\(A=3^{100}-3^{99}+3^{98}-...+3^2-3+1\)

\(3A=3\left(3^{100}-3^{99}+3^{98}-...+3^2-3+1\right)\)

\(3A=3^{101}-3^{100}+3^{99}-...+3^3-3^2+3\)

\(3A+A=\left(3^{101}-3^{100}+3^{99}-...+3^3-3^2+3\right)+\left(3^{100}-3^{99}+3^{98}-...+3^2-3+1\right)\)

\(4A=3^{101}+1\)

\(A=\dfrac{3^{101}+1}{4}\)

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

a.

\(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)-x^4+y^4=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4-x^4+y^4\)

\(=\left(x^4-x^4\right)+\left(y^4-y^4\right)+\left(x^3y-x^3y\right)+\left(xy^3-xy^3\right)+\left(x^2y^2-x^2y^2\right)=0\)

b.

\(\left(2-x\right)\left(1+2x\right)+\left(1+x\right)-\left(x^4+x^3-5x^2-5\right)=2+4x-x-2x^2+1+x-x^4-x^3+5x^2+5\)

\(=-x^4-x^3+\left(5x^2-2x^2\right)+\left(4x-x+x\right)+\left(1+2+5\right)=-x^4-x^3+3x^2+4x+8\)

c.

\(\left(x^2-7\right)\left(x+2\right)-\left(2x-1\right)\left(x-14\right)+x\left(x^2-2x-22\right)+35=x^3+2x^2-7x-14-2x^2+28x+x-14+x^3-2x^2-22x+35\)

\(=\left(x^3+x^3\right)+\left(2x^2-2x^2\right)+\left(28x-22x-7x+x\right)+\left(35-14\right)=2x^3+21\)

Câu c mik sửa lại chút nhé, ở dấu bằng thứ hai

\(\left(x^3+x^3\right)+\left(2x^2-2x^2-2x^2\right)+\left(28x+x-22x-7x\right)+\left(35-14\right)=2x^3-2x^2+21\)

a) \(A=1+2+2^2+2^3+...+2^{99}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{100}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{100}-1-2-2^2-...-2^{99}=2^{100}-1\)

b) \(A=1+2+2^2+...+2^{99}=\left(1+2+2^2+2^3\right)+2^4\left(1+2+2^2+2^3\right)+...+2^{96}\left(1+2+2^2+2^3\right)\)

\(=15+2^4.15+...+2^{96}.15=15\left(1+2^4+...+2^{96}\right)\)

\(=3.5\left(1+2^4+...2^{96}\right)\) chia hết cho 3 và 5

c) \(A=1+2+2^2+...+2^{99}\)

\(=1+2\left(1+2+2^2\right)+...+2^{97}\left(1+2+2^2\right)\)

\(=1+2.7+...+2^{97}.7=1+7\left(2+...+2^{97}\right)\) chia 7 dư 1

=> A không chia hết cho 7

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

Đặt :

\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{99}}\)

\(\Leftrightarrow2A=3+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{98}}\)

\(\Leftrightarrow2A-A=\left(3+\dfrac{1}{2}+....+\dfrac{1}{2^{98}}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{2^{99}}\right)\)

\(\Leftrightarrow A=2-\dfrac{1}{2^{99}}\)

Vậy..

d: \(\dfrac{1}{27}:\left(-\dfrac{1}{3}\right)^2+75\%\cdot\left(-\dfrac{2^2}{3}\right)\)

\(=\dfrac{1}{27}:\dfrac{1}{9}+\dfrac{3}{4}\cdot\dfrac{-4}{3}\)

\(=\dfrac{1}{3}-1\)

\(=-\dfrac{2}{3}\)

Lời giải:

Gọi vế trái là $A$

$2A=\frac{2}{2^2}+\frac{2}{4^2}+\frac{2}{6^2}+...+\frac{2}{2022^2}$

Xét số hạng tổng quát:

$\frac{2}{n^2}$. Ta sẽ cm $\frac{2}{n^2}< \frac{1}{(n-1)n}+\frac{1}{n(n+1)}(*)$

$\Leftrightarrow \frac{2}{n^2}< \frac{n+1+n-1}{n(n-1)(n+1)}$

$\Leftrightarrow \frac{2}{n^2}< \frac{2}{(n-1)(n+1)}$

$\Leftrightarrow \frac{2}{n^2}< \frac{2}{n^2-1}$ (luôn đúng)

Thay $n=2,4,...., 2022$ vào $(*)$ ta có:

$\frac{2}{2^2}< \frac{1}{1.2}+\frac{1}{2.3}$

$\frac{2}{4^2}< \frac{1}{3.4}+\frac{1}{4.5}$

.......

Suy ra: $2A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2022.2023}$

$2A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2022}-\frac{1}{2023}$

$2A< 1-\frac{1}{2023}< 1$

$\Rightarrow A< \frac{1}{2}$