tìm số nguyên âm a biết:
a^2 -[3/5]^2=1/1.2+1/2.7+1/7.5+1.5/13+1/13.8+1/8.19+1/19.11+1/11.25
a2+ ( 3/5)2= 1/1.2+1/2.7+1/7.5+1/5.13+1/13.8+1/8.19+1/19.11+1/11.25
tìm số nguyên âm a
1.Tìm số nguyên âm a, biết:
a2 = (2\5)2 =1\1.2+1\2.7+1\7.5+1\5.13+1\13.8+1\8.19+1\19.11+1\11.25
TÌm số nguyên âm a, biết
a2-(3/5)^2=\(\frac{3}{5}=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{1.13}+\frac{1}{13.8}+\frac{1}{8.19}+\frac{1}{19.11}+\frac{1}{11.25}\)
a mũ 2 -(3/5) mũ 2=1/1.2+1/2.7+1/7.5+1/5.13+1/13.8+1/8.19+1/19.11+1/11.25
Giải:
\(A^2-\left(\dfrac{3}{5}\right)^2=\dfrac{1}{1.2}+\dfrac{1}{2.7}+\dfrac{1}{7.5}+\dfrac{1}{5.13}+\dfrac{1}{13.8}+\dfrac{1}{8.19}+\dfrac{1}{19.11}+\dfrac{1}{11.25}\)
Gọi: \(B=\dfrac{1}{1.2}+\dfrac{1}{2.7}+\dfrac{1}{7.5}+\dfrac{1}{5.13}+\dfrac{1}{13.8}+\dfrac{1}{8.19}+\dfrac{1}{19.11}+\dfrac{1}{11.25}\)
\(B=\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+\dfrac{1}{13.16}+\dfrac{1}{16.19}+\dfrac{1}{19.22}+\dfrac{1}{22.25}\right):\dfrac{1}{2}\) \(B=\left[\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{19.22}+\dfrac{3}{22.25}\right)\right]:\dfrac{1}{2}\)
\(B=\left[\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{25}\right)\right]:\dfrac{1}{2}\)
\(B=\left[\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{25}\right)\right]:\dfrac{1}{2}\)
\(B=\left[\dfrac{1}{3}.\dfrac{24}{25}\right]:\dfrac{1}{2}\)
\(B=\dfrac{8}{25}:\dfrac{1}{2}\)
\(B=\dfrac{16}{25}\)
\(\Rightarrow A^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\)
\(A^2=\dfrac{16}{25}+\dfrac{9}{25}\)
\(A^2=1\)
\(\Rightarrow A^2=1^2\) hoặc \(A^2=\left(-1\right)^2\)
\(A=1\) hoặc \(A=-1\)
Chúc bạn học tốt!
a^2- (3/5)^2= 1/1.2+1/2.7+1/7.5+1/5.13+1/13.8+1/8.19+1/19.11+1/11.25
a)Tìm số nguyên âm a,biết:
a^2-(3/5)^2=1/1.2+1/2.7+1/7.5+1/5.13+1/13.8+1/8.19+1/19.11+1/11.25
b)So sánh: C=1.3.5.7....99 với D=51/9.52/2.53/2....100/2
Giúp mình với mình đang cần gấp
a2 - 3/52 = 1/1.2 + 1/2.7 +1/7.5 + 1/5.13 + 1/13.8 + 1/8.19 + 1/19.11 + 1/11.25
\(a^2-\frac{3}{5^2}=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}+\frac{1}{19.11}+\frac{1}{11.25}\)
\(a^2-\frac{3}{5^2}=2.\left(\frac{1}{2.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}+\frac{1}{19.22}+\frac{1}{22.25}\right)\)
\(a^2-\frac{3}{5^2}=2.\frac{1}{3}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{22}-\frac{1}{25}\right)\)
\(a^2-\frac{3}{5^2}=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{25}\right)\)
=> \(a^2-\frac{3}{25}=\frac{2}{3}.\frac{23}{50}=\frac{23}{75}\)
=> \(a^2=\frac{23}{75}+\frac{3}{25}=\frac{32}{75}\)
=> \(a=\sqrt{\frac{32}{75}}\)(Nếu thế thì đây phải là đề của lớp 7 chứ nhỉ)
Tìm a nguyên âm , biết :
a2 - (3/5)2= 1 / 1.2 + 1/2.7 + 1/7.5 + 1/5.13+ 1/13.8+1/8.19+1/19/11+1/11.25
A= 1 trên 1.2+1 trên 2.7+1 trên 7.5+......+1 trên 97.100