35.273 + 32.35
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
34 . 71 - 34 . 29
34 . ( 71 - 29 )
34 . -441
= -35721
(33 . 52 - 24 - 16 ) . 13
= ( 27 . 25- 16 - 16 ) . 13
= ( 675 - 16 -16 ) . 13
= 643. 13
= 8359
35.237 +33.35
=35 . ( \(237+33\) )
= 35 . 270
= 9450
23 . 42 + 23 . 84 - 40
=23 . ( 16 + 84 ) - 40
= 8 . 100 - 40
= 800 - 40
= 760
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{32.35}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{32}-\frac{1}{35}\)
\(A=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)
\(\Rightarrow x+\frac{6}{35}=-\frac{2}{7}\Rightarrow x=-\frac{2}{7}-\frac{6}{35}=-\frac{16}{35}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{32\cdot35}\)
\(3A=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{32\cdot35}\)
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{35}\)
\(3A=\frac{1}{2}-\frac{1}{35}\)
\(3A=\frac{33}{70}\)
\(A=\frac{11}{70}\)
Hok tốt !
![](https://rs.olm.vn/images/avt/0.png?1311)
Dat A=3/5.8 +3/8.11 +.........+3/32.35
A=1/5-1/8+1/8-1/11+1/11-1/14+.............+1/32-1/35
A=1/5-1/35
A=6/35
=>x+6/35=-29/35
=>x=-29/35-6/35
=>x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1: 19;-34;-56 là thứ tự giảm dần của ba số này
Câu 2:
a: \(\left(-17\right)+13=-\left(17-13\right)=-4\)
b: \(135\cdot3^2-3^2\cdot35\)
\(=3^2\left(135-35\right)\)
\(=9\cdot100=900\)
Câu 3:
\(20=2^2\cdot5;60=2^2\cdot5\cdot3\)
=>\(ƯCLN\left(20;60\right)=2^2\cdot5=20\)
\(a\inƯC\left(20;60\right)\)
=>\(a\inƯ\left(20\right)\)
mà a là số nguyên tố
nên \(a\in\left\{2;5\right\}\)
Câu 1
Theo thứ tự giảm dần : 19;-34;-56
Câu 2
(-17)+13= - 4
135.32-32.35
= 32. ( 135 . 35 )
= 32 - 4725
= - 4692
Câu 3
20 = 22 . 5
60 = 22 .5 . 3
=>ƯCLN ( 20;60)= 22 . 5 = 20
a ∈ ƯC ( 20;50)
=> a ∈ Ư ( 20 )
Mà a là số nguyên tố
Nên a ∈ { 2;5}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2^x.2^2.2^2=2^3\Rightarrow2^{x+2+2}=2^3\Rightarrow x+4=3\Rightarrow x=-1\)-1
\(\left(5^2+3^2\right).x+\left(5^2-3^2\right).x-40.x=10^2\)
\(\Rightarrow x.\left(5^2+3^2+5^2-3^2-40\right)=10^2\)
\(\Rightarrow x.\left(2.5^2-40\right)=10^2\)
\(\Rightarrow x.10=10^2\Rightarrow x=10^2:10\Rightarrow x=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{29.32}+\frac{1}{32.35}\)
\(3A=\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{32.35}\)
\(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{32}-\frac{1}{35}\)
\(3A=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)
\(A=\frac{6}{35}.\frac{1}{3}=\frac{2}{35}\)
\(B=\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{30.35}\)
\(5B=\frac{5}{5.10}+\frac{5}{10.15}+....+\frac{5}{30.35}\)
\(5B=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{30}-\frac{1}{35}\)
\(5B=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)
\(B=\frac{6}{35}.\frac{1}{5}=\frac{6}{175}\)
\(\frac{\left(\frac{6}{5}-\frac{4}{9}\right).\left(3^{22}+4.3^{18}\right)}{9^8.\left(x-2\right)}=\frac{\frac{2}{35}}{\frac{6}{175}}\)
\(\frac{\frac{34}{45}.3^{18}.\left(3^4+4\right)}{\left(3^2\right)^8.\left(x-2\right)}=\frac{5}{3}\)
\(\frac{\frac{34}{45}.85.3^{18}}{3^{16}.\left(x-2\right)}=\frac{5}{3}\)
\(\frac{\frac{578}{9}.3^2}{x-2}=\frac{5}{3}\)
\(\frac{578}{x-2}=\frac{5}{3}\)
\(\Rightarrow578.3=5x-10\)
\(\Rightarrow1734+10=5x\)
\(\Rightarrow x=\frac{1744}{5}=348,8\)
nếu có gì sai mấy bạn sửa nhé
n
35.273 + 32.35
=35(273+3^2)
=35.2457
=85995