A = 1 + 3 + 3² + 3³ + ... + 3⁶⁰

3A = 3 + 3² + 3³ + 3⁴ + ... + 3⁶¹

3A - A = (3 + 3² + 3³ + 3⁴ + ... + 3⁶¹) - (1 + 3 + 3² + 3³ + ... + 3⁶⁰)

2A = 3⁶¹ - 1

\(A=\frac{3^{61}-1}{2}\)

3A=3+3²+3³+3⁴+...+3⁶⁰+3⁶¹

3A - A=(3+3²+3³+3⁴+...+3⁶⁰+3⁶¹) - (1+3+3²+3³+....+3⁶⁰)

2A=3⁶¹-1

A  =\(\frac{3^{61}-1}{2}\)

\(3A=3+3^2+3^3+...+3^{61}\)

\(3A-A=3+3^2+3^3+...+3^{61}-1-3-3^2-...-3^{60}\)

\(2A=3^{61}-1\)

\(A=\frac{3^{61}-1}{2}\)

Chúc em học tốt^^

Đặt \(A=2^{17}-2^{16}-2^{15}-...-2^2-2-1\) ta có : 

\(A=2^{17}-\left(2^{16}+2^{15}+...+2+1\right)\)

Đặt \(B=2^{16}+2^{15}+...+2+1\) ta có : 

\(2B=2^{17}+2^{16}+...+2^2+2\)

\(2B-B=\left(2^{17}+2^{16}+...+2^2+2\right)-\left(2^{16}+2^{15}+...+2+1\right)\)

\(B=2^{17}-1\)

\(\Rightarrow\)\(A=2^{17}-B=2^{17}-\left(2^{17}-1\right)=2^{17}-2^{17}+1=1\)

Vậy \(A=1\)

Chúc bạn iu họk tốt :3 

bài này không biết làm á

em mk phá !

\(\frac{-1997\cdot1996+1}{-1995\cdot\left(-1997\right)+1996}\)

\(=\frac{-1997\cdot\left(1995+1\right)+1}{1995\cdot1997+1996}\)

\(=\frac{-1997\cdot1995+\left(-1997\right)+1}{1995\cdot1997+1996}\)

\(=\frac{-1997\cdot1995+\left(-1996\right)}{1995\cdot1997+1996}\)

\(=-1\)

\(\frac{-1997.1996+1}{-1995.\left(-1997\right)+1996}\)

\(=\frac{-1997.1996+1}{1995.1997+1996}\)

\(=\frac{-1997.\left(1995+1\right)+1}{1995.1997+1996}\)

\(=\frac{-1997.1995+-1997+1}{1995.1997+1996}\)

\(=\frac{1995.-1997+-1996}{1995.1997+1996}\)

\(=\frac{-\left(1995.1997+1996\right)}{1995.1997+1996}\)

\(=-1\)

3/5;2/7;3/7;5/7

đúng thì k

cảm ơn

Phân số sau khi rút gọn là :

\(\frac{2}{7}\);\(\frac{5}{7}\);\(\frac{3}{5}\);\(\frac{3}{7}\).

Sắp xếp theo thứ tự tăng dần là :

\(\frac{2}{7}\);\(\frac{3}{7}\);\(\frac{3}{5}\);\(\frac{5}{7}\).

Kik mik nha ! Hok tốt !

2. \(2xy-4y+2y^2\)

 \(=2y\left(x-2+y\right)\)

\(\sqrt{117,5^2-26,5^2}-1440=-202475\)

\(\sqrt{146,5^2-109,5^2+27,256=}-11816494\) 

bạn lê nhat phuong oi sai rồi

câu đc  của bạn phương thiếu căn ở 1440

câu a đc 108

câu b đc128

\(\dfrac{6}{\sqrt{2}-\sqrt{3}+3}=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{8-2\sqrt{6}}=\dfrac{3\left(\sqrt{2}-\sqrt{3}-3\right)\left(4+\sqrt{3}\right)}{13}\)

\(A=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)

\(=\dfrac{x-\sqrt{x}-2-\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)

\(=-\sqrt{x}\cdot\left(\sqrt{x}+1\right)\)

\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{\sqrt{x}+1}\left(đk:x\ne1,x\ge0\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+\sqrt{x}-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

ĐKXĐ: \(x\ne1,x\ge0\)

\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{\sqrt{x}+1}=\)\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}-\dfrac{2\sqrt{x}}{x-1}-\dfrac{\sqrt{x}-1}{x-1}=\)\(\dfrac{x+\sqrt{x}-2\sqrt{x}-\sqrt{x}+1}{x-1}=\)\(\dfrac{x-2\sqrt{x}+1}{x-1}=\)\(\dfrac{(\sqrt{x}-1)^2}{x-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)