Giúp mình gấp
73-x/98-x=1/6
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73-x/98-x=1/6
<=>73-x:98-x=1/6
<=>(73-98):x =1/6
<=>-25:x =1/6
<=> x =-25 : 1/6
<=> x =-150
Vậy x=-150
Sửa đề : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+....+\frac{\left(x+1\right)-x}{x\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{\left(x+1\right)}=\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{1}-\frac{98}{100}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{50}\)
\(\Leftrightarrow x=50-1=49\)
Sửa đề: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
(=) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{98}{100}\)
(=)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{98}{100}\)
(=)\(1-\frac{1}{x+1}=\frac{98}{100}\)
(=)\(\frac{1}{x+1}=1-\frac{98}{100}\)
(=)\(\frac{1}{x+1}=\frac{1}{50}\)=> \(x+1=50\)
\(x=50-1\)
\(x=49\)
T_i_c_k cho mình nha,thanks you so much!
số cần tìm là :
73/ 98 - 1/6 = 85/147
Đ / S : ......
Nhớ K mik nha
\(A=2\times4+4\times6+...+98\times100\)
\(A=2.2+2.12+...+2.149\)
\(A=2.\left(2+12+...+149\right)\)
Tự làm nốt( dấu x = dấu chấm )
\(=\left(2+4+6+...+98\right)\left(6-6\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\right)\)
=0
Ta có:
$A=\dfrac{1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+98)}{1\times98+2\times97+3\times96+\cdots+98\times1}$
Tử số:
$1+(1+2)+(1+2+3)+\cdots+(1+2+\cdots+98)$
$=1+2+2^2+?$
Cách dễ nhất là viết:
$1+2+\cdots+k=\dfrac{k(k+1)}2$
Nên tử số:
$=\dfrac{1\cdot2}{2}+\dfrac{2\cdot3}{2}+\cdots+\dfrac{98\cdot99}{2}$
$=\dfrac{1}{2}(1\cdot2+2\cdot3+\cdots+98\cdot99)$
$=\dfrac{1}{2}\cdot\dfrac{98\cdot99\cdot100}{3}$
$=161700$
Mẫu số:
$1\cdot98+2\cdot97+3\cdot96+\cdots+98\cdot1$
$=1(99-1)+2(99-2)+\cdots+98(99-98)$
$=99(1+2+\cdots+98)-(1^2+2^2+\cdots+98^2)$
$=99\cdot\dfrac{98\cdot99}{2}-\dfrac{98\cdot99\cdot197}{6}$
$=161700$
Do đó:
$A=\dfrac{161700}{161700}=1$
Vậy $A=1$.