Cho A = 1-$\frac{3}{4}$ +($\frac{3}{4}$ )$^{2}$ -($\frac{3}{4}$ )$^{3}$ +($\frac{3}{4}$ )$^{4}$ -…-($\frac{3}{4}$ )$^{2009}$ +($\frac{3}{4}$ )$^{2010}$
Chứng tỏ A không phải là số nguyên
Cho A = 1-$\frac{3}{4}$ +($\frac{3}{4}$ )$^{2}$ -($\frac{3}{4}$ )$^{3}$ +($\frac{3}{4}$ )$^{4}$ -…-($\frac{3}{4}$ )$^{2009}$ +($\frac{3}{4}$ )$^{20
By Adalyn
Giải thích các bước giải:
$A=1-\dfrac{3}{4}+(\dfrac{3}{4})^2-(\dfrac{3}{4})^3+(\dfrac{3}{4})^4-…-(\dfrac{3}{4})^{2009}+(\dfrac{3}{4})^{2010}$
$\rightarrow A.(\dfrac{3}{4})^2 = (\dfrac{3}{4})^2-(\dfrac{3}{4})^2+(\dfrac{3}{4})^4-(\dfrac{3}{4})^5+….-(\dfrac{3}{4})^{2011}+(\dfrac{3}{4})^{2012}$
$\rightarrow A.(\dfrac{3}{4})^2-A=-(\dfrac{3}{4})^{2011}+(\dfrac{3}{4})^{2012}-(1-\dfrac{3}{4})$
$\rightarrow A((\dfrac{3}{4})^2-1)=(\dfrac{3}{4})^{2011}(\dfrac{3}{4}-1)+(\dfrac{3}{4}-1)$
$\rightarrow A(\dfrac{3}{4}-1)(\dfrac{3}{4}+1)=((\dfrac{3}{4})^{2011}+1)(\dfrac{3}{4}-1)$
$\rightarrow A(\dfrac{3}{4}+1)=(\dfrac{3}{4})^{2011}+1$
$\rightarrow A.\dfrac{7}{4}=(\dfrac{3}{4})^{2011}+1$
$\rightarrow A=\dfrac{4}{7}((\dfrac{3}{4})^{2011}+1)$
$\rightarrow A$ không là số nguyên
Đáp án:
A=1−34+(34)2−(34)3+(34)4−...−(34)2009+(34)2010A=1−34+(34)2−(34)3+(34)4−…−(34)2009+(34)2010
→A.(34)2=(34)2−(34)2+(34)4−(34)5+....−(34)2011+(34)2012→A.(34)2=(34)2−(34)2+(34)4−(34)5+….−(34)2011+(34)2012
→A.(34)2−A=−(34)2011+(34)2012−(1−34)→A.(34)2−A=−(34)2011+(34)2012−(1−34)
→A((34)2−1)=(34)2011(34−1)+(34−1)→A((34)2−1)=(34)2011(34−1)+(34−1)
→A(34−1)(34+1)=((34)2011+1)(34−1)→A(34−1)(34+1)=((34)2011+1)(34−1)
→A(34+1)=(34)2011+1→A(34+1)=(34)2011+1
→A.74=(34)2011+1→A.74=(34)2011+1
→A=47((34)2011+1)→A=47((34)2011+1)
→A→A không là số nguyên
Giải thích các bước giải: