thực hiện phép tính một cách hợp lý:
(1+2+3+..........+2009+2010)*(1176*72+15*1176-1176*87)=?
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\(M=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+\frac{4.5}{2}+...+\frac{48.49}{2}+\frac{49.50}{2}=\frac{1.2+2.3+3.4+4.5+...+48.49+49.50}{2}\)
Dặt\(A=1.2+2.3+3.4+...+48.49+49.50\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+48.49.3+49.50.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+48.49.\left(50-47\right)+49.50.\left(51-48\right)\)
\(=1.2.3+\left(2.3.4-1.2.3\right)+\left(3.4.5-2.3.4\right)+...+\left(48.49.50-47.48.49\right)+\left(49.50.51-48.49.50\right)\)
\(=\left(1.2.3+2.3.4+3.4.5+...+48.49.50+49.50.51\right)-\left(1.2.3+2.3.4+...+47.48.49+48.49.50\right)\)
\(\Rightarrow3A=49.50.51=124950\Rightarrow A=41650\)
Mà\(M=\frac{A}{2}\)
\(\Rightarrow M=\frac{41650}{2}=20825\)
Vậy M=20825
Mk đặt =M cho ngắn gọn
Bài 3:
= 1- 1/2 + 1/2 -1/3 +...+ 1/98 -1/99
= 1- 1/99
= 98/99
Bài 4:
= 1/2*3 + 1/3*4 + 1/4*5 +...+ 1/10*11
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/10 - 1/11
= 1/2 - 1/11= 9/22
Ta đặt A=\(-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(\Rightarrow A=-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)= \(-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
= - \(\left(1-\dfrac{1}{10}\right)=-\left(\dfrac{10-1}{10}\right)=-\dfrac{9}{10}\)
Ta có: \(-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)
\(=-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}\right)\)
\(=-\left(-\dfrac{1}{10}+1\right)\)
\(=-\left(1-\dfrac{1}{10}\right)\)
\(=-\left(\dfrac{10}{10}-\dfrac{1}{10}\right)=-\dfrac{9}{10}\)
Câu 1:
(-13) + (-9) + 257 + (-3)
= -22 + 257 + (-3)
= 235 + (-3)
= 232
Câu 2:
a) (88 - 35) - (28 + 35)
= 88 - 35 - 28 - 35
= (88 - 28) - (-35 - 35)
= 60 + 70 = 130
Câu a này ko tính nhanh đc nhỉ
b) (1176 - 102) - (102 + 1176)
= 1176 - 102 - 102 - 1176
= -102 - 102 = -204
c) (576 - 319) - (76 - 319)
= 576 - 319 - 76 + 319
= 576 - 76 = 500
d) (49 + 73) + (2010 - 49 - 73)
= 49 + 73 + 2010 - 49 - 73
= (49 - 49) + (73 - 73) + 2010
= 2010
Câu 3:
a,b) đã có ng làm
c) -18 - (2 - x) = 4
=> 2 - x = -18 - 4
=> 2 - x = -22
=> x = 2 + 22
=> x = 24
Vậy x = 24.
1.a) 4.(12+88) = 4.100=400
b)72- 6 .[2.8:8 ]=72-6.[16:8]=72-6 .2= 49-12=37 ( tui ko chắc lắm )
2.a) x=26-11=15
b) (4x-15)3=53
4x-15=5
4x= 5+15
4x=20
vậy x = 20
Bài 2:
a: Ta có: x+11=26
nên x=15
b: Ta có: \(\left(4x-15\right)^3=125\)
\(\Leftrightarrow4x-15=5\)
hay x=5
= (1+2+3+..........+2009+2010)*(1176*[72+15-87])
=(1+2+3+...........+2009+2010)*(1176*0)
=(1+2+3+...........+2009+2010)*0
=0
\(=\left(1+2+3+...+2010\right)\cdot\left(1176\cdot72+15\cdot1176-1176\cdot87\right)\)
\(=\left(1+2+3+...+2010\right)\cdot\left(1176\cdot72-1176\cdot\left(87-15\right)\right)\)
\(=\left(1+2+3+...+2010\right)\cdot\left(1176\cdot72-1176\cdot72\right)\)
\(=\left(1+2+3+...+2010\right)\cdot0=0\)