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1) 125.(-24) + 24.225.
\(=24\left(-125+225\right)=24.100=2400.\)
2) 512.(2 – 128) – 128. (-512).
\(=512\left(2-128+128\right)=512.2=1024.\)
a) \(48.19+48.15+134.52\)
\(=48\left(19+15\right)+6968\)
\(=48.34+6968\)
\(=1632+6968\)
\(=8600.\)
b) \(27.121-87.27+73.34\)
\(=27\left(121-87\right)+73.34\)
\(=27.34+73.34\)
\(=34\left(27+73\right)\)
\(=34.100\)
\(=3400.\)
c) \(125.98-125.46-52-25\)
\(=125\left(98-46\right)-52-25\)
\(=125.52-52-25\)
\(=52\left(125-1\right)-25\)
\(=52.124-25\)
\(=6448-25\)
\(=6423.\)
d) \(136.23+136.17-40.36\)
\(=136\left(23+17\right)-40.36\)
\(=136.40-40.36\)
\(=40\left(136-36\right)\)
\(=40.100\)
\(=4000.\)
e) \(66.25+5.66+66.14+33.94\)
\(=33\left(2.50+5.2+14.2+94\right)\)
\(=33\left(50+10+28+94\right)\)
\(=33.182\)
\(=6006.\)
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 !
(-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
