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a) 210 -5.x= 200
<=> 5.x = 210 - 200
<=> 5.x = 10
<=> x = 10 : 5
<=> x = 2
Vậy x = 2
b) 210 - 5 . ( x-10 ) = 200
<=> 5. (x-10) = 210 - 200
<=> 5.(x-10) = 10
<=> x -10 = 10 : 5
<=> x - 10 = 2
<=>x = 12
Vậy x = 12
d)350 : x + 10 = 20
<=> 350 : x = 20 - 10
<=> 350 : x = 10
<=> x = 350 : 10
<=> x = 35
Vậy x = 35
e)36 : ( x - 5 ) = 22
<=> 36 : ( x-5) = 4
<=> x - 5 = 36 : 4
<=> x - 5 = 9
<=> x = 9 + 5
<=> x = 14
Vậy x =14
e) [3 . (70-x) + 5 ] : 2 = 46
<=> [3 . (70-x) + 5 ] = 46 . 2
<=> [3 . (70-x) + 5 ] = 92
<=> 210 - 3.x + 5 = 92
<=>210- 3.x = 92 - 5
<=> 210 - 3.x = 87
<=> 3.x = 210 - 87
<=> 3.x = 123
<=> x = 123 : 3
<=> x = 41
Vậy x = 41
a)210-5x=200
5x=210-200
5x=10
x=10:2
x=5
B)210-5(x-10)=200
5(x-10)=210-200
5(x-10)=10
x-10=10:5=2
x=2+10=12
c)450:[41-(2x-5)]=3 mũ 2 .5
450:[41+5-2x]=45
41+5-2x=450:45
41+5-2x=10
2x=36
x=18
D)210:x-10=20
210:x=20+10
210:x=30
x=210:30
x=7
E)(5x-39).7+3=80
(5x-39).7=77
5x-39=11
5x=50
x=10
F(2x+1):7=2 mũ 2 nhân 3 mũ 2
(2x+1):7=36
2x+1=252
2x=251
x=251/2
\(a,210-5x=200\)
\(5x=10\)
\(x=2\)
Vậy \(x=2\)
\(b,210-5\left(x-10\right)=200\)
\(5\left(x-10\right)=10\)
\(x-10=2\)
\(x=12\)
Vậy \(x=12\)
\(c,450\div\left[41-\left(2x-5\right)\right]=3^2.5\)
\(450\div\left[41-\left(2x-5\right)\right]=45\)
\(41-\left(2x-5\right)=10\)
\(2x-5=31\)
\(2x=36\)
\(x=18\)
Vậy \(x=18\)
\(d,350\div x+10=20\)
\(350\div x=10\)
\(x=35\)
Vậy \(x=35\)
\(e,36\div\left(x-5\right)=2^2\)
\(36\div\left(x-5\right)=4\)
\(x-5=9\)
\(x=14\)
Vậy \(x=14\)
\(f,\left[3\left(70-x\right)+5\right]\div2=46\)
\(3\left(70-x\right)+5=92\)
\(3\left(70-x\right)=87\)
\(70-x=29\)
\(x=41\)
Vậy \(x=41\)
a)210-5x=200 b)210-5(x-10)=200
5x=210-200=10 5(x-10)=210-200=10
x=10:5=2 x-10 =10:5=2 c)45:[41-(2x-5)]=32.5 x =2+10=12
45:[41-(2x-5)] =9.5=45 d)350:x+10=20
41-(2x-5)=45:45=1 350:x =20-10=10
2x-5 =41-1=40 x =350:10=35
2x =40+5=45
x =45:2=22,5
e)36:(x-5)=22 f)[3.(70-x)+5]:2=46
36:(x-5)=4 [3.(70-x)+5] =46.2=92
x-5 =36:4=9 3.(70-x) =92-5=87
x =9+5=14 70-x =87:3=29
x =70-29=41
a) (x+10)(2y-5) = 143
=> (x+10);(2y-5) thuộc Ư(143)={-1,-143,1,143}
\(\orbr{\begin{cases}x+10=-143\\2y-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-153\\y=2\end{cases}}\)
\(\orbr{\begin{cases}x+10=-1\\2y-5=-143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-11\\y=-69\end{cases}}\)
\(\orbr{\begin{cases}x+10=1\\2y-5=143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-9\\y=74\end{cases}}\)
\(\orbr{\begin{cases}x+10=143\\2y-5=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=133\\y=3\end{cases}}\)
Vậy ta có các cặp x,y thõa mãn : (-153,2);(-11,-69);(-9,74);(113,3)
b) x+(x+1)+(x+2)+..+(x+30)=1240
=> (x+x+x+...+x)+(1+2+3+...+30)=1240
=> 31x+465=1240
31x = 1240-465
31x = 775
x = 775 : 31
x= 25
c) 1+2+3+...+x=210
\(\frac{\left(x-1\right)}{1}+1=x\)
=> \(\frac{\left(x+1\right).x}{2}=210\)
(x+1)x = 210:2
(x+1)x = 105
chắc ko có x thõa mãn
d) 2+4+6+...+2x=210
=> 2(1+2+3+...+x)=210
1+2+3+..+x= 210:2 = 105
\(\frac{\left(x-1\right)}{1}+1\) = x
\(\frac{\left(x+1\right).x}{2}=105\)
(x+1)x = 105:2
(x+1)x = 52,5
ko có x thõa mãn đề bài
a, x + 10 và 2y - 5 thuộc Ư(143) = {1;-1;143;-143}
| x + 10 | 1 | -1 | 143 | -143 |
| 2y - 5 | 143 | -143 | 1 | -1 |
| x | -9 | -11 | 133 | -153 |
| y | 74 | -69 | 3 | 2 |
b, x+(x+1)+(x+2)+........+(x+30) = 1240
=> x+x+1+x+2+...+x+30=1240
=> 31x+(1+2+...+30) = 1240
=> 31x + 465 = 1240
=> 31x = 775
=> x = 25
c, 1+2+...+x=210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x(x+1) = 420
Mà 420 = 20.21
=> x = 20
d, 2+4+...+2x = 210
=> 2(1+2+...+x) = 210
=> \(\frac{2x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 210
Mà 210 = 14.15
=> x = 14
e, 1+3+5+...+(2x-1) = 225
=> \(\frac{\left[\left(2x-1\right)+1\right].x}{2}=225\)
=> \(\frac{2x^2}{2}=225\)
=> x2 = \(\left(\pm15\right)^2\)
=> x = 15 hoặc x = -15
a) x = 2.
b) x = 7.
c) x= 12.
d) x= 45.
e) x = 18.
f) x = 10.