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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) \(3.5^2+15.2^2-26\div2\)
= 3.25 + 15.4 - 13
= 75 + 60 - 13
= 135 - 13
= 122
b) \(5^3.2-100\div4+2^3.5\)
= 125.2 - 25 + 8.5
= 250 - 25 + 40
= 225 + 40
= 265
c)\(6^2\div9+50.2-3^3.33\)
= 36 : 9 + 100 - 9.33
= 4 + 100 - 297
= 104 - 297
= -193
d)\(3^2.5+2^3.10-81\div3\)
= 9.5 + 8.10 - 27
= 45 + 80 - 27
= 125 - 27
= 98
e) \(5^{13}\div5^{10}-25.2^2\)
= 53 - 25.4
= 125 - 100
= 25
f) \(20\div2^2+5^9\div5^8\)
= 20 : 4 + 5
= 5 + 5
= 10
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: =(21-11)+(22-12)+(23-13)+(24-14)
=10+10+10+10
=40
b: =(34-14)+(35-15)+(36-16)+(37-17)
=20+20+20+20
=80
(-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
a) 80- (4.52 - 3.23)
= 80- ( 208 -69 )
=80+139 quy tắc đổi dấu trừ tước dấu ngoặc
= 219
e: Ta có: \(2448:\left[119-\left(23-6\right)\right]\)
\(=2448:\left(119-23+6\right)\)
\(=2448:102=24\)
Giải;
1) \(347.222-222.\left(216+184\right):8\)
\(=347.222-222.400:8\)
\(=347.222-222.50\)
\(=222.\left(347-50\right)\)
\(=222.297\)
\(=65934\)
2) \(132-\left[116-\left(132-128\right).22\right]\)
\(=132-\left[116-4.22\right]\)
\(=132-\left[116-88\right]\)
\(=132-28\)
\(=104\)
3) \(16:\left\{400:\left[200-\left(37+46.3\right)\right]\right\}\)
\(=16:\left\{400:\left[200-\left(37+138\right)\right]\right\}\)
\(=16:\left\{400:\left[200-175\right]\right\}\)
\(=16:\left\{400:25\right\}\)
\(=16:16\)
\(=1\)
4) \(\left\{184:\left[96-124:31\right]-2\right\}.3651\)
\(=\left\{184:\left[96-4\right]-2\right\}.3651\)
\(=\left\{184:92-2\right\}.3651\)
\(=\left\{2-2\right\}.3651\)
\(=0.3651\)
\(=0\)
5) \(46-\left[\left(16+71.4\right):15\right]-2\)
\(=46-\left[\left(16+284\right):15\right]-2\)
\(=46-\left[300:15\right]-2\)
\(=46-20-2\)
\(=24\)
6) \(3^3.18+72.4^2-41.18\)
\(=18.\left(27-41\right)+72.16\)
\(=18.-14+1152\)
\(=-252+1152\)
\(=900\)
Giải: (tiếp)
7) \(\left(56.46-25.23\right):23\)
\(=\left(2576-575\right):23\)
\(=2001:23\)
\(=87\)
8) \(\left(28.54+56.36\right):21:2\)
\(=\left(1512+2016\right):21:2\)
\(=3528:21:2\)
\(=84\)
9) \(\left(76.34-19.64\right):\left(38.9\right)\)
\(=\left(2584-1216\right):342\)
\(=1368:342\)
\(=4\)
10) \(\left(2+4+6+...+100\right).\left(36.333-108.111\right)\)
\(=\left(2+4+6+...+100\right).\left(11988-11988\right)\)
\(=\left(2+4+6+...+100\right).0\)
\(=0\)
11) \(\left(5.4^{11}-3.16^5\right):4^{10}\)
\(=5.4^{11}:4^{10}-3.16^5:4^{10}\)
\(=5.4-3.1\)
\(=20-3\)
\(=17\)
12) \(7256.4375-725:3650+4375.7255\)
\(=4375.\left(7256+7255\right)-\dfrac{29}{146}\)
\(=4375.14511-\dfrac{29}{146}\)
\(=63485624,8\)
Câu 12 ko chắc!
\(5\cdot2^2\cdot2^3-4\left(5^8\cdot5^6\right)\)
\(=5\cdot2^{2+3}-4\left(5^{8+6}\right)=5\cdot2^5-4\cdot5^{14}\)
\(=5\cdot32-4\cdot6103515625=160-4\cdot6103515625\)
\(=160-24414062500=24414062340\)