(x+1) + (x+3) + (x+5) +....+ (x+39) =2020
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(f'\left(x\right)=0\) có 2 nghiệm bội lẻ \(x=2019\) và \(x=2021\) nên hàm có 2 cực trị
![](https://rs.olm.vn/images/avt/0.png?1311)
B=5+2(x-2019)2020
Vì (x-2019)2020 ≥0
=>5+(x-2019)2020 ≥5
Để B đạt Min
=>x-2019=0
=>x=2019
Vậy MinB=5 <=>x=2019
![](https://rs.olm.vn/images/avt/0.png?1311)
3x+(7/2020+6/2020+5/2020)+3
=3x+9/1010+3030/1010
=3x+3039/1010
vậy giá trị của biểu thức đại số trên là 3x+3039/1010
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 2025
(x + x + x + x + x) + (1 + 2 + 3 + 4 + 5) = 2025
5x + 15 = 2025
5x = 2025 - 15
5x = 2010
x = 2010 : 5
x = 402
b) 5 * x - x = 2020
5 * x - x * 1 = 2020
x * (5 - 1) = 2020
x * 4 = 2020
x = 2020 : 4
x = 505
mong bạn tick
a) ( x + 1 ) + ( x + 2) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 2025
\(\left(x+x+x+x+x\right)+\left(1+2+3+4+5\right)=2025\)
\(5x+15=2025\)
\(5x=2025-15\)
\(5x=2010\)
\(x=2010:5\)
\(x=402\).
![](https://rs.olm.vn/images/avt/0.png?1311)
1,
a,\(2020-\left(249+2020\right)+\left(249-573\right)\)
\(=2020-249-2020+249-573\)
\(=-573\)
b,\(\left|-257\right|+\left(-3\right)^0-\left(18+257\right)\)
\(=257+1-18-257\)
\(=1-18=-17\)
\(c,25.\left(85-47\right)-85.\left(47+25\right)\)
\(=25.85-25.47-47.85+85.25\)
\(=85.\left(25-47+25\right)-25.47\)
\(=85.3-25.47\)
\(=-920\)
2,
\(a,15-5.\left(x+2\right)=-30\)
\(=>5.\left(x+2\right)=15+30=45\)
\(=>x+2=\frac{45}{5}=9\)
\(=>x=7\)
\(b,\left(x+2\right)^2+5=105\)
\(=>\left(x+2\right)^2=100\)
\(=>\left(x+2\right)^2=10^2\)
\(=>x+2=10\)
\(=>x=8\)
\(c,\left|2x-5\right|-\left(-6\right)=11\)
\(=>\left|2x-5\right|=11-6=5\)
\(=>\orbr{\begin{cases}2x-5=5\\2x-5=-5\end{cases}}\)
\(=>\orbr{\begin{cases}2x=5-5=0\\2x=-5+5=0\end{cases}=>x=0}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow x=0;x=\pm5\)
2, \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+9\right)=0\Leftrightarrow x=-9;x=1\)
3, \(6x\left(x-2\right)=x-2\Leftrightarrow\left(6x-1\right)\left(x-2\right)=0\Leftrightarrow x=\frac{1}{6};x=2\)
4, \(7\left(x-2020\right)^2-x+2020=0\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)
\(\Leftrightarrow\left(x-2020\right)\left[7\left(x-2020\right)-1\right]=0\Leftrightarrow x=2020;x=\frac{14141}{7}\)
5, \(x^2-10x=-25\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
6, \(x^2-2x-3=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow x=-1;x=3\)
\(1,\)
\(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x^2-25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)
\(2,\)
\(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow x^2-x+9x-9=0\)
\(\Leftrightarrow x\left(x-1\right)+9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)
\(3,\)
\(6x\left(x-2\right)=x-2\)
\(\Leftrightarrow6x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{6}\end{cases}}\)
\(4,\)
\(7\left(x-2020\right)^2-x+2020=0\)
\(\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)
\(\Leftrightarrow\left(x-2020\right)[7\left(x-2020\right)-1]=0\)
\(\Leftrightarrow\left(x-2020\right)[7x-14141]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\7x=14141\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{14141}{7}\end{cases}}\)
\(5,\)
\(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
\(6,\)
\(x^2-2x-3=0\)
\(\Leftrightarrow x^2-3x+x-3=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x+3)(x+5)=0
=>x+3=0 hoặc x+5=0
=>x=-3 hoặc -5
b) (x-1).5-1=0
=>5x-5-1=0
=>5x-6=0
=>5x=6
=>x=6/5
c)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/\(-2020+23+x=-2020\\ \Leftrightarrow23+x=-2020+2020\\ \Leftrightarrow23+x=0\\ \Leftrightarrow x=0-23\\ \Leftrightarrow x=-23\)
Vậy...
2/\(2x-35=25\\ \Leftrightarrow2x=60\\ \Leftrightarrow x=30\)
Vậy...
3/\(3x+17=26\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\)
Vây...
4/\(\left|\text{x}-1\right|=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)
Vậy...
5/ \(-17.\left|x\right|=-34\\ \Leftrightarrow\left|x\right|=2\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
VẬy...
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)
\(=4x^2-20x+25-4x^2+20x\)
=25
b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)
\(=16-9x^2+9x^2+6x+1\)
=6x+17
c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)
\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)
=1
d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)
\(=\left(2021x-2020-2020x+2021\right)^2\)
\(=\left(x+1\right)^2\)
\(=x^2+2x+1\)
(x+1)+(x+3)+9x+5)+....+(x+39)=2020
Tách ra ta được :
A=x+x+x+....+x
B=1+3+5+....+39
=>A+B=2020
- Khoảng cách mỗi số hạng là 2
- Số các số hạng của A(hay B) là : (39-1):2+1=20(số)
- Tổng của 20 số đó : (39+1)x20:2=400
Vậy ta có :
20x+400=2020
20x=1620
x=81
Vậy:......
#H