a) 1/5 - (1/2 + 3/4 ) : 5/2

= 1/5 - ( 1/4 + 3/4 ) : 5/2

=1/5 - 1 : 5/2

= 1/5 - 1 . 2/5

= 1/5 - 2/5

= -1/5

b) 1,5 . (1/3 - 2/3)

=3/2 . ( -1/3)

=-1/2

c) 9/10 . 23/11 - 1/11 . 9/10 + 9/10

= 9/10 . ( 23/11 - 1/11 ) + 9/10

= 9/10 . 1 + 9/10

= 9/10 + 9/10

= 18/10 = 9/5

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)

\(A=1-\frac{1}{2^{2016}}< 1\)

i

\(a^3\left(a-1\right)+2\left(a-1\right)=\left(a-1\right)\left(a^3+2\right)=0\)\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a^3+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=1\\a^3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=1\\\sqrt[3]{-2}\end{matrix}\right.\)

\(a^3\left(a-1\right)+2\left(a-1\right)=0\\ \Leftrightarrow\left(a-1\right)\left(a^3+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}a-1=0\\a^3+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=1\\a=\sqrt[3]{-2}\end{matrix}\right.\)

Nếu học căn bậc 3 rồi thì bạn lấy bài bạn Sky Sơn Tùng nhá!

Ta có :

\(1+a+a^2+....+a^{63}\)

\(=\left(1+a\right)+a^2\left(1+a\right)+....+a^{62}\left(1+a\right)\)

\(=\left(1+a\right)\left(1+a^2+a^4+....+a^{62}\right)\)

\(=\left(1+a\right)\left[\left(1+a^2\right)+a^4\left(1+a^2\right)+.....+a^{60}\left(1+a^2\right)\right]\)

\(=\left(1+a\right)\left(1+a^2\right)\left(1+a^4+....+a^{60}\right)\)

.....

\(=\left(1+a\right)\left(1+a^2\right).....\left(1+a^{32}\right)\)

Có \(\left(1+a\right)\left(1+a^2\right)...\left(1+a^{32}\right)=\frac{\left(a-1\right)\left(a+1\right)\left(a^2+1\right)...\left(a^{32}+1\right)}{a-1}\)

\(=\frac{\left(a^2-1\right)\left(a^2+1\right)...\left(a^{32}+1\right)}{a-1}\)

\(...\)

\(=\frac{\left(a^{32}-1\right)\left(a^{32}+1\right)}{a-1}\)

\(=\frac{a^{64}-1}{a-1}\)

\(=\frac{\left(a-1\right)\left(a^{63}+a^{62}+...+a^2+a+1\right)}{a-1}\)

\(=a^{63}+a^{62}+...+a^2+a+1\)

Vậy ...

ta có (a-1)(1+a+a²+......+a⁶³)=a⁶⁴-1

        (a-1)(a+1)(a²+1)....(a³²+1)=a⁶⁴-1

trần thùy duy bạn phải chừng minh rằng nó đúng khi a=1

Số số hạng của vế trái là :

  (a - 1) : 1 + 1 = a (số hạng)

Suy ra : \(\dfrac{\left(a+1\right)a}{2}=4095\), từ đó :

\(\left(a+1\right)a=4095\) x \(2=8190\)

Ta có :  90 x 91 = 8190 nên a = 90

                           Đ/s : 90