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a)73+86+968+914+3032
= (86 + 914) +(968 + 3032) + 73
= 1000 + 4000 + 73
= 5073
b)341.67+341.16+659.83
=341 (67 + 16) + 659 * 83
= 341 * 83 + 659 * 83
= (341+ 659) * 83
= 1000 * 83
= 830000
c)252-84:21+7
= 252 - 2 +7
=257
d)4.8.25.125.27
= (8 * 125) (4* 25) * 27
= 1000 * 100 * 27
= 2700000
e)34+35+36+37-24-25-26-27
= (34-24) + (35-25) + (36-26) + (37-27)
= 10 + 10 + 10 + 10
= 40
g)1-2+3-4+...-98+99
= (1 + 99) + (-2-98) + .....+ (-48 - 52) + (49 + 51) - 50
= 100 - 50
= 50
h)1-4+7-10+...-100+103
Bài 2:Tìm số nguyên x,biết:
l)3636:(12.x-91)=36
3636 : (12x - 91) = 36
12x - 91 = 101
12x = 192
x= 16
m)(x:23+45).67=8911
x: 23 + 45 = 133
x : 23 = 88
x= 88/23
n)[(6.x-39):7].4=12
6x - 39 = 3
6x = 42
x=7
p)128-3(x+4)=23
3(x+4) = 105
x+4= 35
x= 31
q)[(4.x+28).3+55]:5=35
(4x + 28)*3 +55 = 175
(4x + 28) * 3 = 120
4x + 28 = 40
4x = 12
x=3
r)123-5.(x+4)=38
5(x+4) = 85
x+4 = 17
x= 13
t)11-(53+x)=97
53 + x = -86
x= -139
i)23.75+25.23+180
= 23(75+25) + 180
= 2300 + 180
= 2480
`a,`
`31/23-(7/32+8/23)`
`=31/23-7/32-8/23`
`=(31/23-8/23)-7/32`
`=1-7/32=25/32`
`b,`
`38/45-(8/45-17/51-3/11)`
`=38/45-8/45+17/51+3/11`
`= (38/45-8/45)+17/51+3/11`
`=2/3+17/51+3/11`
`=1+3/11=14/11`
`c,`
`(1/3+12/67+13/41)-(79/67-28/41)`
`= 1/3+12/67+13/41-79/67+28/41`
`= 1/3+(12/67-79/67)+(13/41+28/41)`
`= 1/3+(-1)+1=1/3+(-1+1)=1/3+0=1/3`
`d,`
`1/5+(-1/6)+1/7+(-1/8)+1/9+1/8+(-1/7)+1/6+(-1/5)`
`= (1/5+ -1/5)+(-1/6+1/6)+(1/7+ -1/7)+(-1/8 +1/8)+1/9`
`= 0+0+0+0+1/9=1/9 .`
\(a\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)\)
\(=\frac{31}{23}-\frac{7}{23}-\frac{8}{23}\)
\(=\frac{16}{23}\)
Hot
tốt
!
a)\(\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)=\frac{31}{23}-\frac{8}{23}+\frac{7}{23}=1+\frac{7}{23}=1\frac{7}{23}\)
b)\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)=\left(\frac{13}{41}+\frac{28}{41}\right)+\left(\frac{79}{67}-\frac{12}{67}\right)-\frac{1}{3}=1+1-\frac{1}{3}\)
\(=2-\frac{1}{3}=\frac{5}{3}\)
Bài 1:
1) Ta có: \(\left(-12\right)+6\cdot\left(-3\right)\)
\(=-12-18\)
=-30
2) Ta có: \(\left(36-2020\right)+\left(2019-136\right)-27\)
\(=36-2020+2019-136-27\)
\(=1-100-27\)
\(=-126\)
3) Ta có: \(\left(144-97\right)-\left(244-197\right)\)
\(=144-97-244+197\)
\(=-100+100=0\)
4) Ta có: \(\left(-24\right)\cdot13-24\cdot\left(-3\right)\)
\(=-24\cdot13+24\cdot3\)
\(=24\cdot\left(-13+3\right)\)
\(=24\cdot\left(-10\right)=-240\)
5) Ta có: \(54+55+56+57+58-\left(64+65+66+67+68\right)\)
\(=54+55+56+57+58-64-65-66-67-68\)
\(=\left(54-64\right)+\left(55-65\right)+\left(56-66\right)+\left(57-67\right)+\left(58-68\right)\)
\(=\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)\)
=-50
6) Ta có: \(24\cdot\left(16-5\right)-16\cdot\left(24-5\right)\)
\(=24\cdot16-24\cdot5-16\cdot24+16\cdot5\)
\(=-24\cdot5+16\cdot5\)
\(=5\cdot\left(-24+16\right)\)
\(=-5\cdot8=-40\)
7) Ta có: \(47\cdot\left(23+50\right)-23\cdot\left(47+50\right)\)
\(=47\cdot23+47\cdot50-23\cdot47-23\cdot50\)
\(=47\cdot50-23\cdot50\)
\(=50\cdot\left(47-23\right)\)
\(=50\cdot24=1200\)
8) Ta có: \(\left(-31\right)\cdot47+\left(-31\right)\cdot52+\left(-31\right)\)
\(=-31\cdot\left(47+52+1\right)\)
\(=-31\cdot100=-3100\)
Bài 2:
1) Ta có: \(-17-\left(2x-5\right)=-6\)
\(\Leftrightarrow-17-2x+5+6=0\)
\(\Leftrightarrow-2x-6=0\)
\(\Leftrightarrow-2x=6\)
hay x=-3
Vậy: x=-3
2) Ta có: \(10-2\left(4-3x\right)=-4\)
\(\Leftrightarrow10-8+6x+4=0\)
\(\Leftrightarrow6x+6=0\)
\(\Leftrightarrow6x=-6\)
hay x=-1
Vậy: x=-1
3) Ta có: \(-12+3\left(-x+7\right)=-18\)
\(\Leftrightarrow-12-3x+21+18=0\)
\(\Leftrightarrow-3x+27=0\)
\(\Leftrightarrow-3x=-27\)
hay x=9
Vậy: x=9
4) Ta có: \(-45:\left[5\cdot\left(-3-2x\right)\right]=3\)
\(\Leftrightarrow5\cdot\left(-3-2x\right)=-15\)
\(\Leftrightarrow-2x-3=-3\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: x=0
5) Ta có: x(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3\right\}\)
6) Ta có: (x-2)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-4\right\}\)
7) Ta có: \(x\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-1;3\right\}\)
Bài 1:
1) Ta có: (−12)+6⋅(−3)(−12)+6⋅(−3)
=−12−18=−12−18
=-30
2) Ta có: (36−2020)+(2019−136)−27(36−2020)+(2019−136)−27
=36−2020+2019−136−27=36−2020+2019−136−27
=1−100−27=1−100−27
=−126
Tớ chcs cậu học thật giỏi nha !