\(1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)               

\(\left(1-1\right)-\left(\frac{1}{2}+\frac{1}{2}\right)+\left(2-2\right)-\left(\frac{2}{3}+\frac{1}{3}\right)+\left(3-3\right)-\left(\frac{3}{4}+\frac{1}{4}\right)-1\)

\(-1-1-1+4=1\)

MIK XIN LỖI BN NHA VÌ ĐÁNH MÁY HƠI LÂU NHA !! CHÚC BN HOK TỐT NHAA

với giải thích hộ mik số trên có chia hết cho 13 ko và có là số chính phương không ạ

Đặt biểu thức trên là A , ta có :

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

\(\Rightarrow3A=3+3^2+3^3+3^4+...+3^{99}\)

\(\Leftrightarrow3A-A=\left(3+3^2+3^3+3^4+...+3^{99}\right)-\left(1+3+3^2+3^3+...+3^{98}\right)\)

\(\Leftrightarrow2A=3^{99}-1\)

\(\Leftrightarrow A=\frac{3^{99}-1}{2}\)

Đặt \(A=1+3+3^2+...+3^{98}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{99}\)

\(\Rightarrow3A-A=.......................=3^{99}-1\)

\(\Rightarrow2A=3^{99}-1\).\(\Rightarrow A=\frac{3^{99}-1}{2}\)

\(=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{2021\cdot2022:2}\)

\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2021\cdot2022}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)

\(=2\cdot\dfrac{505}{1011}=\dfrac{1010}{1011}\)

\(9,\Leftrightarrow x+1=8\Leftrightarrow x=7\\ 10,\Leftrightarrow3-2x=-8\Leftrightarrow-2x=-11\Leftrightarrow x=\dfrac{11}{2}\)

9. \(\sqrt[3]{x+1}=2\left(ĐK:x\ge-1\right)\)

<=> x + 1 = 2³

<=> x + 1 = 8

<=> x = 7 (TM)

10. \(\sqrt[3]{3-2x}=-2\left(ĐK:x\le\dfrac{3}{2}\right)\)

<=> 3 - 2x = (-2)³

<=> 3 - 2x = -8

<=> -2x = -11

<=> \(x=\dfrac{11}{2}\left(loại\right)\)

Vậy nghiệm của PT là \(S=\varnothing\)

\(1,2\sqrt{27}+5\sqrt{12}-3\sqrt{48}\\ =2.3\sqrt{3}+5.2\sqrt{3}-3.4\sqrt{3}\\ =6\sqrt{3}+10\sqrt{3}-12\sqrt{3}\\ =4\sqrt{3}\)

\(2,\sqrt{147}+\sqrt{75}-4\sqrt{27}\\ =7\sqrt{3}+5\sqrt{3}-4.3\sqrt{3}\\ =7\sqrt{3}+5\sqrt{3}-12\sqrt{3}\\ =\sqrt{3}\left(7+5-12\right)\\ =0\)

\(3,3\sqrt{2}\left(4-\sqrt{2}\right)+3\left(1-2\sqrt{2}\right)^2\\ =3\sqrt{2}.\left(4-\sqrt{2}\right)+3\left(1-4\sqrt{2}+8\right)\\ =12\sqrt{2}-6+3-12\sqrt{2}+24\\ =21\)

\(4,2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\\ =2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\\ =\sqrt{5}\left(2-5-4+11\right)\\ =4\sqrt{5}\)

1: =6căn 3+10căn 3-12căn 3=4căn 3

2: =7căn 3+5căn 3-12căn 3=0

3: =12căn 2-6+3(9-4căn 2)

=12căn 2-6+27-12căn 2=21

4: =2căn 5-5căn 5+4căn 5+9 căn 5

=10căn 5

1. \(S=1+3+3^2+3^3+........+3^{2019}+3^{2020}\)

\(\Rightarrow3S=3+3^2+3^3+3^4+........+3^{2020}+3^{2021}\)

\(\Rightarrow3S-S=3^{2021}-1\)

\(\Rightarrow2S=3^{2021}-1\)

\(\Rightarrow S=\frac{3^{2021}-1}{2}\)

2. \(\left(3x-2\right)^3=64\)

\(\Leftrightarrow\left(3x-2\right)^3=4^3\)

\(\Leftrightarrow3x-2=4\)

\(\Leftrightarrow3x=6\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

[3x-2]^3=64

Ta có:64=4^3    

Suy ra:3x-2=4

           3x   =4+2

          3x=6

         x=6:3

         x=2

Đặt: \(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..+\dfrac{1}{100}\)

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

\(\Rightarrow\dfrac{1}{A}=2+3+4+5+...+100\)

\(\Rightarrow\dfrac{1}{A}=\left(100+2\right)\cdot99:2\)

\(\Rightarrow\dfrac{1}{A}=5049\)

\(\Rightarrow A=\dfrac{1}{5049}\)

1/5049