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a: =25x100-150=2500-150=2350
c: \(=520:\left\{515\cdot25\right\}\)
=104/2575
1, 69. (-74) - 31 . 74
= (-69) . 74 - 31 . 74
= 74(-69 - 31)
= 74(-100)
= -7400
2, (-2020) . 35 - 2020 . 65
= 2020 . (-35) - 2020 . 65
= 2020(-35 - 65)
= 2020(-100)
= -202000
3, -23 . 59 - 159 . 23
= 23 . (-59) - 159 . 23
= 23(-59 - 159)
= 23 . (-218)
= -5014
4, -147 . 47 + (-47)(-47) - 300
= -147 . 47 + 47 . 47 - 300
= 47(-147 + 47) - 300
= 47 . (-100) - 300
= -4700 - 300
= -5000
5, (-72) . 30 - 31 . (-30) + 3 + 30
= (-72) . 30 + 31 . 30 + 3 + 30
= 30(-72 + 31 + 1) + 3
= 30 . (-40) + 3
= -1200 + 3
= -1197
6, (-82)(-55) + (-82) + (-46)(-82)
= (-82)(-55 + 1 - 46)
= (-82)(-100)
= 8200
7, (-41) . 135 + 135 . (-24) - 35 . (-65)
= 135(-41 - 24) - 35(-65)
= 135(-65) - 35(-65)
= (-65)(135 - 35)
= -65(100)
= -6500
10, 125 . (-88)
= 1000 . \(\frac{1}{8}\) . (-88)
= 1000 . (-11)
= -11000
Câu 8 với 9 mình ko hiểu nên ko giúp bn được
Chúc bạn học tốt!
1) $(-35)\cdot18+28\cdot35$
$=35\cdot(-18)+28\cdot35$
$=35\cdot(-18+28)$
$=35\cdot10$
$=350$
2) $53\cdot29-47\cdot(-29)$
$=53\cdot29-47\cdot(-1)\cdot29$
$=53\cdot29+47\cdot29$
$=29\cdot(53+47)$
$=29\cdot100$
$=2900$
3) $125\cdot(-24)+24\cdot225$
$=(-125)\cdot24+24\cdot225$
$=24\cdot(-125+225)$
$=24\cdot100$
$=2400$
4) $26\cdot(-121)-121\cdot(-36)$
$=-121\cdot[26+(-36)]$
$=-121\cdot(26-36)$
$=-121\cdot(-10)$
$=121\cdot10$
$=1210$
5) $65\cdot(-25)+25\cdot(-23)$
$=-65\cdot25+25\cdot(-23)$
$=25\cdot(-65-23)$
$=25\cdot(-88)$
$=-2200$
6) $237\cdot(-26)+26\cdot137$
$=-237\cdot26+26\cdot137$
$=26\cdot(-237+137)$
$=26\cdot(-100)$
$=-2600$
7) $30\cdot(-125)+25\cdot30$
$=30\cdot(-125+25)$
$=30\cdot(-100)$
$=-3000$
8) $(-37)\cdot69+(-31)\cdot37$
$=37\cdot(-69)+37\cdot(-31)$
$=37\cdot(-69-31)$
$=37\cdot(-100)$
$=-3700$
9) $31\cdot72-31\cdot70-31\cdot2$
$=31\cdot(72-70-2)$
$=31\cdot(2-2)$
$=31\cdot0$
$=0$
10) $(-12)\cdot47+(-12)\cdot52+(-12)$
$=-12\cdot(47+52+1)$
$=-12\cdot100$
$=-1200$
$\text{#}Toru$
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
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-47 . 69 + 31 . ( -47 ) - ( -155 ) + 5 . ( -32)
= -47. ( 69 + 31 ) +155 + 5. ( -9 )
= -47 . 100 + 155 -45
= -4700 + 155 + -45
= -4950
-47.69+31.(-47)-(-155)+5.(32)
=-47.(69+31)-(-155)+5.9
=47.100-(-155)+45
=4700-(-110)
=4810