b) 2003+a×15=2003+10×3×5
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A) 95.(a-75)=95.(100-75)
=>a-75=100-75
vậy a =100
B) 2003+a.15=2003.10.3.5
a.15=10.3.5
![](https://rs.olm.vn/images/avt/0.png?1311)
2003 + a x 15 = 2003 + 10 x 3
2003 + a x 15 = 2003 + 10 x 3
a x 15 = 10 x 3
= 30
a = 30 : 15
a = 2
\(\Rightarrow2003+15a=2003+30\)
\(\Rightarrow2003-2003+15a=30\)
\(\Rightarrow15a=30\)
\(vây\)\(a=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=11+14+17+...+62+65\)
Số số hạng của \(A\)là
\(\left(65-11\right)\div3+1=19\)(số hạng)
Tổng của \(A\)là:
\(\left(11+65\right)\times19\div2=722\)
Đáp số: 722
\(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(B=\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+.....+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\times\frac{1}{2}\)
\(B=\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{9}-\frac{1}{11}\right)\times\frac{1}{2}\)
\(B=\left(1-\frac{1}{11}\right)\times\frac{1}{2}\)
\(B=\frac{10}{11}\times\frac{1}{2}\)
\(B=\frac{5}{11}\)
\(C=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+\frac{3}{154}\)
\(C=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)
\(C=\left(\frac{3}{2}-\frac{3}{5}+\frac{3}{5}-\frac{3}{8}+....+\frac{3}{11}-\frac{3}{14}\right)\div3\)
\(C=\left(\frac{3}{2}-\frac{3}{14}\right)\div3\)
\(C=\frac{9}{7}\div3\)
\(C=\frac{3}{7}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(B=1-\frac{1}{11}\)
\(B=\frac{10}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(15-x\right)+\left(x-12\right)=7-\left(x-5\right)\)
=>7-x+5=15-x+x-12
=>12-x=3
hay x=9
b: \(\Leftrightarrow x-\left\{57-\left[42-23-x\right]\right\}=13-\left\{47+25-32+x\right\}\)
\(\Leftrightarrow x-\left\{57-19+x\right\}=13-\left\{40+x\right\}\)
=>x-38-x=13-40-x
=>-27-x=-38
=>x+27=38
hay x=11
e: \(x^2+3x+9⋮x+3\)
\(\Leftrightarrow x\left(x+3\right)+9⋮x+3\)
\(\Leftrightarrow x+3\in\left\{1;-1;9;-9;3;-3\right\}\)
hay \(x\in\left\{-2;-4;6;-12;0;-6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (35 + 75) + (345 - 35 - 75)
= 35 + 75 + 345 - 35 - 75 ( quy tắc mở ngoặc đổi dấu)
= (35 - 35) + ( 75 - 75) + 345
= 345
b) (2003 - 79 + 15) - (-79 + 15)
= 2003 - 79 + 15 + 79 - 15 (quy tắc mở ngoặc đổi dấu)
= 2003 + ( 79 - 79) + (15 - 15)
= 2003
c) -(515 - 80 + 91) - (2003 + 80 - 91)
= -515 + 80 - 91 - 2003 - 80 + 91 ( quy tắc mở ngoặc đổi dấu)
= (80 - 80) + (91 - 91) + ( -2003 - 515)
= -2518
d) 4575 + 46 - 4573 + 35 - 16 -5
= (4575 - 4573) + (46 - 16) + (35 - 5)
= 2 + 30 + 30
= 62
Chúc học tốt!!
a) ( 35 + 75 ) + ( 345 - 35 - 75 )
= 35 + 75 + 345 - 35 - 75
= ( 35 - 35 ) + ( 75- 75 ) + 345
= 0 + 0 + 345
= 345
![](https://rs.olm.vn/images/avt/0.png?1311)
a) -2003 + ( -21 + 75 + 2003) = ( -2003 + 2003) + ( -21 + 75 )
= 0 + 54 = 54
b) 942 - 2567 + 2563 - 1942 = ( 942 - 1942 ) + ( -2567 + 2563 )
= -1000 + ( -4)
= - 1004
![](https://rs.olm.vn/images/avt/0.png?1311)
a 95 x { a - 75 } = 95 x { 100 - 75 }
=>a-75=100-75
vậy a=100
b } 2003 + a x 15 = 2003 x 10 x 3 x5
a x 15 = 10 x 3 x5
a x 15 = 10x(3x5)
a x 15=10x15
Vậy a=10
a = 10
hok tốt
2003+a*15=2153
a*15=150
a=10