2021^2^0^1^0
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(A=\left(99+1\right)\times50\div2=2500\).
\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\)
\(\Leftrightarrow50x+2500=0\)
\(\Leftrightarrow x=-50\).
![](https://rs.olm.vn/images/avt/0.png?1311)
a, - 1,2 + (- 0,8) + 0,25 + 5,75 - 2021
= - (1,2 + 0,8) + (0,25 + 5,75) - 2021
= - 2 + 6 - 2021
= 4 - 2021
= - 2017
b, - 0,1 + \(\dfrac{16}{9}\) + 11,1 - \(\dfrac{20}{9}\)
= (11,1 - 0,1) - (\(\dfrac{20}{9}\) - \(\dfrac{16}{9}\))
= 11 - \(\dfrac{4}{9}\)
= \(\dfrac{95}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 5. Kết quả của phép tính 20200 +2021.(20002021 : 20002021 )+20210
A.2020
B.2023
C.1
D.0
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=3\left|2x-\dfrac{3}{2}\right|+2021^0=3\left|2x-\dfrac{3}{2}\right|+1\ge1\)
\(minA=1\Leftrightarrow2x=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{4}\)
b) \(B=2\left|x-6\right|+3\left(2y-1\right)^2+2021^0=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\)
\(minB=1\Leftrightarrow\) \(\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)
\(A=3\left|2x-\dfrac{3}{2}\right|+1\ge1\\ A_{min}=1\Leftrightarrow2x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{4}\\ B=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Leftrightarrow\left(2x-1\right)\left(x-2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2021\end{matrix}\right.\\ c,\Leftrightarrow4x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ d,\Leftrightarrow\left(3x+7-x-1\right)\left(3x+7+x+1\right)=0\\ \Leftrightarrow\left(2x+6\right)\left(4x+8\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2^0+2^1+...+2^{2021}\)
\(\Rightarrow2A=2^1+2^2+...+2^{2022}\)
\(\Rightarrow2A-A=2^1+2^2+...+2^{2022}-2^0-2^1-...-2^{2021}=2^{2022}-2^0=2^{2022}-1\)
A = 20 + 21 + ... + 22021
2A = 2(20+21+...+22021)
2A = 21 + 22 + ... + 22022
A = ( 2^1 + 2^2 +...+2^2022 ) - ( 2^0 + 2^1 + ...+2^2021 )
A = ( 2^1 - 2^1 ) + ( 2^2 - 2^2 ) + .... + (2^2021 - 2^2021 ) + 2^2022 - 2^0
A = 0 + 0 +....+0 + 2^2022 - 2^0
A = 2^2022 - 2^0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{1}{3}+\dfrac{1}{4}:2-1=\dfrac{1}{3}+\dfrac{1}{2}-1=\dfrac{5}{6}-1=-\dfrac{1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)
\(\Rightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow ab^2+a^2b+ac^2+a^2c+bc^2+b^2c+2abc=0\)
\(\Leftrightarrow ab^2+a^2b+ac^2+bc^2+a^2c+abc+b^2c+abc=0\)
\(\Leftrightarrow\left(a+b\right)ab+c^2\left(a+b\right)+bc\left(a+b\right)+ac\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(c^2+ab+bc+ac\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
Vậy ta có các trường hợp: \(a=-b,c=0\)hoặc \(b=-c,a=0\)hoăc \(a=-c,b=0\).
Với từng trường hợp ta đều có đpcm.
Đáp án = 2021