a) (x+1)+(x+2)+(x+3)+...+(x+100)=5750
b)1+2+3+...+x=2014
c)1+3+5+...+x=1600
d)x+(x+1)+(x+2)+...+(x+30)=1240
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15 tháng 4 2022
\(∘backwin\)
\(a ) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750\)
\( ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 \)
\( 100 x + ( 1 + 100 ) ×100 : 2 = 5750\)
\(100 x + 5050 = 5750\)
\( 100 x = 5750 − 5050\)
\(100 x = 700\)
\(x = 700 : 100\)
\(x = 7\)
\(b,\) \(B=\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020}+2021\)
\( B < 1 -\)\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)
\(B<1-\)\(\dfrac{1}{2021}\)
\(B<\)\(\dfrac{2020}{2021}\)
\(\dfrac{2020}{2021}< 1\)
\(B<1\)
TL
15 tháng 4 2022
a) (x+1) +(x+2 ) + ...+(x+100)=5750
= 100x + (1+2+3+...+100) = 5750
=100x + 5050 = 5750
--> 100x = 5750-5050=700
--> x=7
PT
Phạm Trần Hoàng Anh
CTVHS
22 tháng 7 2021
`(x+1) + (x+2) + ... + (x+100) = 5750`
Số số ngoặc trong phép tính là:
`(100 - 1) : 1 + 1 = 100` (ngoặc)
`=> 100x + (1+2+3+...+100) = 5750`
`=> 100x + ((100 + 1) . 100 : 2) = 5750`
`=> 100x + 5050 = 5750`
`=> 100x = 200`
`=> x = 2`
`(x+1) . (2y-5) = 143`
`=> (2y-5) ∈ Ư(143)`
mà `2y-5 lẻ`
`=> 2y-5 ∈ {-1;-11;1;11} => y = {2;-3;3;8}`
mà `y ∈ N => y = {2;3;8}`
`=> x+1 ∈ {-143;143;13}`
`=> x ∈ {-144;142;12}`
mà `x ∈ N => x ∈ {142;12}`
Vậy `(x;y) = (142;3);(12;8)`
(Chúc bạn học tốt)
4 tháng 8 2018
\(a,-12\left(x-5\right)+7\left(3-x\right)=5\)
\(-12x+60+21-7x=5\)
\(-12x-7x+81=5\)
\(-19x=5-81\)
\(-19x=-76\)
\(x=-76:\left(-19\right)\)
\(x=4\)
\(Vậyx=4\)
\(b,30\left(x+2\right)-6\left(x-5\right)-24x=100\)
\(30x+60-6x-30-24x=100\)
\(30x-6x-24x+60-30=100\)
\(0x+30=100\)
\(\Rightarrow Vôlý\)
Vậy không có giá trị nào của x thỏa mãn đề bài.
\(c,-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-1-\frac{1}{2}x-\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-\frac{1}{2}x-1-\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-\frac{11}{2}x-\frac{2}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-\frac{2}{3}+\frac{5}{6}=\frac{3}{2}x+\frac{11}{2}x\)
\(-\frac{4}{6}+\frac{5}{6}=\frac{14}{2}x\)
\(\frac{1}{6}=7x\)
\(x=\frac{1}{6}:7\)
\(x=\frac{1}{6}.\frac{1}{7}\)
\(x=\frac{1}{42}\)
\(Vậyx=\frac{1}{42}\)
\(d,-3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(-3x+\frac{3}{2}-5x-3=-x+\frac{1}{5}\)
\(-3x-5x+\frac{3}{2}-3=-x+\frac{1}{5}\)
\(-8x+\frac{3}{2}-\frac{6}{2}=-x+\frac{1}{5}\)
\(-8x-\frac{3}{2}=-x+\frac{1}{5}\)
\(-\frac{3}{2}-\frac{1}{5}=-x+8x\)
\(\frac{15}{10}-\frac{2}{10}=7x\)
\(7x=\frac{13}{10}\)
\(x=\frac{13}{10}:7\)
\(x=\frac{13}{10}.\frac{1}{7}\)
\(x=\frac{13}{70}\)
\(Vậyx=\frac{13}{70}\)
N
10 tháng 7 2017
A. \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)
\(x+1+x+2+....+x+100=5750\)
\(100x+\left(1+2+3+.......+100\right)=5750\)
\(100x+5050=5750\)
\(100x=700\)
\(x=700:100=7\)
B. x+(1+2+......+100) = 2000
x + 5050 = 2000
x = 2000 - 5050
x= -3050
C. ( x-1 )+(x-2)+......+( x - 100 ) = 50
x-1+x-2+.........+x-100 = 50
100x + ( -1-2-........-100 ) = 50
100x + ( - 5050 ) = 50
100x = 50 + 5050
100 x = 5100
x = 5100 : 100
x = 51
DP
11 tháng 7 2017
A . \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=\frac{700}{100}=7\)
B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)
\(x+\frac{\left(100+1\right).100}{2}=2000\)
\(x+5050=2000\)
\(\Rightarrow x=2000-5050=-3050\)
C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)
\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)
\(100x-5050=50\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
TD
23 tháng 1 2017
Bài 1: Tìm x
a) x . (x + 3) = 0
=> \(\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=0-3\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
b) (x -1) (x2 - 1) = 0
=> \(\orbr{\begin{cases}x-1=0\\x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0+1\\x^2=0+1\left(bỏ\right)\end{cases}}\)
=> x = 1
Bài 2: Tìm x, biết
a) -12(x - 5) + 7(3 - x) = 5
-12x - (-12 . 5) + 7 . 3 - 7x = 5
-12x + 60 + 21 - 7x = 5
-12x - 7x = 5 - 21 - 60
-19x = -76
x = -76 : (-19)
x = 4
5 tháng 8 2016
2: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+1\right)-12\)
Đặt \(x^2+x+1=a\)ta có
\(a\left(a+1\right)-12=a^2+a-12=a^2+4a-3a-12=a\left(a+4\right)-3\left(a+4\right)=\left(a+4\right)\left(a-3\right)\)
Thay \(a=x^2+x+1\)ta được
\(\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)Kl...
3. \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+7+8\right)+15\)
Đặt \(x^2+8x+7=a\) Ta có
\(a\left(a+8\right)+15=a^2+8a+15=a^2+5a+3a+15=a\left(a+5\right)+3\left(a+5\right)=\left(a+5\right)\left(a+3\right)\)
Thay \(a=x^2+8x+15\)ta được
\(\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x^2+6x+2x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x+6\right)\left(x+2\right)\left(x^2+8x+10\right)\)
23 tháng 11 2017
1.
A = x2 + x4 + x6 + ... + x100 ( 50 số hạng )
A = ( -1 )2 + ( -1 )4 + ( -1 )6 + ... + ( -1 )100
A = 1 + 1 + 1 + ... + 1
A = 50
2.
| x - 1/3 | + 4/5 = | (-3,2) + 2/50 |
| x - 1/3 | + 4/5 = 3,16
| x - 1/3 | = 2,36
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2,36\\x-\frac{1}{3}=-2,36\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{202}{75}\\x=\frac{-152}{75}\end{cases}}\)