a) \(xy+3x=5y-2\)
\(\Leftrightarrow x\left(y+3\right)=5y-2\)
\(\Leftrightarrow x=\frac{5y-2}{y+3}\)
\(\Leftrightarrow x=\frac{5\left(y+3\right)-17}{y+3}\)
\(\Leftrightarrow x=5-\frac{17}{y+3}\)
Do x nguyên, y nguyên nên y+3 là Ư(17)

Ta có bảng:
 

Vậy (x;y) là (6;-20);(22;-4);(-12;-2);(4;14)  

b) \(\Leftrightarrow\frac{\frac{99.100.101}{3}}{100x^2+\frac{99.100}{2}}=\frac{6666}{131}\Rightarrow x=\pm4\)

cau b hoi tat nhi

dễ ghê chiều làm cho giờ off

Bài 1:

a) Ta có: \(\frac{2^8\cdot4\cdot13+2^7\cdot8\cdot65}{2^9\cdot39}\)

\(=\frac{2^8\cdot4\cdot13+2^8\cdot4\cdot13\cdot5}{2^9\cdot39}\)

\(=\frac{2^{10}\cdot13\left(1+5\right)}{2^9\cdot13\cdot3}=\frac{6}{3}=2\)

b) Đặt \(A=4+2^2+2^3+2^4+...+2^{20}\)

Ta có: \(A=4+2^2+2^3+2^4+...+2^{20}\)

\(\Rightarrow2A=2^3+2^3+2^4+...+2^{21}\)

Ta có: \(2A-A=2^3+2^{21}-2^2-2^2=8+2^{21}-8=2^{21}\)

hay \(A=2^{21}\)

Vậy: \(4+2^2+2^3+2^4+...+2^{20}=2^{21}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)=1\)

\(\Leftrightarrow3x+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)=1\)

\(\Leftrightarrow3x+\frac{3}{2}=1\)

\(\Leftrightarrow3x=-\frac{1}{2}\)

\(\Leftrightarrow x=-\frac{1}{2}\div3=-\frac{1}{6}\)

Sửa đề \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)

\(\Leftrightarrow x=99\)

a) => ( x + 1/2 ) . 3 = 1

=> 3x + 3/2 = 1

=> 3x = 1 - 3/2

=> 3x = -1/2

=> x = -1/2 : 3 = -1/6

Câu a:

3\(\frac12\)\(x\) - \(\frac{x}{2}\) + \(x\) = 3,5 : 1\(\frac15\)

3\(x\) + \(\frac12x-\frac12x+x\) = 3,5 : 6/5

(3\(x+x\)) + (1/2\(x\) - 1/2\(x\)) = 3,5 x 5/6

4\(x\) = \(\frac{35}{12}\)

\(x\) = 35/12 : 4

\(x=\frac{35}{48}\)

Vậy \(x\) = 35/48

Câu b:

(x - 3/1.2) + (x - 3/2.3) + ...+ (x - 3/99.100) = 1

(x + x + ..+ x) - (3/1.2 + 3/2.3 + ..+ 3/99.100) = 1

(x+ x + ... + x) - 3.(1/1.2 + 1/2.3 + ... + 1/99.100) = 1

Xét dãy số: 1; 2; 3;..; 99 dãy số trên có 99 số hạng vậy

Vế trái bằng:

99x - 3.(1/1 - 1/2 + 1/2 - 1/3+ ... + 1/99 - 1/100)

99x - 3.(1/1 - 1/100)

99x - 3.99/100 = 1

99x = 1 + 3.99/100

99x = 1 + 297/100

99x = 397/100

x = 397/100 : 99

x = 397/9900

Vậy: x = 397/9900

đổi k ko,mk hứa sẽ k lại(nếu ko làm chó!!!!!!!!!!!!!)

Bài 1: <Cho là câu a đi>:

a. \(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\) 

\(\rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\) 

\(\rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{49}{50}\) 

\(\rightarrow1-\frac{1}{x+1}=\frac{49}{50}\) 

\(\rightarrow\frac{1}{x+1}=1-\frac{49}{50}=\frac{1}{50}\) 

\(\rightarrow x+1=50\rightarrow x=49\) 

Vậy x = 49.

Ta có : 

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{99^2}{99.100}\)

\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1.2.2.3.3.4.....99.100}\)

\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1^2.2^2.3^2.4^2.....99^2}.\frac{1}{100}\)

\(=\)\(\frac{1}{100}\)