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a) [ 316 – ( 25 . 4 + 16 )] : 8 – 24
=( 316 – 116 ) : 8 – 24 = 200 ∶ 8 – 24 = 25 – 24 = 1
b) | -15| + (-27) + 8 + | - 23|
= 15 – 27 + 8 + 23 = 19
c) 5 8 : 5 6 + 2 2 . 3 3 - 2010 0 = 5 2 + 4 . 27 – 1 = 25 + 108 – 1 = 132
a) 5.22 + (x + 3) = 52
5.4 + (x + 3) = 25
20 + (x + 3) = 25
x + 3 = 25 – 20
x + 3 = 5
x = 5 – 3 = 2
b) 23 + (x – 32) = 53 - 43
8 + (x – 9) = 125 – 64
8 + (x – 9) = 61
x – 9 = 61 – 8
x – 9 = 53
x = 53 + 9 = 62
a) \(5.2^2+\left(x+3\right)=5^2\)
\(x+3=5^2-5.2^2\)
\(x+3=25-20\)
\(x+3=5\)
\(x=2\)
b) \(2^3+\left(x-3^2\right)=5^3-4^3\)
\(8+\left(x-9\right)=125-64\)
\(x-9=53\)
\(x=62\)
(-17) + 42 +(-183) + 58
=[(-17)+(-183)] + (42+58)
=(-200) + 100
=(-100)
cho mik rùi mik giải tiếp
( -17 ) + 42 + ( -183 ) + 58
= ( -17 ) + ( -183 ) + ( 42 + 58 )
= -200 + 100
= -100
125 ( -21 ) + 21 . 225
= 125 ( -21 ) - ( -21 ) 225
= -21 ( 125 - 225 )
= -21 . ( -100 )
= 2100
37 ( 13 - 5 ) - 13 ( 37 - 5 )
= 37.13 -37.5 - 13.37 - 13.5
= ( 37.13 - 13.37 ) - ( 37.5 + 13.5 )
= 0 - 5 ( 37 + 13 )
= 0 - 5. 50
= -250
25 - 25 \(\orbr{ }\)23 + 7( -59 + 56 ) + \(|\)-12\(|\):( -4 )
25 -25 ( 23 + 7 . -3 ) + 12 : (-4)
25 - 25 . -2 + -3
= 25 -( -50 ) + ( -3 )
= 72
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 !
a) \(227+50+23=\left(227+23\right)+50=250+50=300\)
b) \(135+360+65+40=\left(135+65\right)+\left(360+40\right)=200+400=600\)
c) \(1+2+3+4+5+...+97+98+99+100\)
\(=\left(100+1\right)+\left(99+2\right)+...+\left(50+51\right)\)
\(=101+101+101+...+101\)
\(=101\cdot50\)
\(\Leftrightarrow5050\)
d) \(115\cdot13-13\cdot15=13\cdot\left(115-15\right)=13\cdot100=1300\)
e) \(50-49+48-47+...+4-3+2-1\)
\(=\left(50-49\right)+\left(48-47\right)+...+\left(2-1\right)\)
\(=1+1+1+1+..+1\)
\(=1\cdot25\)
\(=25\)
f) \(30\cdot40\cdot50\cdot60=10\cdot3+10\cdot4+10\cdot5+10\cdot6\)
\(=10\cdot10\cdot10\cdot10\cdot3\cdot4\cdot5\cdot6\)
\(=10000\cdot360\)
\(=3600000\)
g) \(27\cdot36+27\cdot64=27\cdot\left(36+64\right)=27\cdot100=2700\)
h) \(5\cdot2^2-18:3=5\cdot4-18:3=20-6=14\)
i) \(13\cdot17-256:16+14:7-2021^0\)
\(=13\cdot17-4^4:4^2+2-1\)
\(=13\cdot17-16+2-1\)
\(=13\cdot17-17\)
\(=17\cdot\left(13-1\right)\)
\(=204\)
j) \(7^2-36:3=49-12=37\)
a) – 1 < x < 49 18 => x ∈ {0;1;2}
b) − 1 24 < x 24 < 1 14 => x = 0
c) 2 3 ≤ x ≤ 26 5 => x ∈ {1;2;3;4;5}
d) 0 < x ≤ 1 => x = 1
2522