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

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

dễ ghê chiều làm cho giờ off

đề chưa đầy đủ

à đề thiếu tổng các giá trị tuyệt đối ở trên =100x

What wrong with you ?

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

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

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

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

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

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

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

\(\Leftrightarrow3\left(x+\frac{1}{2}\right)=1\)

\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{3}-\frac{1}{2}\)

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

Vì GTTĐ luôn lớn hơn hoặc bằng 0 với mọi x

\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\ge0\)

\(\Rightarrow100x\ge0\)

\(\Rightarrow x\ge0\)

Từ điều kiện trên ta có :

\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{99\cdot100}=100x\)

\(50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)

\(50x=1-\frac{1}{100}\)

\(50x=\frac{99}{100}\)

\(x=\frac{99}{5000}\)

Do \(\left|a\right|\ge0\forall a\) nên:

\(A=\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\forall x\)

\(\Leftrightarrow100x\ge0\) hay \(x\ge0\)

Do vậy ta có: \(A=\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\) ( 50 chữ số x)

\(\Leftrightarrow A=50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)

\(\Leftrightarrow50x+\left(1-\frac{1}{100}\right)=100x\Leftrightarrow50x+\frac{99}{100}=100x\)

\(\Leftrightarrow50x=\frac{99}{100}\Leftrightarrow x=\frac{99}{100.50}=\frac{99}{5000}\)

tưởng là có 99x lận mà

a, \(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)

\(\Rightarrow\frac{1}{2^x}+\frac{1}{2^x}\cdot\frac{1}{16}=17\)

\(\Rightarrow\frac{1}{2^x}\left(1+\frac{1}{16}\right)=17\)

\(\Rightarrow\frac{1}{2^x}\cdot\frac{17}{16}=17\)

\(\Rightarrow\frac{1}{2^x}=17:\frac{17}{16}=\frac{1}{16}=\frac{1}{2^4}\)

=> x = 4

b, Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;....;\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow100x\ge0\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+...+x+\frac{1}{99.100}=100x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\)

\(\Rightarrow99x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=100x\)

\(\Rightarrow100x-99x=1-\frac{1}{100}\)

\(\Rightarrow x=\frac{99}{100}\)