2x+3+2x+1=1280
Những câu hỏi liên quan
21 tháng 7 2017 lúc 15:53
tìm x :
a) (x+3)+(x+5)+.......+(x+53)=754
b)9+11+13+...+(2x+1)=1280
f)(3x-9).(x-7)=0
a, (x+3)+(x+5)+....+(x+53)=754
=> [(x+53)-(x+3):2+1].[(x+53)+(x+3)]:2=754
=> 26.(2x+56):2=754
=> 26.(2x+56)=1508
=> 2x+56= 58
=> 2x = 2
=> x=2:2
=>x=1
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
22 tháng 7 2017 lúc 14:17
a)9+11+13+...+(2x+1)=1280
mấy bạn giúp mk nha , mk tick cho
Theo công thức tính tổng , có :
\(\frac{\left\{\left[\left(2x+1\right)-9\right]:2+1\right\}.\left[\left(2x+1\right)+9\right]}{2}=1280\)
\(\frac{\left\{\left[2x-8\right]:2+1\right\}.\left[2x+10\right]}{2}=1280\)
\(\frac{\left\{x-4+1\right\}.2.\left[x+5\right]}{2}=1280\)
\(\left\{x-3\right\}.\left(x+5\right)=1280\)
\(x^2+5x-3x-15=1280\)
\(x^2+2x=1295\)
\(x\left(x+2\right)=1295\)
\(\Rightarrow x=35\)
Đúng 0
Bình luận (0)
21 tháng 7 2017 lúc 16:12
tìm x :
a) (x+3)+(x+5)+.......+(x+53)=754
b)9+11+13+...+(2x+1)=1280
f)(3x-9).(x-7)=0
ai nhanh mk tick cho nha
(3x-9).(x-7)=0 TH1;3x-9=0 TH2;x-7=0 3x=0+9 x=0+7 3x=9 x=7 x=3 Vậy x có 2 giá trị x=3 và x=7
Đúng 0
Bình luận (0)
Rút gọn :
1. (2x-5)(3x+1)-(x-3)^2+(2x+5)^2-(3x+1)^3
2. (2x-1)(2x+1)-3x-2)(2x+3)-(x-1)^3+(2x+3)^3
3. (x-2)(x^2+2x+4)-(3x-2)^3+(3x-4)^2
4. (7x-1)(8x+2)-(2x-7)^2-(x-4)^3-(3x+1)^3
5. (5x-1)(5x+1)-(x+3)(x^2-3x+9)-(2x+4)^2-(3x-4)^2+(2x-5)^3
6. (4x-1)(x+2)-(2x+5)^2-(3x-7)^2+(2x+3)^3=(3x-1)^3
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
Đúng 0
Bình luận (0)
Tính (rút gọn )
1, 2x(3x-1)-(2x+1)(x-3)
2, 3(x^2-2x)-(4x+2)(x-1)
3, 3x(x-5)-(x-2)^2 -(2x+3)(2x-3)
4, (2x-3)^2+(2x-1) (x+4)
1) `2x(3x-1)-(2x+1)(x-3)`
`=6x^2-2x-2x^2+6x-x+3`
`=4x^2+3x+3`
2) `3(x^2-3x)-(4x+2)(x-1)`
`=3x^2-9x-4x^2+4x-2x+2`
`=-x^2-7x+2`
3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`
`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`
`=3x^2-15x-x^2+4x-4-4x^2+9`
`=-2x^2-11x+5`
4) `(2x-3)^2+(2x-1)(x+4)`
`=4x^2-12x+9+2x^2+8x-x-4`
`=6x^2-5x+5`
Đúng 1
Bình luận (0)
d) (3x – 5)(7 – 5x) – (5x + 2)(2 – 3x) = 4 g) 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) =0 j) (2x – 1)(3x + 1) – (4 – 3x)(3 – 2x) = 3 k) (2x + 1)(x + 3) – (x – 5)(7 + 2x) = 8 m) 2(3x – 1)(2x + 5) – 6(2x – 1)(x + 2) = - 6
g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Đúng 0
Bình luận (1)
Bài I. Rút gọn các biểu thức sau:
a) 3x(2x+1)+ (2x - 3)(x+1),
b) x(3x - 2)2 + 3(x-2)(x+2)
c) (2x+1)(4x² - 2x+1)-2x(2x+3)(2x - 3)-(x-3)²
a: Ta có: \(3x\left(2x+1\right)+\left(2x-3\right)\left(x+1\right)\)
\(=6x^2+3x+2x^2+2x-3x-3\)
\(=8x^2+2x-3\)
Đúng 0
Bình luận (0)
1) |2x - 1| 52) |2x - 1| |x + 5|3) |3x + 1| x - 2 4) |3 - 2x| x + 2 5) |2x - 1| 5 - x 6) |- 3x| x - 2 7) |2 - 3x| 2x + 1 8) |2x - 1| + |4x ^ 2 - 1| 0 9) (2x + 5)/(x + 3) + 1 4/(x ^ 2 + 2x - 3) - (3x - 1)/(1 - x) 10) (x - 1)/(x + 3) - x/(x - 3) (7x - 3)/(9 - x ^ 2) 11) 5 + 96/(x ^ 2 - 16) (2x - 1)/(x + 4) + (3x - 1)/(x - 4) 12) (2x)/(2x - 1) + x/(2x + 1) 1 + 4/((2x - 1)(2x + 1)) 13) (x + 2)/(x - 2) - 1/x 2/(x ^ 2 - 2x) 14) x/(2x - 6) + x/(2x + 2) (2x + 4)/(x ^ 2 - 2x - 3)
Đọc tiếp
1) |2x - 1| = 5
2) |2x - 1| = |x + 5|
3) |3x + 1| = x - 2
4) |3 - 2x| = x + 2
5) |2x - 1| = 5 - x
6) |- 3x| = x - 2
7) |2 - 3x| = 2x + 1
8) |2x - 1| + |4x ^ 2 - 1| = 0
9) (2x + 5)/(x + 3) + 1 = 4/(x ^ 2 + 2x - 3) - (3x - 1)/(1 - x)
10) (x - 1)/(x + 3) - x/(x - 3) = (7x - 3)/(9 - x ^ 2)
11) 5 + 96/(x ^ 2 - 16) = (2x - 1)/(x + 4) + (3x - 1)/(x - 4)
12) (2x)/(2x - 1) + x/(2x + 1) = 1 + 4/((2x - 1)(2x + 1))
13) (x + 2)/(x - 2) - 1/x = 2/(x ^ 2 - 2x)
14) x/(2x - 6) + x/(2x + 2) = (2x + 4)/(x ^ 2 - 2x - 3)
(2x-1).(4x2+2x+1)-(3+2x).(3+2x).(3+2x)+36.(x-1).(x-1)
(2x-3)^2+(2x-1)(2x+1)-(2x+1)(4x^2-2x+1)+8x^3(x-1)=13