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\(x+2x+3x+4x+.....+2011x=2012.2013\)
\(1x+2x+3x+4x+.....+2011x=2012.2013\)
\(1x+2x+3x+4x+.....+2011x=4050156\)
\(x(1+2+3+4+.....+2011)=4050156\)
\(x.2011.(2011+1):2=4050156\)
\(x.2023066=4050156\)
\(x=4050156:2023066\)
\(x=2...............\)
Ta có
x + 2x + 3x + 4x + 5x + ... + 2011x = 2012.2013
x + 2x + 3x + 4x + 5x + ... + 2011x = 4050156
x(2 + 3 + 4 + ... + 2011) = 4050156
x.2023066 = 40501156
x = 40501156 : 2023066
x = 20,...
\(\dfrac{2\text{x}-1}{3}=\dfrac{3\text{x}+1}{4}\)
\(\Leftrightarrow=\dfrac{4\left(2\text{x}-1\right)}{12}=\dfrac{3\left(3\text{x}+1\right)}{12}\)
\(\Leftrightarrow8\text{x}-4=9\text{x}+3\)
\(\Leftrightarrow8\text{x}-9\text{x}=3+4\)
\(\Leftrightarrow-x=7\)
\(\Leftrightarrow x=-7\)
Theo đề, ta có: \(\dfrac{1+2x}{18}=\dfrac{1+4x}{34}\)
\(\Leftrightarrow34\left(1+2x\right)=18\left(1+4x\right)\)
\(\Leftrightarrow34+68x=18+72x\)
\(\Leftrightarrow34-18=72x-68x\)
\(\Leftrightarrow16=4x\)
\(\Leftrightarrow x=4\)
Khi \(x=4\) vào ta có: \(\dfrac{1+4.4}{34}=\dfrac{1+6.4}{2y^2}\Leftrightarrow\dfrac{1}{2}=\dfrac{25}{2y^2}\)
\(\Leftrightarrow2y^2=50\)
\(\Leftrightarrow y^2=50\)
\(\Leftrightarrow y=\pm5\)
a)2x-35=15
2x=15+35
2x=50
x=50:2
x=25
b)3x+17=2
3x=2-17
3x=-15
x=-15:3
x=-5
c)|x-1|=0
=>x-1=0=>x=0-1
=>x=-1
a)2x-35=15
2x=15+35
2x=50
x=50:2
x=25
b)3x+17=2
3x=2-17
3x=-15
x=-15:3
x=-5
c)|x-1|=0
x-1=0
x=0+1
x=1
\(x+2x+3x+4x+35=-65\Rightarrow (1+2+3+4)x=-65-35\Rightarrow 10x = -100 \Rightarrow x=-10\)
264+12x=1800
12x=1536
x=128
(x+73):19=321
x+73=6099
x=6026
2x+3x+4x=18
9x=18
x=2
10<2x<20
=>5<x<10
x=6,7,8,9
\(264+12x=1800\)
\(\Leftrightarrow12x=1536\)
\(\Leftrightarrow x=128\)
==============
\(\left(x+73\right):19=321\)
\(\Leftrightarrow x+73=6099\)
\(\Leftrightarrow x=6026\)
===============
\(2x+3x+4x=18\)
\(\Leftrightarrow x.\left(2+3+4\right)=18\)
\(\Leftrightarrow x.9=18\)
\(\Leftrightarrow x=2\)
================
\(10< 2x< 20\)
..................................
\(2x=12\Leftrightarrow x=6\)
\(2x=14\Leftrightarrow x=7\)
\(2x=16\Leftrightarrow x=8\)
\(2x=18\Leftrightarrow x=9\)
\(x\in\left\{6;7;8;9\right\}\)
12x = 1800 - 264
12x = 1536
x = 1536 : 12
x = 128
( x + 73 ) : 19 = 321
( x + 73 ) = 321 x 19
x + 73 = 6099
x = 6099 - 73
x = 6026
2x + 3x + 4x = 18
x . ( 2 + 3 + 4 ) = 18
x . 9 = 18
x = 18 : 9
x = 2
10 < 2x < 20
x = 6 ; 7 ; 8 ; 9
Ta có : \(\frac{2x+5}{x+1}=\frac{2x+2+3}{x+1}=\frac{2\left(x+1\right)+3}{x+1}=2+\frac{3}{x+1}\)
Vì 2 \(\inℤ\Rightarrow\frac{3}{x+1}\inℤ\Rightarrow3⋮x+1\Rightarrow x+1\inƯ\left(3\right)\Rightarrow x+1\in\left\{1;3;-1;-3\right\}\)
=> \(x\in\left\{0;2-2;-4\right\}\)
Để \(\frac{3x-1}{2x-1}\inℤ\Rightarrow3x-1⋮2x-1\Rightarrow2\left(3x-1\right)⋮2x-1\Rightarrow6x-2⋮2x-1\)
=> \(6x-3+1⋮2x-1\Rightarrow3\left(2x-1\right)+1⋮2x-1\)
Vì \(3\left(2x-1\right)⋮2x-1\)
=> \(1⋮2x-1\Rightarrow2x-1\inƯ\left(1\right)\Rightarrow2x-1\in\left\{1;-1\right\}\Rightarrow x\in\left\{1;0\right\}\)
\(\frac{2x+5}{x+1}=\frac{2\left(x+1\right)+3}{x+1}=2+\frac{3}{x+1}\)
Để phân số nguyên => \(\frac{3}{x+1}\)nguyên
=> \(3⋮x+1\)
=> \(x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
=> \(x=\left\{0;-2;2;-4\right\}\)
\(\frac{3x-1}{2x-1}\)
Để phân số nguyên => \(3x-1⋮2x-1\)
=> \(2\left(3x-1\right)⋮2x-1\)
=> \(6x-2⋮2x-1\)
\(\Rightarrow3\left(2x-1\right)+1⋮2x-1\)
\(\Rightarrow1⋮2x-1\)
\(\Rightarrow2x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Rightarrow x=\left\{1;0\right\}\)
=\(\frac{4026}{2011}\)