353^2+94.353+47^2
146^2-112.146+56^2
Những câu hỏi liên quan
2+5+11+...+47+56 . Tìm tổng
Ta có :
5 = 2 + 3
11 = 5 + 3 . 2
20 = 11 + 3 . 3
32 = 20 + 3 . 4
47 = 32 + 3 . 5
65 = 47 + 3 . 6
Vậy : 2 + 5 + 11 + 20 + 32 + 47 + 56
= 2 + ( 11 + 32 + 27 ) + 20 + ( 5 + 65 )
= 2 + 90 + 20 + 70
= 182
Bỏ ngoặc rồi tính: 1. (56+47)-(56-53+40) 2. (34+59)-(34-41-100) 3. (109+34)-(109-66+10)
1.\(=56+47-56+53-40\)
\(=\left(56-56\right)+\left(47+53\right)-40\)
\(=0+100-40=60\)
2.\(34+59-34+41+100\)
\(=\left(34-34\right)+\left(59+41\right)+100\)
\(=0+100+100=200\)
3.\(=109+34-109+66-10\)
\(=\left(109-109\right)+\left(34+66\right)-10\)
\(=0+100-10=90\)
Đúng 2
Bình luận (0)
Tính:
\(\left(2\frac{1}{47}-9\frac{3}{56}+7\frac{6}{121}\right):\left(\frac{5}{47}-\frac{15}{56}+\frac{30}{121}\right)\)
Bài 2:Bỏ ngoặc rồi tính hợp lí
1.-(-66)+34+(-73)-27+99
2.54-(45-30+54)-30+5
3.42-56+33-(33-56+42)
4.50-(47+50-18)+(47-18)
5.10-(28-30+4)
6.-(-73)+(44-94+27)+94
1: =66+34-73-27+99=99
2: =54-45+30-54-30+5=-40
3: =42-56+33-33+56-42=0
4: =50-47-50+18+47-18=0
5: =10-28+30-4=2-4+10=12-4=8
Đúng 0
Bình luận (0)
\(\text{1.-(-66)+34+(-73)-27+99}.\)
\(=66+34-73-27+99=100-100+99=-99.\)
\(\text{2.54-(45-30+54)-30+5}.\)
\(54-45+30-54-30+5=\left(54-54\right)+\left(-45+5\right)+\left(30-30\right)=-40.\)
\(\text{3.42-56+33-(33-56+42)}.\)
\(=42-56+33-33+56-42=\left(42-42\right)+\left(-56+56\right)+\left(33-33\right)=0.\)
\(\text{4.50-(47+50-18)+(47-18)}.\)
\(=50-47-50+18+47-18=\left(50-50\right)+\left(-47+47\right)+\left(18-18\right)=0.\)
\(\text{5.10-(28-30+4)}.\)
\(=10-28+30-4=10+30-20-8-4=20-12=8.\)
\(\text{6.-(-73)+(44-94+27)+94}.\)
\(=73+44-94+27+94=\left(73+27\right)+\left(-94+94\right)+44=100+44=144.\)
Đúng 0
Bình luận (0)
B1.Thực hiện phép tính (tính nhanh nếu có thể)
1)195-(2.33 - 100 : 52)
2)-47.43 + (-47).56 + (-47)
1) 195-(2.3\(^3\) - 100 : 5\(^2\))
= 195-(2.27 - 100 : 25)
= 195-(54 - 4)
= 195-50
= 145
2) -47.43 + (-47).56 + (-47)
= -47.43 + (-47).56 + (-47).1
= -47.(43+56+1)
= -47.100
= -4700
1) Ta có: \(195-\left(2\cdot3^3-\frac{100}{5^2}\right)\)
\(=195-\left(54-4\right)=195-50=145\)
2) Ta có: \(-47\cdot43+\left(-47\right)\cdot56+\left(-47\right)\)
\(=-47\left(43+56+1\right)=-47\cdot100=-4700\)
Tìm số nguyên x, biết:a)
5
6
x
−
1
x
b)
1
2
x
+
1
3
x
c)
3
x
+
2
5...
Đọc tiếp
Tìm số nguyên x, biết:
a) 5 6 = x − 1 x
b) 1 2 = x + 1 3 x
c) 3 x + 2 = 5 2 x + 1
d) 5 8 x − 2 = − 4 7 − x
1-2-2-1-2-1-2-1-2-1--45-4356-42-------32423432---5-45-54535-435-34-5-35-2-5-4-52-5-54-353--3-5--5-34-5-3--4-47-87-8-787-9-76-8--34-54-6----34-3-31-4-31-423-34-1-134-41-4232-431-321------------546-3----------------2315-413-513465146-456-4-164-316361-4363411111111111111=
-4.3634111e+18
HT
@acquybemon
Tính hợp lí:
1) -(-66)+34+(-73)-27+99
2) 54-(45-30+54)-30+5
3)42-56+33-(33+56+42)
4)50-(47+50-18)+(47-18)
5)|25-24-10|-(28-30+4)
Tính bằng cách hợp lí (nếu có thể):
a
,
(
-
17
+
91
)
-
(
91
-
-
17
-
2011
)
b
,
4
7
+
5
6
:
5
-
0
,
375
.
(...
Đọc tiếp
Tính bằng cách hợp lí (nếu có thể):
a , ( - 17 + 91 ) - ( 91 - - 17 - 2011 ) b , 4 7 + 5 6 : 5 - 0 , 375 . ( - 2 ) 2 c , ( - 7 11 : 5 12 + - 4 12 : 5 12 ) . 2 - 5 3
Bài 1. Tính hợp lý1) (–12) +6.(–3) 2) (36 -2020) + (2019 -136) – 273) (144 – 97) – (244 – 197) 4) (–24).13 – 24.( –3)5) 54+55+56+57+58-(64+65+66+67+68) 6) 24(16 – 5) – 16(24 – 5) 7) 47.(23 + 50) – 23.(47 + 50)8) (-31). 47 + (-31). 52 + (-31)Bài 2: Tìm số nguyên x, biết: 1)-17-(2x-5)-62) 10-2(4-3x)-43)-12+3(-x+7)-184)-45:[5.(-3-2x)]35) x.(x+3)06) (x-2).(x+4)07) x.(x+1).(x-3)0
Đọc tiếp
Bài 1. Tính hợp lý
1) (–12) +6.(–3)
2) (36 -2020) + (2019 -136) – 27
3) (144 – 97) – (244 – 197)
4) (–24).13 – 24.( –3)
5) 54+55+56+57+58-(64+65+66+67+68)
6) 24(16 – 5) – 16(24 – 5)
7) 47.(23 + 50) – 23.(47 + 50)
8) (-31). 47 + (-31). 52 + (-31)
Bài 2: Tìm số nguyên x, biết:
1)-17-(2x-5)=-6
2) 10-2(4-3x)=-4
3)-12+3(-x+7)=-18
4)-45:[5.(-3-2x)]=3
5) x.(x+3)=0
6) (x-2).(x+4)=0
7) x.(x+1).(x-3)=0
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: \(x\in\left\{0;-1;3\right\}\)
Đúng 0
Bình luận (0)
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 !