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\(M=\frac{1}{10}+\frac{4}{20}+\frac{9}{30}+\frac{16}{40}+...+\frac{81}{90}\)
\(M=\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+...+\frac{9}{10}\)
\(M=\frac{\left(9+1\right)\cdot\left(9-1+1\right):2}{10}\)
\(M=\frac{10\cdot9:2}{10}=4,5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =-64(32+68)=64x100=-6400
b: =76(-54-12+36)=76x(-30)=-2380
c: =19x50+38x5=950+190=1140
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 5.3x-3=45
=>3x-3=9
=>3x-3=32
=>x-3=2
=>x=5
b) 50 ⋮ x
=>x\(\inƯ\left(50\right)=\left\{\pm1;\pm2;\pm5;\pm10;\pm25\pm50\right\}\)
c) x ϵ B(8) và 64 ≤ x \(\le\)100
B(8)={0;8;16;....;56;64;71;80;88;96;104 ...}
mà 64\(64\le x\le100\)
thì x\(\in\left\{64;72;80;88;96\right\}\)
d) x ϵ Ư(36) và 2< x < 15
![](https://rs.olm.vn/images/avt/0.png?1311)
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 !
![](https://rs.olm.vn/images/avt/0.png?1311)
chia làm 2 phần:
36x28+36x82
+
64x69+69x41
=
36 x (28+82)
+
69 x (64+41)
=
36 x 110
+
69 x 105
=
(36 x 100) + (36 x 10)
+
(69 x 100) + (69 x 5)
=
3600 + 360
+
6900 + 345
=
3960
+
7245
đáp số:
11205
![](https://rs.olm.vn/images/avt/0.png?1311)
36 . 64 + 18 . 64 + 36 . 54
=(36+18)x64+36x54
=54x64+36x54
=(64+36)x54
=100x54
=5400
36. 64 + 18 . 64 + 36 . 54
= ( 36 + 18 ) . 64 + 36 . 54
= 54 . 64 + 36 . 54
= ( 54 + 36 ) . 54
= 100 .54
=5400
(-36)+(-50)+(-64)
= -(36+64+50)
= -(100+50)
= -150