(x+1) + (x+2) +....+(x+20) giai nhanh nhak minh gap lam
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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 \(x=\frac{10}{-176}\).
TA CÓ:\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}\)
\(=1-\frac{1}{8}=\frac{7}{8}\)
( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x ( 1212 x 27 - 2727 x 12 )
= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x ( 32724 -32724 )
= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x 0
= 0