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\(\text{Đ}\text{ặt}:A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}\)
\(A=\frac{100}{101}:2=\frac{50}{101}\)
\(\Rightarrow\frac{1}{3}x.x=\frac{50}{101}\)
\(x.\left(\frac{1}{3}.1\right)=\frac{50}{101}\)
\(x.\frac{1}{3}=\frac{50}{101}\)
$x=\frac{50}{101}:\frac{1}{3}=\frac{150}{101}$
\(.\frac{1}{3}x.x=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\frac{1}{3}xx=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(\frac{1}{3}xx=\frac{1}{2}.\left(\frac{100}{101}\right)\)
\(\frac{1}{3}xx=\frac{50}{101}\)
\(x.x=\frac{150}{101}\)
còn lại tự tính
Đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(A=\frac{50}{101}\)
b) \(\frac{2^{10}+3^{31}+2^{40}+3^6}{2^{11}\cdot3^{31}+2^{41}\cdot3^6}=\frac{2^{10}+2^{40}}{2^{11}+2^{41}}\)
\(\frac{2^{10}+2^{40}}{2^{11}+2^{41}}=\frac{1}{2}\)
=1/2x(1/1.3+1/3.5+...+1/99.101)
=1/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)
=1/2.(1-1/101)
=1/2.100/101
=50/101
chúc bạn học tốt
\(A=xemlai\) chưa hưa hiểu Quy luật
\(B=\frac{\left(n.\left(n+2\right)+1\right)}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(B=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.5}...\frac{98.98}{97.99}\frac{99.99}{98.100}\frac{100.100}{99.101}\\\)
\(B=\frac{2.100}{1.101}=\frac{200}{101}\)
1/2[2/1.3+2/3.5+2/5.7+.........+2/x(x+2)]=16/34
2/1.3+2/3.5+2/5.7+......+2/x(x+2)=16/34:1/2=16/17
1/1-1/3+1/3-1/5+1/5-1/7+.....+1/x-1/x+2=16/17
1-1/x+2=16/17
1/x+2=1-16/17=1/17
suy ra:x+2=17
x=17-2
x=15
\(S=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
X+(1/1.3+1/3.5+1/5.7+...+1/99.101)=100
X+(2/1.3+2/3.5+2/5.7+...+2/99.101)=100
X+(1 -1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)=100
X+(1-1/101)=100
X+100/101=100
X=100-100/101
X=10000/101
ĐK : 51x \(\ge0\Rightarrow x\ge0\)
Với \(x\ge0\)thì \(x+\frac{1}{1.3}>0;x+\frac{1}{3.5}>0;...;x+\frac{1}{99.101}>0\)
Khi đó : \(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+\left|x+\frac{1}{5.7}\right|+...+\left|x+\frac{1}{99.101}\right|=51x\)
<=> \(x+\frac{1}{1.3}+x+\frac{1}{3.5}+x+\frac{1}{5.7}+....+x+\frac{1}{99.101}=51x\)(50 hạng tử x ở VT)
<=> \(50x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}=51x\)
<=> \(x=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\right)\)
<=> \(x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
<=> \(x=\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{50}{101}\)
Vậy x = 50/101
\(A=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{4}{9}-\frac{1}{5}=\frac{11}{45}\)
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{9}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
\(S=\frac{4}{9}-\frac{1}{5}\)
\(S=\frac{11}{45}\)
Ta có : 1,12(32) = 1,12 + 0,0032
Mà 0,0032 = 32/9990
Nên : 1,12(32) = 28/25 + 32/9990 = 556/495
Nhập vào máy : Sích mak
công thức (2n - 1) ( 2n + 1) x chạy từ 1 đến 15 ok
C1:
Ta có
x+10%x-10%x=297
=>x=297
C2:
S=Đề bài...
=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)
=\(\frac{1}{1}-\frac{1}{101}\)
=\(\frac{100}{101}\)
#hoctot
Câu 2:
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\)
\(\Rightarrow2S=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}\)
\(\Rightarrow2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}+\frac{1}{101}\)
\(\Rightarrow2S=1-\frac{1}{101}\)
\(\Rightarrow2S=\frac{100}{101}\)
\(\Rightarrow S=\frac{100}{101}:2=\frac{100}{101}.\frac{1}{2}=\frac{50}{101}\)