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(x-15)+(x-14)+(x-13)+....+(x-1)+19+20=0
(x-15)+(x-14)+(x-13)+....+(x-1)+19 = 0 - 20
(x-15)+(x-14)+(x-13)+....+(x-1)+19 = -20
(x-15)+(x-14)+(x-13)+....+(x-1) = (-20) - 19
(x-15)+(x-14)+(x-13)+....+(x-1) = -39
<=> có 15 cặp như vậy
=> (x+x+x+...+x) - (15+14+13+...+1) = -39
=> 15x - 120 = -39
15x = (-39) + 120
15x = 81
x = 81 : 15
x = 5,4
* = 1 ; 2 ; 3 ; 4 5 ; 6 ; 7 ; 8 ; 9 ; 0
b/ 120 - x : 4 = 34 : 311
120 - x : 4 = 37
120 - x : 4 = 2187
x : 4 = 120 - 2187
x : 4 = -2067
=> x = -8268
a) 3*2 có tận cùng là 2 nên chia hết cho 2
vậy * = 0;1;2 ... 9
b) 120 - x : 4 = \(3^4:3^{11}\)
120 - x : 4 = \(-\left(3^7\right)\)
x : 4 = 120 - \(\left[-\left(3^7\right)\right]\)
x : 4 = 2307
x = 2307 x 4
x = 9228
Theo bài ra ta có:
|x+\(\frac{1}{2}\)|\(\ge\)0
|x+\(\frac{1}{6}\)|\(\ge\)0
............................
|x+\(\frac{1}{110}\)|\(\ge\)0
\(\Rightarrow\)|x+\(\frac{1}{2}\)|+|x+\(\frac{1}{6}\)|+...+|x+\(\frac{1}{110}\)|\(\ge\)0
\(\Rightarrow\)11.x\(\ge\)0
\(\Rightarrow\)x\(\ge\)0
\(\Rightarrow\)x dương.
Khi đó:|x+\(\frac{1}{2}\)|+|x+\(\frac{1}{6}\)|+...+|x+\(\frac{1}{110}\)|=11.x
\(\Rightarrow\)x+\(\frac{1}{2}\)+x+\(\frac{1}{6}\)+...+x+\(\frac{1}{110}\)=11.x
\(\Rightarrow\)27.x+\(\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)=11x
\(\Rightarrow\)\(\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)=-16x
\(\Rightarrow\)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)=-16x
\(\Rightarrow\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)=-16x
\(\Rightarrow\)\(\frac{10}{11}\)=-16x
\(\Rightarrow\)\(\frac{10}{-176}=x\)
Vậy \(x=\frac{10}{-176}\).
\(a,-12.\left(x-5\right)+7.\left(-x+3\right)=5\)
\(-12x+60-7x+21=5\)
\(-19x+81=5\)
\(-19x=5-81\)
\(-19x=-76\)
\(x=4\)
\(b,30.\left(x+2\right)-6.\left(x-5\right)-24.x=100\)
\(30x+60-6x+30-24x=100\)
\(0x+90=100\)
\(0x=100-90\)
\(0x=10\)
=> ko có giá trị nào thõa mãn x
x.30+(1+2+3+......+29+30)=795
x.30+465=795
x.30=795-465
x.30=330
x=330:30
x=11
\(\text{(x+1)+(x+2)+(x+3)+...+(x+29)+(x+30)=795}\)
Số số hạng là:
(30-1):1+1 = 30 ( số hạng )
=> \(\text{x+1+x+2+x+3+...+x+29+x+30=795}\)
Đặt A = 1+2+3+...+30
A = \(\left(\left(30+1\right)\cdot30\right):2\)
A = 465
=> x+x+x+...+x+1+2+3+...+30=795
30x + 465 = 795
30x = 795 - 465 = 330
x = 330 : 30 = 11
Vậy x là 11
đề bài là gì vậy bạn uiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Thui mình giải đại nhennnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
=(x+x+x)+(1/2+2/3+3/6)
=3x+(3/6+4/6+3/6)
=3x+15
Hết ùiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Bye nhennnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn