z=-x\).

-Đến đây thôi bạn, câu hỏi sai rồi ạ.

\(\left(2021-x\right)\left(x+2021\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2021\\x=-2021\end{matrix}\right.\)

\(\Leftrightarrow x=\pm2021\)

\(\Rightarrow2021-x=0\) hoặc \(x+2021=0\)
\(\Rightarrow x=2021\) hoặc \(x=-2021\)
\(\Rightarrow x\in\left\{2021;-2021\right\}\)

giải chi tiết giúp mình nhé

a) \(x\left(x+2021\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+2021=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2021\end{cases}}\).

b) \(\left(x-2020\right)\left(x+2021\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2020=0\\x+2021=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-2021\end{cases}}\).

c) \(\left(x-2021\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2021=0\\x^2+1=0\end{cases}}\Leftrightarrow x=2021\).

d) \(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\)

Xét tổng: \(A=1+3+5+...+99\)

Số số hạng của dãy số là: \(\frac{99-1}{2}+1=50\).

Tổng của dãy là: \(A=\left(99+1\right)\times50\div2=2500\).

\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\)

\(\Leftrightarrow50x+2500=0\)

\(\Leftrightarrow x=-50\).

a) \(5\left(x+3\right)-2x\left(3+x\right)=0\\ \Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\5-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

b) \(4x\left(x-2021\right)-x+2021=0\\ \Leftrightarrow4x\left(x-2021\right)-\left(x-2021\right)=0\\ \Leftrightarrow\left(4x-1\right)\left(x-2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=0\\x-2021=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=2021\end{matrix}\right.\)

Bạn tự kết luận cả 2 câu giúp mình nhé.

a: \(5\left(x+3\right)-2x\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(5-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

b: Ta có: \(4x\left(x-2021\right)-x+2021=0\)

\(\Leftrightarrow\left(x-2021\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow4x^2\left(x+2021\right)-\left(x+2021\right)=0\\ \Leftrightarrow\left(x+2021\right)\left(4x^2-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\\x=-2021\end{matrix}\right.\)

Lời giải:

$4x^2(x+2021)-x-2021=0$

$\Leftrightarrow 4x^2(x+2021)-(x+2021)=0$

$\Leftrightarrow (x+2021)(4x^2-1)=0$

$\Leftrightarrow (x+2021)(2x-1)(2x+1)=0$

$\Rightarrow x+2021=0$ hoặc $2x-1=0$ hoặc $2x+1=0$

$\Leftrightarrow x=-2021$ hoặc $x=\pm \frac{1}{2}$

  2021 x 2020 + 21 + 2000 - 2021 x 2020

= 2021 x (2020-2020) +21 +2000

= 2021 x 0 + 21 + 2000

= 0 + 21 + 2000

= 2021

Cách này làm nhanh hơn cách của em nè. Chúc em học tốt nha <3

e cảm ơn ạ