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Tìm x, biết:
a.
75883-(31200+x)=999
31200+x =75883-999
31200+x= 74884
x = 74884-31200
x = 43684
b.
24 : ( x +1 ) = 8
x+1=24:8
x+1=3
x=3-1
x=2
c.
1+2+3+...+x=55
Dãy trên có số số hạng là:
(x−1):1+1=x−1+1=x
Ta có:
(x+1).x:2=55
(x+1).x=55.2
(x+1).x=110
(x+1).x=11.10
⇒x=10
a, ---> 31200 + x = 75883 - 999
---> 31200 + x = 74884
---> x = 74884 - 31200 = 43684
b, ---> x + 1 = 3
---> x = 2
c, Số số hạng là : x
Tổng = ( x + 1 ) * x : 2 = 55
---> x ( x + 1 ) = 110
---> 10 ( 10 + 1 ) = 110
---> x = 10
xin tiick
1)
\(\left(a\right)37+397+3997+39997\)
\(=40-3+400-3+4000-3+40000-3\)
\(=\left(40+400+4000+40000\right)-\left(3+3+3+3\right)\)
\(=44440-12=44428\)
\(\left(b\right)298+2998+29998+299998\)
\(=300-2+3000-2+30000-2+300000-2\)
\(=\left(300+3000+30000+300000\right)-\left(2+2+2+2\right)\)
\(=333300-8=333296\)
\(\left(c\right)9+99+999+9999+99999\)
\(=10-1+100-1+1000-1+10000-1+100000-1\)
\(=\left(10+100+1000+10000+100000\right)-\left(1+1+1+1+1\right)\)
\(=111110-5=111105\)
2)
\(\left(a\right)\left(2+4+6+...+2002+2004+2006\right)-\left(1+3+5+...+2001+2003+2005\right)\)
\(=\left(2-1\right)+\left(4-3\right)+\left(6-5\right)+...+\left(2002-2001\right)+\left(2004-2003\right)+\left(2006-2005\right)\)
\(=1+1+1+...+1+1+1\)( 1003 số 1 )
\(=1003\)
\(\left(b\right)88-87+86-85+84-83+...+6-5+4-3+2-1\)
\(=\left(88-87\right)+\left(86-85\right)+\left(84-83\right)+...+\left(6-5\right)+\left(4-3\right)+\left(2-1\right)\)
\(=1+1+1+...+1+1+1\)( 44 số 1 )
\(=44\)
\(\left(c\right)100-98+96-94+92-90+...+12-10+8-6+4-2\)
\(=\left(100-98\right)+\left(96-94\right)+\left(92-90\right)+...+\left(12-10\right)+\left(8-6\right)+\left(4-2\right)\)
\(=2+2+2+...+2+2+2\) ( 25 số 2 )
\(=50\)
3)
\(\left(a\right)360-357+354-351+348-345+...+312-309+306-303+300-297\)
\(=\left(360-357\right)+\left(354-351\right)+\left(348-345\right)+...+\left(312-309\right)+\left(306-303\right)+\)\(\left(300-297\right)\)
\(=3+3+3+3+3+3+3+3+3+3+3=33\)
\(\left(b\right)2006-1-2-3-4-...-47-48-49-50\)
\(=2006-\left(1+2+3+4+...+47+48+49+50\right)\)
\(=2006-\frac{\left(50+1\right)\left[\left(50-1\right)+1\right]}{2}\)
\(=2006-1275=731\)
\(\left(c\right)280-276+272-268+264-260+...+216-212+208-204+200-196\)
\(=\left(280-276\right)+\left(272-268\right)+\left(264-260\right)+...+\left(216-212\right)+\left(208-204\right)+\)\(\left(200-196\right)\)
\(=4+4+4+4+4+4+4+4+4+4+4=44\)
\(3\dfrac{2}{5}\cdot1\dfrac{4}{7}=\dfrac{17}{5}\cdot\dfrac{11}{7}=\dfrac{187}{35}\)
a)34-2:(3/5-1/12)
=32:31/60
=1920/31
b)12 1/3-(3 3/4+4 3/4)
=4-(9/4+3)
=4-(-3/4)
=4+3/4
=-1,25
đề = \(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)( áp dụng c.thức tính tổng )
= ..........
= 2 .( \(\frac{1}{2}-\frac{1}{51}\))
= dễ
Khoảng cách : `1`
Số số hạng là : `(999-1):1+1=999`
Tổng là : `(999+1) xx 999 : 2=499500`
sos