\(\frac{1+2^4+2^8+.....+2^{20}}{2^4+2^8+.......+2^{24}}=\frac{1+2^4+2^8+....+2^{20}}{2^4\cdot\left(1+2^4+2^8+...+2^{20}\right)}\)

\(=\frac{1}{2^4}=\frac{1}{16}\)

Ta nhận thấy mẫu của biểu thức trên là:

              x²⁶+x²⁴+x²²+...+x²+1=(x²⁶+x²²+...+x²)+(x²⁴+x²⁰+...+x⁴+1)

            =x²(x²⁴+x²⁰+...+x¹⁶+...+1)+(x²⁴+x²⁰+...+x⁴+1)

            =(x²⁴+x²⁰+...+1)(x²+1)

Như vậy\(\frac{x^{24}+x^{20}+x^{16}+...+1}{\left(x^{24}+x^{20}+...+1\right)\left(x^2+1\right)}\)=\(\frac{1}{x^2+1}\)

Tự hỏi tự trả lời

học giỏi vclllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll.......

Ta có: \(\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)

\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{26}+x^{22}+...+x^2\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)

\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^2\left(x^{24}+x^{20}+...+1\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)

\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{24}+x^{20}+x^{16}+...+1\right)\left(x^2+1\right)}\)

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

 =x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)=x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)

=x24+x20+x16+...+x4+1(x24+x20+x16+...+1)(x2+1)

1. 10 phần 14 = 5/7

2. 5 phần 15 = 1/3

3. 14 phần 22 = 7/11

4. 2 phần 8 = 1/4

5. 4 phần 24= 1/6

6. 2 phần 10 = 1/5

1. \(\frac{10}{14}=\frac{10:2}{14:2}=\frac{5}{7}\)

2. \(\frac{5}{15}=\frac{5:5}{15:5}=\frac{1}{3}\)

3. \(\frac{14}{22}=\frac{14:2}{22:2}=\frac{7}{11}\)

4. \(\frac{2}{8}=\frac{2:2}{8:2}=\frac{1}{4}\)

5. \(\frac{4}{24}=\frac{4:4}{24:4}=\frac{1}{6}\)

6. \(\frac{2}{10}=\frac{2:2}{10:2}=\frac{1}{5}\)

1)10/14=5/7

2)5/15=1/3

3)14/22=7/11

4)2/8=1/4

5)4/24=1/6

6)2/10=1/5

tick cho mk nha bạn

b: \(=\left(x^2+3x+1-3x+1\right)^2=\left(x^2+2\right)^2\)

1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)

\(=x^2+4x+4+x^2-6x+9\)

\(=2x^2-2x+13\)

2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=-2x+7\)

3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)

\(=x^2-25-x^2-10x-25\)

=-10x-50

4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)

\(=x^2-6x+9-x^2+16\)

=-6x+25

5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)

\(=y^2-6y+9-y^2+6y-9\)

=0

6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)

\(=4x^2+12x+9-4x^2+9\)

=12x+18

1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)

\(=x^2+4x+4+x^2-6x+9\)

\(=2x^2-2x+13\)

2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=\left(-2x+7\right)\cdot1\)

\(=-2x+7\)

3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)

\(=x^2-25-x^2-10x-25\)

\(=-10x-50\)