(12+77+34+23+88+45+55+66) x (45.200-1808x25)
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\(=\left[66+34+77+23+12+88+45+55\right]\times\left[1808\times\left(25-25\right)\right]\\ =0\)
(12 + 77 + 34 + 23 + 88 + 45 + 55 + 66) x (45200 – 1808 x 25)
=(12 + 77 + 34 + 23 + 88 + 45 + 55 + 66) x (45200-45200)
=(12 + 77 + 34 + 23 + 88 + 45 + 55 + 66) x 0
=0
Giải:
( 12 + 77 + 34 + 23 + 88 + 45 + 55 + 66 ) . ( 45200 - 1808 .25 )
=( 12 + 77 + 34 + 23 + 88 + 45 + 55 + 66 ) . ( 45200 - 45200 )
=( 12 + 77 + 34 + 23 + 88 + 45 + 55 + 66 ) . 0
=0
Học tốt!!!
(12+77+34+23+88+45+55+66)* (45200-1808*25)
=(12+77+34+23+88+45+55+66)* (45200-45200)
=(12+17+34+23+88+45+55+66)* 0
=0
(45+75)*66+(75+45)*34
=(45+75)*(66+34)
=120*100
=12000
11+22+33+44+55+66+77+88+99:11
=11+22+33+44+55+66+77+88+9
=(11+9)+(22+88)+(33+77)+(44+66)+55
=20+110*3+55
=20+330+55
=405
Ta có :
a, (45 + 75) x 66 + (75 + 45) x 34
= (45 + 75) x (66 + 34)
= 120 x 100
= 12000
11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99 : 11
= 33 + 77 + 121 + 165 + 9
= 405
11+89=100 66+34=100
22+78=100 77+23=100
33+67=100 88+12=100
44+56=100 99+1=100
55+45=100
\(\dfrac{x-9}{55}+\dfrac{x-10}{66}=\dfrac{x-11}{77}+\dfrac{x-12}{88}\\ \Leftrightarrow\left(\dfrac{x-9}{55}+\dfrac{1}{11}\right)+\left(\dfrac{x-10}{66}+\dfrac{1}{11}\right)-\left(\dfrac{x-11}{77}+\dfrac{1}{11}\right)-\left(\dfrac{x-12}{88}+\dfrac{1}{11}\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-9}{55}+\dfrac{5}{55}\right)+\left(\dfrac{x-10}{66}+\dfrac{6}{66}\right)-\left(\dfrac{x-11}{77}+\dfrac{7}{77}\right)-\left(\dfrac{x-12}{88}+\dfrac{8}{88}\right)=0\)
\(\Leftrightarrow\dfrac{x-4}{55}+\dfrac{x-4}{66}-\dfrac{x-4}{77}-\dfrac{x-4}{88}=0\)
\(\Leftrightarrow\left(x-4\right)\left(\dfrac{1}{55}+\dfrac{1}{66}-\dfrac{1}{77}-\dfrac{1}{88}\right)=0\\ \Leftrightarrow x=4\left(vì.\dfrac{1}{55}+\dfrac{1}{66}-\dfrac{1}{77}-\dfrac{1}{88}\ne0\right)\)
34 < 50 47 > 45 55 < 66
78 > 69 81< 82 44 > 33
72 < 81 95 > 90 77 < 99
62 = 62 61 < 63 88 > 22
Bài 2:
a: \(17-x=3\)
=>\(x=17-3\)
=>x=14(nhận)
b: \(2\cdot\left(x-1\right):3=6\)
=>\(2\left(x-1\right)=6\cdot3=18\)
=>x-1=18/2=9
=>x=9+1=10(nhận)
c: \(x+\left(-2\right)=\left(-11\right)+7\)
=>\(x-2=-4\)
=>\(x=-4+2=-2\left(loại\right)\)
d: \(\left(x-1\right)^2-5=20\)
=>\(\left(x-1\right)^2=25\)
=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=-4\left(loại\right)\end{matrix}\right.\)
Câu 3:
a: Đặt *=a
\(\overline{57a3}⋮9\)
=>\(5+7+a+3⋮9\)
=>\(a+15⋮3\)
mà 0<=a<=9
nên a=3
=>*=a
b: \(A=123\cdot7+8+9\)
123*7 là số lẻ
9 là số lẻ
=>123*7+9 chia hết cho 2
mà 8 chia hết cho 2
nên \(A=123\cdot7+9+8⋮2\)
\(123\cdot7⋮3;9⋮3;8⋮̸3\)
=>\(A=123\cdot7+9+8⋮̸3\)
c: \(B=3\cdot5\cdot7+10^{50}\)
\(=5\cdot3\cdot7+5\cdot5^{49}\cdot2^{49}\)
\(=5\left(3\cdot7+5^{49}\cdot2^{49}\right)⋮5\)
=>B là hợp số