Đặt \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\)

\(\Rightarrow2A=2+1+\frac{1}{2}+...+\frac{1}{2^9}\)

\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(\Rightarrow A=2-\frac{1}{2^{10}}\)

đặt \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)

\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^9}\)

\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(A=2-\frac{1}{2^{10}}\)

Đặt A =1 +  \(\frac{1}{2}\)+\(\frac{1}{2^2}\)...+\(\frac{1}{2^{10}}\)

Ta có : A =1 +  \(\frac{1}{2}\)+\(\frac{1}{2^2}\)...+\(\frac{1}{2^{10}}\)

=>     \(\frac{1}{2}\)A =  \(\frac{1}{2}\)+\(\frac{1}{2^2}\)...+\(\frac{1}{2^{10}}\)+ \(\frac{1}{2^{11}}\)

=> A - \(\frac{1}{2}\)A=    (     1 +  \(\frac{1}{2}\)+\(\frac{1}{2^2}\)...+\(\frac{1}{2^{10}}\) )   -  ( \(\frac{1}{2}\)+\(\frac{1}{2^2}\)...+\(\frac{1}{2^{10}}\)+ \(\frac{1}{2^{11}}\))

=>  \(\frac{1}{2}\)A = 1 -  \(\frac{1}{2^{11}}\)

=>  \(\frac{1}{2}\)A= \(\frac{2^{11}-1}{2^{11}}\)

=> A = \(\frac{2^{11}-1}{2^{10}}\)

Vậy A = \(\frac{2^{11}-1}{2^{10}}\)

a) a^3+b^3

 =(a+b).(a^2-ab+b^2)

=S.(a^2+2ab+b^2-3ab)

=S.(a+b)^2-3ab

=S.S^2-3P

=S^3-3P

c) C=a^2+ab+b^2

=a^2+2ab+b^2-ab

=(a+b)^2-ab

=S^2-P

b) B=a^2+B^2

=A^2+2ab+b^2-2ab

=(a+b)^2 -2ab

=S^2-2P

a: Xét tứ giác AEMF có 

\(\widehat{MEA}=\widehat{MFA}=\widehat{FME}=90^0\)

Do đó: AEMF là hình chữ nhật

a)Tứ giác AEMF có :

\(\widehat{MEA}=\widehat{MFA}=\widehat{FME}=90^0\)

=>AEMF là hình chữ nhật

\(\frac{7}{3}\)\(+\frac{1}{2}\)\(+\frac{-3}{70}\)\(=\frac{293}{105}\)
\(\frac{5}{12}\)\(+\frac{3}{-16}\)\(+\frac{3}{4}\)\(=\frac{47}{48}\)
\(\frac{5}{3}\)\(+\left(7+\frac{-5}{3}\right)=\frac{5}{3}\)\(+\frac{-5}{3}\)\(+7=0+7=7\)
\(\frac{-7}{31}\)\(+\left(\frac{24}{17}+\frac{7}{31}\right)=\left(\frac{-7}{31}+\frac{7}{31}\right)+\frac{24}{17}=0+\frac{24}{17}\)\(=\frac{24}{17}\)
\(\frac{3}{7}\)\(+\left(\frac{-1}{5}+\frac{-3}{7}\right)=\left(\frac{3}{7}+\frac{-3}{7}\right)+\frac{-1}{5}\)\(=0+\frac{-1}{5}\)\(=\frac{-1}{5}\)
Nếu được cho mình xin 1 k đúng ^_^

Bài 1 : 

\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\Rightarrow x=16;y=24;z=30\)

bài 2 : 

Đặt \(x=2k;y=5k\Rightarrow xy=10k^2=10\Leftrightarrow k^2=1\Leftrightarrow k=\pm1\)

Với k = 1 thì x = 2 ; y = 5

Với k = - 1 thì x = -2 ; y = -5

\(a,a^2+b^2=\left(a+b\right)^2-2ab=3^2-2\left(-10\right)=29\\ b,a^2+b^2=\left(a-b\right)^2+2ab=2^2+2\cdot24=52\)

Hộp bánh còn lại là : 4 - 7/4 = 9/4 ( chiếc bánh )

Đ/s : 9/4 chiếc bánh