Cho mình bổ sung tí : b = 100x nhé 

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

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

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

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

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

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

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

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

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

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

Ta có :

\(\begin{cases}\left|x+\frac{1}{1.2}\right|\ge0\\\left|x+\frac{1}{2.3}\right|\ge0\\...\\\left|x+\frac{1}{99.100}\right|\ge0\end{cases}\)\(\left(\forall x\right)\)

\(\Rightarrow100x>0\)

=> x > 0

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

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

\(\Rightarrow100x+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=100x\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=0\)

Dễ thấy VT \(\ne\)VP

=> \(x\in\varnothing\)

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

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

=> \(100x\ge0\Rightarrow x\ge0\)

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

=> 99x + \(\frac{99}{100}\) = 100x

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

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

Ta có

\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|=100x\)
\(\left|x+x+...x\right|+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=100x\)
\(\left|99x\right|+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)
\(\left|99x\right|+\left(\frac{1}{1}-\frac{1}{100}\right)=100x\)
\(\left|99x\right|+\frac{99}{100}=100x\)
Sau đó tự biến đổi nha! Mik chỉ giải tới đó thôi vì mới lớp 6 à!

\(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{99\cdot100}\right|=100x\)

có :

\(\left|x+\frac{1}{1\cdot2}\right|;\left|x+\frac{1}{2\cdot3}\right|;\left|x+\frac{1}{3\cdot4}\right|;...;\left|x+\frac{1}{99\cdot100}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\ge0\)

\(\Rightarrow100x\ge0\)

\(\Rightarrow x\ge\frac{0}{100}\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\)

\(=x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+x+\frac{1}{3\cdot4}+...+x+\frac{1}{99\cdot100}\)

bước này tự lm tp