tìm x thuộc z
a. | x - ( -7) | -2 = -19
b. | x - (-13) | - ( -6) =25
c. | x - ( -31) | - ( -100) = ( -81) + 181
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a: =>x=-7/6+5/8=-13/24
b: =>x=-14/25-3/4=-131/100
c: \(x=\dfrac{-33}{26}:\dfrac{-9}{13}=\dfrac{33}{26}\cdot\dfrac{13}{9}=\dfrac{11}{3}\cdot\dfrac{1}{2}=\dfrac{11}{6}\)
d: \(x=\dfrac{4}{9}:\dfrac{5}{3}=\dfrac{4}{9}\cdot\dfrac{3}{5}=\dfrac{12}{45}=\dfrac{4}{15}\)
1) = 35 . 18 - 35 . 18 = 0
2) = 45 - 5 . 12 + 5 . 9 = 45 - 60 + 45 = 30
3) = 24 . 16 - 24 . 5 - 16 . 24 - 16 . 5
= (24 . 16 - 16 . 24) - 5 (24 - 16)
= 0 - 5 . 8 = 0 - 40 = -40
4) = 29 . 19 - 29 . 13 - 19 . 29 - 19 . 13
= (29 . 19 - 29 . 19) - 13 (29 - 19)
= 0 - 13 . 10 = 0 - 130 = -130
5) = 31 . (-18) + 31 . (-81) - 31 . 1
= 31 [(-18) + (-81) - 1]
= 31 . (-100) = -3100
6) = (-12) . 47 + (-12) . 52 + (-12). 1
= (-12) . (47 + 52 + 1)
= (-12) . 100 = -1200
7) = 13 . 45 - 3 . 45
= 45 (13 - 3)
= 45 . 10 = 450
8) = (-48) + 48 . (-78) + 48 . (-21)
= (-48) . 1 + 48 . (-78) + 48 . (-21)
= 48 . (-1) + 48 . (-78) + 48 . (-21)
= 48 . [(-1) + (-78) + (-21)]
= 48 . (-100) = -4800
1) 35x18-5x7x18
= 35x18-35x18
= 35x(18-18)
= 35x0=0
3) 24.(16-5)-16.(24-5)
= 24.16-24.5-16.24-16.5
=[24.16-16.24]-[24.5-16.5]
= [ 24. (16-16)]- [(24-16).5]
= [24.0]-[8.5]
= 0-40=-40
4) 29.(19-13)-19.(29-13)
=29.19-29.13-19.29-19.13
=[ 29.19-19.29]-[ 29.13-19.13]
= [ 29.(19-19)]-[(29-19).13]
=[29.0]-[10.13]
= 0-130= -130
5) 31.(-18)+31.(-81)-31
= 31.(-18)+31.(-81)-31.1
= 31.[(-18)+(-81)-1]
=31.(-100)
= -3100
6);7);8 tương tự 4);5);6)
a)
\(=\frac{7\cdot7\cdot8\cdot8\cdot9\cdot9\cdot10\cdot10\cdot11\cdot11}{6\cdot8\cdot7\cdot9\cdot8\cdot10\cdot9\cdot11\cdot10\cdot12}\)
\(=\frac{7\cdot11}{6\cdot12}\)
\(=\frac{77}{72}\)
b)
\(=1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)
\(=6+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=6+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=6+\frac{1}{2}-\frac{1}{8}\)
\(=6+\frac{3}{8}\)
\(=\frac{51}{8}\)
Chia thành...a và b nhé.
Bg
a)Ta có: \(\frac{49}{48}.\frac{64}{63}.\frac{81}{80}.\frac{100}{99}.\frac{121}{120}\)
= \(\frac{49.64.81.100.121}{48.63.80.99.120}\)
= \(\frac{7.7.8.8.9.9.10.10.11.11}{6.8.7.9.8.10.9.11.10.12}\)
= \(\frac{7.11}{6.12}\) (chịt tiêu trên dưới)
= \(\frac{77}{72}\)
b) Ta có: \(\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}\)
Có 6 số hạng (đếm)
= \(1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)
= \(1+1+...+1+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
= \(1.6+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
= \(6+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
= \(6+\frac{1}{2}-\frac{1}{8}\)
= \(\frac{13}{2}-\frac{1}{8}\)
= \(\frac{51}{8}\)
Hơi dài....
a) Ta có: -7<x<-1
mà \(x\in Z\)
nên \(x\in\left\{-6;-5;-4;-3;-2\right\}\)
Vậy: \(x\in\left\{-6;-5;-4;-3;-2\right\}\)
b) Ta có: -3<x<3
mà \(x\in Z\)
nên \(x\in\left\{2;1;0;1;2\right\}\)
Vậy: \(x\in\left\{2;1;0;1;2\right\}\)
c) Ta có: \(-1\le x\le6\)
mà \(x\in Z\)
nên \(x\in\left\{-1;0;1;2;3;4;5;6\right\}\)
Vậy: \(x\in\left\{-1;0;1;2;3;4;5;6\right\}\)
d) Ta có: \(-5\le x< 6\)
mà \(x\in Z\)
nên \(x\in\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
Vậy: \(x\in\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
a: Bạn ghi lại đề nha bạn
b: \(30\left(x+2\right)-6\left(x-5\right)-24x=100\)
=>\(30x+60-6x+30-24x=100\)
=>\(\left(30x-6x-24x\right)+\left(60+30\right)=100\)
=>0x=100-90=10(vô lý)
c: \(\left(x-7\right)\left(x+3\right)< 0\)
TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)
=>-3<x<7
mà x nguyên
nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)
d: -1<2x-1<4
=>\(-1+1< 2x< 4+1\)
=>0<2x<5
=>0<x<2,5
mà x nguyên
nên \(x\in\left\{1;2\right\}\)
a. | x - ( -7) | -2 = -19
| x - (-7) | = -19 + 2
| x - (-7) | = -17
vô lí => đề sai
b. | x - (-13) | - ( -6) =25
| x - (-13) | = 25 + (-6)
| x - (-13) | = 19
=> x - (-13) = 19 hoặc x - (-13) = -19
x = 19 + (-13) x = -19 + (-13)
x = 6 x = -32
c. | x - ( -31) | - ( -100) = ( -81) + 181
| x - (-31) | - (-100) = 100
| x - (-31) | = 100+ (-100)
| x - (-31) | = 0
=> x - (-31) = 0
x = 0 + (-31)
x = -31
a.|x-(-7)|-2=-19
|x+7|-2=-19
|x+7|=-19-2
|x+7|=-21
|x|=-21-7
|x|=-28
x thuộc{28,-28}
b.|x-(-13)|-(-6)=25
|x+31|+6=25
|x+13|=25-6
|x+13|=19
|x|=19-13
|x|=6
x thuộc {6,-6}
c.|x-(-31)|-(-100)=(-81)+181
|x+31|+100=100
|x+31|=100-100
|x+31|=0
|x|=0-31
|x|=-31
x thuộc {31,-31}