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a) (2x - 10)37 = 0
2x - 10 = 0 : 37
2x - 10 = 0
2x = 0 + 10
2x = 10
x = 10 : 2
x = 5
b) 135(34 - x) = 810
34 - x = 810 : 135
34 - x = 6
x = 34 - 6
x = 28
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Bài 5 :
S = 1 + 3 - 5 - 7 + 9 + 11 - ... - 397 - 399
S = 1 + (3 - 5 - 7 + 9) + (11 - 13 - 15 + 17) + ... + (387 - 389 - 391 + 393) + (395 - 397 - 399)
S = 1 + 0 + 0 + ... + 0 + (- 401)
S = 1 - 401
S = - 400
Bài 5
A= 1+3-5-7+9+11-13-15+...-397-399
A= ( 1+3-5-7)+( 9+11-13-15)+...+( 393+395-397-399)
A= -8 -8 -...-8
A = -8.50 ( từ 1 đến 399 có 200 số, chia làm 4 cặp)
A= -400
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a: Tổng các số hạng là:
\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)
Ta có: A+1=2x
\(\Leftrightarrow2x=24311\)
hay \(x=\dfrac{24311}{2}\)
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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{399}{400}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{399}{400}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{399}{400}:2\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{399}{400}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{399}{800}\)
\(\frac{1}{x+1}=\frac{400}{800}-\frac{399}{800}\)
\(\frac{1}{x+1}=\frac{1}{800}\)
\(=>x+1=800\)
\(=>x=800-1=799\)
Vậy x = 799
Ủng hộ mk nha ^_-
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{399}{400}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{399}{400}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{399}{400}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{399}{400}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{200}\)
\(\frac{1}{x+1}=\frac{-299}{200}\)
\(x+1=\frac{-200}{299}\)
\(x=\frac{-499}{299}\)
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a: (x-3)(y+1)=15
=>\(\left(x-3\right)\left(y+1\right)=1\cdot15=15\cdot1=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=3\cdot5=5\cdot3=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)
=>(x-3;y+1)\(\in\){(1;15);(15;1);(-1;-15);(-15;-1);(3;5);(5;3);(-3;-5);(-5;-3)}
=>(x,y)\(\in\){(4;14);(18;0);(2;-16);(-12;-2);(6;4);(8;2);(0;-6);(-2;-4)}
b: Sửa đề:\(m=1+3+3^2+3^3+...+3^{99}+3^{100}\)
\(m=1+3+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(=4+3^2\left(1+3+3^2\right)+3^5\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)
\(=4+13\left(3^2+3^5+...+3^{98}\right)\)
=>m chia 13 dư 4
\(m=1+3+3^2+...+3^{99}+3^{100}\)
\(=1+\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=1+3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{97}\left(1+3+3^2+3^3\right)\)
\(=1+40\left(3+3^5+...+3^{97}\right)\)
=>m chia 40 dư 1
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(x+1)+(x+3)+(x+5)+...+(x+399)=40400
<=> x +1 + x + 3 + x + 5 + ...+ x + 399 = 40400
<=> ( x+x+x+...+x)+(1+3+5+...+399)=40400
<=> 200x + 40000=40400
<=> 200x = 400
<=> x = 2
Vậy...
\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+..+\left(x+399\right)=40400\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+3+5+..+399\right)=40400\)
\(\Leftrightarrow200.x+\left(399+1\right).200:2=40400\) ( 200 số hạng, 200 x )
\(\Leftrightarrow200x+40000=40400\)
\(\Leftrightarrow200x=400\)
\(\Leftrightarrow x=2\)
Vậy x = 2