a. 123 846 + 68 972
b. 345 273 - 56 891
c. 6 980 x 6
d. 56 040 : 4
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x + 34 = 56 - 6
x + 34 = 50
x = 50 - 34
x = 16
y + 56 = 78 + 4
y + 56 = 82
y = 82 - 56
y = 26
x + 32 = 345
x = 345 - 32
x = 313
y + 67 - 6 = 89 - 0
y + 67 - 6 = 89
y + 67 = 89 + 6
y + 67 = 95
y = 95 - 67
y = 28
x + 34 = 56 - 6
x + 34 = 50
x = 50 - 34
x = 16
y + 56 = 78 + 4
y + 56 = 82
y = 82 - 56
y = 26
x + 32 = 345
x = 345 - 32
x = 313
y + 67 - 6 = 89 - 0
y + 67 - 6 = 89
y + 67 = 89 + 6
y + 67 = 95
y = 95 - 67
y = 28
2 + 4 + 6 + 8 + .... + 24 có số số hạng là:
(24 - 2) : 2 + 1 = 12 (số)
tổng là :
(24 + 2) x 12 : 2 = 156
a) 321 + 123 - 23 + 32
= ( 321 + 123 ) - ( 23 + 32 )
= 444 - 55
= 389
b) 345 - 56 + 35
= 345 - ( 56 + 35 )
= 345 - 91
= 254
Trung bình cộng của các số đó là:
( 23+345+56+12) : 4 = 109
Chọn A. 109
123×45+123×55
=123×(45+55)
=123×100
=12300
56×4+56×3+56×2+56
=56×4+56×3+56×2+56×1
=56×(4+3+2+1)
=56×10
=560
[ 123×154-65×123]:89
=[123×(154-65)]:89
= 123×89:89
=123×(89:89)
=123×1
=123
123 x 45 + 123 x 55
= 123 x ( 45 + 55 )
= 123 x 100
= 12300
56 x 4 + 56 x 3 + 56 x 2 + 56
= 56 x ( 4 + 3 + 2 + 1 )
= 56 x 10
= 560
[ 123 x 154 - 65 x 123 ] : 89
= [ 123 x ( 154 - 65 ) ] : 89
= [ 123 x 89 ] : 89
= 123
Học tốt #
B) (-11)+350+16+(-350 )
= - 11 + 350 + 16 - 350 )
= ( - 11 + 16 ) + ( 350 - 350 )
= 5 + 0 =5
C) (-10)+23+(-20)+30
= - 10 + 23 - 20 + 30
= ( - 10 - 20 + 30 ) + 23
= - 30 + 30 + 23
= 0 + 23 = 23
D) (456-123)+123
= 456 - 123 + 123
= 456
E) (-102)-(345-102)
= - 102 - 345 + 102
= ( - 102 + 102 ) - 345
= 0 - 345
= - 345
F) (12-34)-(34-8)
= 12 - 34 - 34 - 8
= (12 - 8 ) + ( -34 - 34 )
= 4 - 68
= 64
G) (12+34-56)-(12-56)
= 12 + 34 - 56 - 12 +56
= ( 12 - 12 ) + ( -56 + 56 ) + 34
= 0 + 0 + 34
= 34
Câu F k giống tính nhanh ____ Có thể là bạn gõ sai đề
@@ Học tốt @@
## Chiyuki Fujito
b ) (-11) + 350 + 16 + (-350)
= (16 - 11 ) + ( 350 - 350 )
= 5 + 0
= 5
c) (-10) + 23 + (-20) + 30
= 23 + ( 30 - 10 - 20 )
= 23 + 0
= 23
d) ( 456 - 123 ) + 123
= 456 - ( 123 - 123 )
= 456 - 0
= 456
e) ( - 102 ) - ( 345 - 102 )
= - 345 - ( 102 - 102 )
= - 345 - 0
= - 345
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.192 819
c: =41880