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\(\left(x+3\right)+\left(x+4\right)+\left(x+5\right)+...+\left(x+22\right)=450\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(3+4+5+...+22\right)=450\) ( 20 số x )
\(\Rightarrow20x+250=450\)
\(\Rightarrow20x=200\)
\(\Rightarrow x=10\)
Vậy \(x=10\)
1, \(\left(3x-6\right)\left(2x-10\right)=0\)
\(\Leftrightarrow3x-6=0or2x-10=0\Leftrightarrow x=3orx=5\)
or là từ '' hoặc ''
2, \(7\left(x+5\right)+10=5x-11\)
\(\Leftrightarrow7x+35+10=5x-11\)
\(\Leftrightarrow7x-5x=-11-10-35\)
\(\Leftrightarrow2x=-56\Leftrightarrow x=-28\)
2:
a: =>x-1/5=2/15
=>x=2/15+3/15=5/15=1/3
b: =>x+7/12=-5/6-2/6=-7/6
=>x=-14/12-7/12=-21/12=-7/4
c: =>x+2/3=-10/3
=>x=-4
d: =>1/4:x=-11/4
=>x=-1/4:11/4=-1/11
e: =>8:x=1,6
=>x=5
1: Ta có: 7x+6(3-x)=27-20+73
\(\Leftrightarrow7x+18-6x=80\)
\(\Leftrightarrow x=80-18=62\)
Vậy: x=62
2: Ta có: \(6x-5\left(x-7\right)=\left(27-514\right)-486-73\)
\(\Leftrightarrow6x-5x+35=27-514-486-73\)
\(\Leftrightarrow x+35=-1046\)
\(\Leftrightarrow x=-1081\)
Vậy: x=-1081
`**x in NN`
`a)x+12 vdots x-4`
`=>x-4+16 vdots x-4`
`=>16 vdots x-4`
`=>x-4 in Ư(16)={+-1,+-2,+-4,+-16}`
`=>x in {3,5,6,2,20}` do `x in NN`
`b)2x+5 vdots x-1`
`=>2x-2+7 vdots x-1`
`=>7 vdots x-1`
`=>x-1 in Ư(7)={+-1,+-7}`
`=>x in {0,2,8}` do `x in NN`
`c)2x+6 vdots 2x-1`
`=>2x-1+7 vdots 2x-1`
`=>7 vdots 2x-1`
`=>2x-1 in Ư(7)={+-1,+-7}`
`=>2x in {0,2,8,-6}`
`=>x in {0,1,4}` do `x in NN`
`d)3x+7 vdots 2x-2`
`=>6x+14 vdots 2x-2`
`=>3(2x-2)+20 vdots 2x-2`
`=>2x-2 in Ư(20)={+-1,+-2,+-4,+-5,+-10,+-20}`
Vì `2x-2` là số chẵn
`=>2x-2 in {+-2,+-4,+-10,+-20}`
`=>x-1 in {+-1,+-2,+-5,+-10}`
`=>x in {0,2,3,6,11}` do `x in NN`
Thử lại ta thấy `x=0,x=2,x=6` loại
`e)5x+12 vdots x-3`
`=>5x-15+17 vdots x-3`
`=>x-3 in Ư(17)={+-1,+-17}`
`=>x in {2,4,20}` do `x in NN`
a) Ta có: \(x+12⋮x-4\)
\(\Leftrightarrow16⋮x-4\)
\(\Leftrightarrow x-4\inƯ\left(16\right)\)
\(\Leftrightarrow x-4\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)
hay \(x\in\left\{5;3;6;2;8;0;12;-4;20;-12\right\}\)
Vậy: \(x\in\left\{0;5;3;6;2;8;20\right\}\)
b) Ta có: \(2x+5⋮x-1\)
\(\Leftrightarrow7⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{2;0;8;-6\right\}\)
Vậy: \(x\in\left\{0;2;8\right\}\)
c) Ta có: \(2x+6⋮2x-1\)
\(\Leftrightarrow7⋮2x-1\)
\(\Leftrightarrow2x-1\inƯ\left(7\right)\)
\(\Leftrightarrow2x-1\in\left\{1;-1;7;-7\right\}\)
\(\Leftrightarrow2x\in\left\{2;0;8;-6\right\}\)
hay \(x\in\left\{1;0;4;-3\right\}\)
Vậy: \(x\in\left\{0;1;4\right\}\)
d) Ta có: \(3x+7⋮2x-2\)
\(\Leftrightarrow6x+14⋮2x-2\)
\(\Leftrightarrow20⋮2x-2\)
\(\Leftrightarrow2x-2\in\left\{1;-1;2;-2;4;-4;5;-5;10;-10;20;-20\right\}\)
\(\Leftrightarrow2x\in\left\{3;1;4;0;6;-2;7;-3;12;-8;22;-18\right\}\)
\(\Leftrightarrow x\in\left\{\dfrac{3}{2};\dfrac{1}{2};2;0;3;-1;\dfrac{7}{2};-\dfrac{3}{2};6;-4;11;-9\right\}\)
Vậy: \(x\in\left\{2;0;3;6;11\right\}\)
e) Ta có: \(5x+12⋮x-3\)
\(\Leftrightarrow27⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;3;-3;9;-9;27;-27\right\}\)
\(\Leftrightarrow x\in\left\{4;2;6;0;12;-6;30;-24\right\}\)
Vậy: \(x\in\left\{4;2;6;0;12;30\right\}\)
`a, (4x+5) : 3 -121:11=14`
`=> (4x+5) : 3 - 11=14`
`=> (4x+5) : 3 =14+11`
`=> (4x+5) : 3 = 25`
`=> 4x+5=25xx3`
`=>4x+5= 75`
`=> 4x=75-5`
`=>4x=70`
`=>x= 70/4`
`=>x= 35/2`
`b, 2+4+6+...+x=2450`
Số lượng của dãy là :
`(x-2)/2 + 1= (x-2)/2 +2/2= x/2`
Tổng số lượng là :
\(\dfrac{\left(x+2\right)\cdot\dfrac{x}{2}}{2}=\dfrac{\dfrac{x^2}{2}+\dfrac{2x}{2}}{2}=\dfrac{x\left(x+2\right)}{4}\)
\(\dfrac{x\left(x+2\right)}{4}=2450\)
\(\Rightarrow x\left(x+2\right)=2450\cdot4\\ \Rightarrow x\left(x+2\right)=9800\)
`=>x=98`
`c,` `1` nhân `32` sao?
`d, 2x+3x=1505`
`=> (2+3)x=1505`
`=> 5x=1505`
`=> x= 1505:5`
`=>x= 301`
a)\(x-15\%x=\frac{1}{3}\)
\(x.\left(1-15\%\right)=\frac{1}{3}\)
\(x.\frac{-280}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{-280}{3}\)
\(x=\frac{-1}{280}\)
Vậy \(x=\frac{-1}{280}\)
b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)
\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)
\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)
\(-\frac{17}{10}x=\frac{-61}{30}\)
\(x=\frac{-61}{30}:\frac{-17}{10}\)
\(x=\frac{61}{51}\)
Vậy \(x=\frac{61}{51}\)
a)\(\frac{x}{5}=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2\times5}{5}=2\)
Vậy .............
b) \(\frac{3}{8}=\frac{6}{x}\)
\(\Leftrightarrow x=\frac{8\times6}{3}=16\)
Vậy ................
c) \(\frac{1}{9}=\frac{x}{27}\)
\(\Leftrightarrow x=\frac{1\times27}{9}=3\)
Vậy ................
d) \(\frac{4}{x}=\frac{8}{6}\)
\(\Leftrightarrow x=\frac{4\times6}{8}=3\)
Vậy ............
e) \(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow3x+4x=20-6\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
Vậy .............
f) \(\frac{x}{-2}=\frac{-8}{x}\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
Vậy ...............
\(\frac{x}{5}=\frac{2}{5}\Rightarrow x=\frac{2\times5}{5}=2\)
\(\frac{1}{9}=\frac{x}{27}\Rightarrow x=\frac{1\times27}{9}=3\)
\(\frac{4}{x}=\frac{8}{6}\Rightarrow x=\frac{8\times6}{8}=6\)
\(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Rightarrow3.\left(x+2\right)=-4.\left(x-5\right)\)
\(\Rightarrow3x+6=-4x+20\)
\(\Rightarrow3x+4x=20-6\)
\(\Rightarrow7x=14\)
\(\Rightarrow x=2\)
TA CÓ :
| X - 4 | = 10
=>\(\orbr{\begin{cases}X-4=10\\X-4=-10\end{cases}}\)
=>\(\orbr{\begin{cases}\\X=-6\end{cases}X=14}\)
a,x=14
b,x=-11
c,x=2