46+72+46+59+46+11
Tính hợp lý
-4 . 32 . (-25) . 11
-86 . 46 - 46 . 27 + 46 . 23
a)
(-4) . 32 . (-25) . 11
= (-4).(-25) . (32.11)
= 100 . 352
= 35200
b)
-86 . 46 - 46 . 27 + 46 . 23
= 46. ( -86 - 27 + 23)
= 46. (-90) = -4140
Chúc em học tốt!!!
a)
(-4) . 32 . (-25) . 11
= (-4).(-25) . (32.11)
= 100 . 352
= 35200
P/s tham khảo nha
b)
-86 . 46 - 46 . 27 + 46 . 23
= 46. ( -86 - 27 + 23)
= 46. (-90) = -4140
P/s tham khảo nha
Bài 1: Thực hiện phép tính (Tính nhanh nếu có thể)
a)5.(–8).2.(–3)
b) (45 – 135 +72 ) - ( 45 +72 )
c) 3.(–5)2 2 + 2.(–5) – 20
d) 86. (-46) + 46. 27 - 46. 41
e) 34.(15 –10) – 15.(34 –10)
Giải hộ mik
a)5.(-8).2.(-3)
=[5.(-2)].[(-8).(-3)]
=-10.24
=-240
b)(45-135+72)-(45+72)
=45-135+72-45-72
=(45-45)+(72-72)-135
=0+0-135
=-135
c)3.(-5)2.2+2.(-5)-20
=3.25.2+2(-5)-20
=75.2+(-10)-20
=150+(-30)
=120
d)86.(-46)+46.27-46.41
=86.(-1).46+46.27-46.41
=-86.46+46.27-46.41
=46(-86+27-41)
=46(-100)
-4600
e)34(15-10)-15(34-10)
=34.15-34.10-15.34+15.10
=(34.15-15.34)+(15.10-34.10)
=[34(15-15)]+[10(15-34)]
=34.0+10(-9)
=0+(-90)
=-90
(10+59)+(27+87)-(48+46)
69+114-94=89
=69+114-94
=89
ĐÂY MÀ LÀ TOÁN LỚP 1 Á ???/
hm....lúc mình học có mấy bài này đâu nhỉ? mà thôi pick đúng cho mình nhé!
(10+59)+(27+87)-(48+46)
=69+114-94
=277
a) 4x^5-3=125
b) 31+45:x=46
c) 72-x:7=59
d) 3^x=81
e) x^5=32
f) (x-1)^4=625
a + 46 voi a =59
a) 59 + 27 : x = 62
b) X x 3 + 16 = 46
27 : x = 62 - 59
27 : x = 3
x = 29 : 3
x = 9
X x 3 = 46 - 16
X x 3 = 30
X = 30 : 3
X = 10
\(a,\Leftrightarrow27:x=3\)
\(\Leftrightarrow x=27:3\)
\(\Leftrightarrow x=9\)
\(b,\Leftrightarrow x\times3=30\)
\(\Leftrightarrow x=10\)
27:x= 62-59
=>27:x=3
=>x=29:3
x=9
x*3+16=46
=>x*3 = 46 - 16
=>X*3 = 30
=>x = 30 : 3
=> x =10
[2.(70 - x) + 72] : 2 = 46
Lời giải:
$[2(70-x)+72]:2=46$
$(140-2x+72):2=46$
$212-2x=46.2=92$
$2x=212-92=120$
$x=120:2=60$
Kết quả của phép tính: |-46|+|-11|+|-143|+(-200)∣−46∣+∣−11∣+∣−143∣+(−200) là
ai tích mình mình tích lại cho
Tính giá trị biểu thức
a, A= 1 - 2002/2003
b, B = 179/30 - ( 59/30 - 3/5)
c, C= (46/5 - 1/11) × 11
a) \(A=1-\dfrac{2002}{2003}\)
\(A=\dfrac{1}{2003}\)
b) \(B=\dfrac{179}{30}-\left(\dfrac{59}{30}-\dfrac{3}{5}\right)\)
\(B=\left(\dfrac{179}{30}-\dfrac{59}{30}\right)-\dfrac{3}{5}\)
\(B=\dfrac{120}{30}-\dfrac{3}{5}\)
\(B=4-\dfrac{3}{5}\)
\(B=\dfrac{17}{5}\)
c) \(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right).11\)
\(C=\dfrac{501}{55}.11\)
\(C=\dfrac{501}{5}\)
a) \(A=1-\dfrac{2002}{2003}=\dfrac{2003}{2003}-\dfrac{2002}{2003}=\dfrac{1}{2003}\)
b) \(B=\dfrac{179}{30}-\left(\dfrac{59}{30}-\dfrac{3}{5}\right)=\dfrac{179}{30}-\dfrac{59}{30}+\dfrac{3}{5}=4+\dfrac{3}{5}=\dfrac{23}{5}\)
c) \(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right)\cdot11=\dfrac{46}{5}\cdot11-\dfrac{1}{11}\cdot11=\dfrac{506}{5}-1=\dfrac{501}{5}\)
a)
\(A=1-\dfrac{2002}{2003}\\ \Rightarrow A=\dfrac{2003}{2003}-\dfrac{2002}{2003}\\ \Rightarrow A=\dfrac{1}{2003}\)
b)
\(B=\dfrac{179}{30}-\left(\dfrac{59}{30}-\dfrac{3}{5}\right)\\ \Rightarrow B=\dfrac{179}{30}-\dfrac{59}{30}+\dfrac{3}{5}\\ \Rightarrow B=\left(\dfrac{179}{30}-\dfrac{59}{30}\right)+\dfrac{3}{5}\\ \Rightarrow B=\dfrac{120}{30}+\dfrac{3}{5}\\ \Rightarrow B=4+\dfrac{3}{5}\\ \Rightarrow B=\dfrac{20}{5}+\dfrac{3}{5}\\ \Rightarrow B=\dfrac{23}{5}\)
c)
\(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right)\cdot11\\ C=\left(\dfrac{506}{55}-\dfrac{5}{55}\right)\cdot11\\ \Rightarrow C=\dfrac{501}{55}\cdot11\\ \Rightarrow C=\dfrac{501}{5}\)