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a) 35 - 102 : 52 + 23 . 7
= 3^5 - 2^2 +2^3 .7
= 2^2.(3^5 + 2.7)
=4.(243 +14)
= 4.257=1028
b) 24 - 5 . 23 - 2 . 33
= 2^3 (2-5)-2. 27
=8. (-3) - 54
= -24 -54
=-78
c) 36 . 57 - 36 . 47 + 63 . 62 - 63 . 52
=36 (57-47) + 63 (62-62)
=36.10+63.10
=10.(36+63)
=10.99
=990
d) 24 . 5 - { 140 - [ 868 - 12 ( 3087 : 72 + 40 ) ] }
=16 . 5 - { 140 - [ 868 - 12 ( 3087 : 49 + 1) ] }
=80 - { 140 - [ 868 - 12 ( 63+ 1) ] }
=80 - { 140 - [ 868 - 12.64 ] }
=80 - { 140 - [ 868 - 768 ] }
=80 - { 140 - 100}
=80 - 40 =40
e) 723 . 542
____________
108
(23. 32 ) 3 .(2.33)2
___________________
22.33
bạn xem có đúng ko nha
29. 36 . 22.36
___________________
22.33
211. 312
___________
22.33
= 28. 39
=256.19683
=5038848
f) 310 . 11 + 310 . 5
__________________
39 . 24
310 . (11 + 5)
__________________
39 . 24
310 . 16
_______
39 . 24
310 . 24
_______
39 . 24
=3
Lời giải:
$22+23-25+27-29+31-33$
$=22+(23-25)+(27-29)+(31-33)$
$=22+(-2)+(-2)+(-2)=22+(-2).3=22-6=16$
a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
a, 125.(-61).(-2)3 .(-1)2n
125.(-61).(-8).1
125.61.8
61.(125.8)
61.1000
61000
a: =34x(-100)=-3400
b: =41(-76-24)=-4100
c: =-24(125-225)=2400
d: =26(-237+137)=-2600
e: =25(-63+23)=25x(-40)=-1000
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 !
Các bạn ghi mỗi đáp số từng câu thôi r sau đó nhắn tin trả lời 4 lần nữa nhắn gì cx đc để mình tick cho
bài 1
tỉ số phần trăm của \(\frac{58}{7}\) và 63 là
\(\frac{58}{7}:63.100=\frac{5800}{441}\)
a) 33 - 102 : 52 + 23 . 7
= 27 - 54 : 52 + 8.7
= 27 - 52 + 56
= 27 - 25 + 56
= 2 + 56
= 58
a) 79
b) 2
c) 990
d) = 80 - {140 - [868-12(64)]}
= 80 - (140-100)
= 80- 40
= 40
mk ko bít bạn có cần trình bày ko nên mk vít kq hoi