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ví dụ
a là 1
b là 2
ta có
1/1 - 1/2 và 1/1x2
= 1/2 và 1/2
khi đó ta thấy 1/2 = 1/2
và 1/1 - 1/2 = 1/1x2
a)Ta có:
A= 1/15+1/35+1/63+1/99+1/143
A= 1/3.5+1/5.7+1/7.9+1/9.11+1/11.13
2A= 2/3.5+2/5.7+2/7.9+2/9.11+2/11.13
2A= 1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
Đơn giản đi ta được:
2A= 1/3-1/13
2A= 10/39
A= 5/39
Vậy A= 5/39
b) Để A và B có giá trị bằng nhau thì:
\(\frac{3}{4}\cdot x+7=\frac{4}{3}\cdot x-35\)
\(7+35=\frac{4}{3}\cdot x-\frac{3}{4}\cdot x\)
\(42=\frac{7}{12}\cdot x\)
\(x=42:\frac{7}{12}\)
\(x=72\)
a) 20,18 x 36 + 63 x 20,18 + 20,18
= 20,18 x ( 36+63+1)
= 20,18 x 100
= 2018
b) \(\frac{a}{b}\times\frac{3}{5}=\frac{1}{5}+\frac{2}{3}\)
\(\frac{a}{b}\times\frac{3}{5}=\frac{13}{15}\)
\(\frac{a}{b}=\frac{13}{15}:\frac{3}{5}\)
\(\frac{a}{b}=\frac{13}{9}\)
a) 20,18.(36+63+1)
= 20,18 . 100
= 2018
a/b . 3/5 =3/15 + 10/15
a/b . 3/5 =13/15
a/b = 13/15 : 3/5
a/b = 13/9
2/
a) \(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+\frac{4}{17\cdot21}\)
\(=\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+....+\frac{1}{17}-\frac{1}{21}\right)\)
\(=1-\frac{1}{21}=\frac{20}{21}\)
b) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot..\cdot\frac{2016}{2017}\)
\(=\frac{1}{2017}\)
c) \(A=2000-5-5-5-..-5\)(có 200 số 5)
\(A=2000-\left(5\cdot200\right)\)
\(A=2000-1000\)
\(A=1000\)
a, \(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}=\frac{1\cdot2\cdot3\cdot4\cdot5}{2\cdot3\cdot4\cdot5\cdot6}=\frac{1}{6}\)
b, 10,3 x 37,5 - 37,5 x 10,2
= 37,5 x ( 10,3 - 10,2 )
= 37,5 x 0,1
= 3,75
a) \(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\)
\(=\frac{1\times2\times3\times4\times5}{2\times3\times4\times5\times6}\)
\(=\frac{1}{6}\)
b) 10,3 x 37,5 - 37,5 x 10,2
= 37,5 x ( 10,3 - 10,2 )
= 37,5 x 0,1
= 3,75
\(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+...+\dfrac{1}{a\times\left(a+4\right)}=\dfrac{50}{609}\)
\(\dfrac{1}{4}\times\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+...+\dfrac{4}{a\times\left(a+4\right)}\right)=\dfrac{50}{609}\)
\(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{a}-\dfrac{1}{a\times4}=\dfrac{50}{609}\div\dfrac{1}{4}\)
\(\dfrac{1}{3}-\dfrac{1}{a\times4}=\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{3}-\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{203}\)
\(a\times4=203\)
\(a=\dfrac{203}{4}\)
\(\dfrac{1}{3\times7}\)+\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\)
4\(\times\)( \(\dfrac{1}{3\times7}\) +\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\)) = \(\dfrac{50}{609}\) \(\times\)4
\(\dfrac{4}{3\times7}\)+ \(\dfrac{4}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{4}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\) \(\times\) 4
\(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\)-\(\dfrac{1}{15}\)+...+\(\dfrac{1}{a}\)-\(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{3}\) - \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{203}\)
a + 4 = 203
\(a\) = 203 - 4
\(a\) = 199
Đáp số: \(a\) = 199
Lời giải:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{a\times b}+a+b=a\times b$
$\frac{a+b}{a\times b}+(a+b)+\frac{1}{a\times b}+1=a\times b+1$
$(a+b)\times (\frac{1}{a\times b}+1)+(\frac{1}{a\times b}+1)=a\times b+1$
$(\frac{1}{a\times b}+1)\times (a+b+1)=a\times b+1$
$\frac{(a\times b+1)\times (a+b+1)}{a\times b}=a\times b+1$
$(a\times b+1)\times (\frac{a+b+1}{a\times b}-1)=0$
$\Rightarrow a\times b+1=0$ hoặc $\frac{a+b+1}{a\times b}=1$
Hiển nhiên $a\times b+1>0$ với $a,b$ là số tự nhiên.
$\Rightarrow \frac{a+b+1}{a\times b}=1$
$\Rightarrow a+b+1=a\times b$
$a\times b-a-b=1$
$a\times (b-1)-(b-1)=2$
$(b-1)\times (a-1)=2=1\times 2=2\times 1$
TH1:
$a-1=2, b-1=1\Rightarrow a=3; b=2$
TH2:
$a-1=1, b-1=2\Rightarrow a=2; b=3$