Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bạn Đúc giúp người kiểu giì đấy :))) , giúp mà không giúp hết à ???
a) 2x + 2020 2021
=> 2x = 2021 - 2020
=> 2x = 1
=> 2x = 20
=> x = 0
b) Ta có :
4x + 14 ⋮ x + 2
=> 4. ( x + 2 ) + 6 ⋮ x + 2
Mà 4 . ( x + 2 ) ⋮ x + 2
=> 6 ⋮ x + 2 => x + 2 ∈ { 1 ; 2 ; 3 ;6 }
=> x ∈ { 0 ; 1 ; 4 } ( do x ∈ N )
c) ( x - 3 )2021 - ( x - 3 )5 = 0
=> ( x - 3 )5 . [ ( 2 - 3 )2016 - 1 ] = 0
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^5=0\\\left(x-3\right)^{2016}-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^{2016}=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x-3\in=\left\{-1;1\right\}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x\in=\left\{2;4\right\}\end{cases}}\)
a) 2x = 2021 - 2020
2x = 1
\(\Rightarrow\)2x = 10
\(\Rightarrow\)x = 0
Ta có : \(\left(2020.x^2+2021\right).\left(x^2-1\right).\left(2.x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2020.x^2+2021=0\\x^2-1=0\\2.x+=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\notinℝ\\x=\pm1\\x=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=1\\x=-1\\x=-\frac{1}{2}\end{cases}}\)
Vậy \(x=\left\{\pm1;-\frac{1}{2}\right\}\)
\(x_1+x_2=x_3+x_4=...=x_{2019}+x_{2020}=2\Rightarrow x_1+x_2+x_3+x_4+...+x_{2019}+x_{2020}=2.1010=2020\)
\(\Rightarrow x_1+x_2+x_3+x_4+...+x_{2019}+x_{2020}+x_{2021}=2020+x_{2021}\)
\(\Rightarrow0=2020+x_{2021}\)
\(\Rightarrow x_{2021}=-2020\)
Vậy \(x_{2021}=-2020\)
a) ( 175 + 325 ) + ( 64 + 536 )
= 500 + 600
= 1100
b) 82 . ( 34 + 63 + 3 )
= 82 . 100
= 8200
c) 4 . 25 . 35.2
= 100 . 70
= 7000
b. 1404 : [118 - (4x + 6)] = 27
118 - (4x + 6) = 52
4x + 6 = 66
4x = 60
x = 15
d) \(5x^2-3x=0\)
\(\Leftrightarrow x\left(5x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\5x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{5}\end{cases}}\)
e) \(3\left(x-1\right)+4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[3-4.\left(x-1\right)\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\3-4\left(x-1\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\4\left(x-1\right)=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x-1=\frac{3}{4}\Rightarrow x=\frac{7}{4}\end{cases}}\)
f) \(2\left(x-2\right)^2=\left(x-2\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2\left(x-2\right)-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x-2=\frac{1}{2}\Rightarrow x=\frac{5}{2}\end{cases}}\)
g) \(\left(x-2020\right)^4=\left(x-2020\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-2020\right)^2=0\\\left(x-2020\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=2019,x=2021\end{cases}}\)
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: \(A=\left(99+1\right)\times50\div2=2500\).
\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\)
\(\Leftrightarrow50x+2500=0\)
\(\Leftrightarrow x=-50\).