3. Thực hiện phép tính: A, 61• 45 + 61 • 23 - 68 • 51 B, 3 • 52 - ( 75 - 4 • 23 ) C,36 : { 22 • 5 - [ 30 - ( 5 - 1 ]2 } D,(12•49 -3 • 22•72):(2020• 2021)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18
![](https://rs.olm.vn/images/avt/0.png?1311)
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
a: \(61\cdot45+61\cdot23-68\cdot51\)
\(=61\left(45+23\right)-68\cdot51\)
\(=68\cdot61-68\cdot51\)
\(=68\left(61-51\right)=68\cdot10=680\)
b: \(3\cdot5^2-\left(75-4\cdot2^3\right)\)
\(=75-75+4\cdot8\)
\(=4\cdot8=32\)
c: \(36:\left\{2^2\cdot5-\left[30-\left(5-1\right)^2\right]\right\}\)
\(=\dfrac{36}{20-30+4^2}\)
\(=\dfrac{36}{-10+16}=\dfrac{36}{6}=6\)
d: \(\left(12\cdot49-3\cdot2^2\cdot7^2\right):\left(2020\cdot2021\right)\)
\(=\dfrac{\left(12\cdot49-12\cdot49\right)}{2020\cdot2021}=0\)