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\(C=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{870}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{29.30}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{29}-\frac{1}{30}\)
\(=1-\frac{1}{30}\)
\(=\frac{29}{30}\)
a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
Bài 1. Tính hợp lý ( nếu có thể ) :
a) ( 509 - 42 ) - ( 509 + 158 )
= 509 - 42 - 509 - 158
=(509 - 509) - ( 42+ 158)
= 0 - 200
= -200
b) - ( 510 - 427 ) - 473 + 510
= -510 +427 -473 +510
= (510-510) + (427-473)
= 0 + (-46)
=-46
c) 1995 - (-2021 + 1994) - 1
= 1995 +2021 -1994 -1
=( 1995 -1994) + (2021-1)
= 1+2020
= 2021
d) 2020 - [ - 1079 -(-1179 + 3020 )]
= 2020 - [ -1079 +1179-3020]
= 2020 +1079 -1179 +3020
=(2020+3020) + (1079 - 1179)
= 5040 +(-100)
=4940
e) 12+(-47).12-6.(-12)
= 12.1 + (-47) .12 -(-6).12
= 12.1 -47 .12 +6 .12
= 12. (1-47+6)
=12.(-40)
=-480
g) 32.(132-247)-132.(32-247)
=32.132 -32.247 - (132.32 - 132.247)
= 32.132 -32.247 - 132.32 + 132.247
= ( 32.132 - 132.32) + ( 132.247 - 32.247)
= 0 + 247.(132-32)
= 247.100
= 24700
i) (-25) .68+(-34).(-250)
= ( -25) . ( -2) . ( -34) + ( -34) . ( -250)
= 50 . (-34) + ( -34) . (-250)
= (-34) . [ 50 + ( - 250) ]
= ( - 34 ) . ( -200)
= 6800
(870 – 1.2).(870 – 2.3).(870 – 3.4) … (870 – 99.100)
Ta có: 870 = 29.30
Nên suy ra: 870 – 29.30 = 29.30 – 29.30 = 0
G = 0.
k cho mik nha, cô mik giảng vậy
a) Ta có: \(14-\left(5-8\right)^3+\left(-3\right)\cdot5\)
\(=14+27-15=26\)
b) Ta có: \(-7\cdot15+7\cdot\left(-35\right)+\left(-1\right)^{2019}\)
\(=-7\left(15+35\right)-1=-350-1=-351\)
c) Ta có: \(-18\cdot32+\left(-18\right)\cdot45+77\cdot\left(-32\right)-77\cdot50\)
\(=-18\left(32+45\right)+77\left(-32-50\right)\)
\(=-18\cdot77+77\cdot\left(-82\right)=77\left(-18-82\right)=77\cdot\left(-100\right)=-7700\)
d) Ta có: \(104-4\cdot\left[-5\cdot8+\left(7-10\right)^3\right]\)
\(=104-4\left[-40-27\right]=104-4\cdot\left(-67\right)=104+268=372\)
a)C= (-124)+ (36 + 124 - 99 ) - ( 136 - 1 ) = (-124) +36 +124 -99 - 136 +1
= -198
b) D = { 115+[ 32 - ( 132 -5 )] } +(-25) +(-25)
= {115+[32-132+5]} +(-25) + (-25)
= {115+(-95)} + (-25)+ (-25)
= 20 +(-25) +(-25) = -30
c)F = [(123 - 17 ) - (123 + 33 ) ] - { 34 - [ 34 + ( 57 -50 ) -7 ] }
= [123 -17 -123 -33 ] - {34- [34 + 57-50 -7]}
= -50 - {34-34} = -50 - 34 +34 = -50
d) mik bo nha
e) E = (-7) +( -2020) -(-7) +2020
= [(-7)-(-7)]+[(-2020) +2020]= 0
g)B= (-2019)-( 29 - 2019) = (-2019) - 29 +2019 = -29
co moi cau d mik ko biet lam :v