\(\frac{x}{8}=\frac{-2}{5}\cdot\frac{3}{16}\)

\(\frac{x}{8}=\frac{-3}{40}\)

\(\Rightarrow x=\frac{8.\left(-3\right)}{40}=\frac{-3}{5}\)

\(\frac{x+4}{5}=\frac{-1}{7}\cdot\frac{14}{15}\)

\(\frac{x+4}{5}=\frac{-2}{15}\)

\(\Rightarrow x+4=\frac{5.\left(-2\right)}{15}=\frac{-2}{3}\)

\(\Rightarrow x=\frac{-14}{3}\)

Hơi căng 

\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+...+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+...+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)

gfvfvfvfvfvfvfv555

\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)

https://www.youtube.com/watch?v=cFZDEMTQQCs

j thế bạn

1.
a) |x + 1| = 5
=> x +1 ∈ {5;-5}
TH1 : x + 1 = 5
          x       = 5 - 1
          x       = 4
TH2 : x + 1 = -5 (Tương tự như trên)
Vậy x ∈ {4;-6}
|x + 1| = 0
=> x + 1 = 0 (Vì 0 không phải là số âm hay dương nên chỉ có 1 TH duy nhất)
     x       = ... (Tự tính)
Vậy x = ...
b) |x| = x
=> x ∈ {x;-x}
Vậy x ∈ {x;-x}
2.
Ta có : |x| = 5 => x ∈ {5;-5}
            |y| = 8 => y ∈ {8;-8}
Thay những số trên vào x + y rồi tính.
- Có quá khó hiểu ?

Thank

1.a) \(\left|x+1\right|=5\)

\(\Rightarrow\orbr{\begin{cases}x+1=5\\x+1=-5\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)

    \(\left|x+1\right|=0\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

b) \(\left|x\right|=x\)

\(\Rightarrow x\in\left\{x;-x\right\}\)

2.

\(\left|x\right|=5\)

\(\Rightarrow x\in\left\{5;-5\right\}\)

\(\left|x\right|=8\)

\(\Rightarrow x\in\left\{8;-8\right\}\)

Tính

Hai số dương cộng : \(5+8=13\)

Hai số âm cộng : \(-5+-8=-14\)

Số dương cộng số âm : \(5+-5=0\) 

                                          \(8+-8=0\)

Hc tốt