1/15+1/21+1/28 +... 2/Xx(X+1)=1/3
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Ta có : A=20/11×13 + 20/13×15 +20/15×17+...+20/53×55
A = 10 ×( 2/11×13+2/13×15+...12/53×55)
A = 10 ×(1/11-1/13+1/13-1/15+1/15-1/17+...+1/53-1/55)
A = 10 × (1/11-1/55)
A =10 × 4/55
A = 8/11
\(\Rightarrow\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{x.\left(x+1\right)}=2.\left(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x.\left(x+1\right)}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}\right)=2.\left(\frac{1}{5}-\frac{1}{x+1}\right)=\frac{2}{5}-\frac{2}{x+1}=\frac{3}{10}\)
=> \(\frac{2}{x+1}\)= \(\frac{1}{10}=\frac{2}{20}\)
=> x +1 = 20 => x = 19
bạn trên sai rồi, nếu đã nhân đôi lên tất cả thì cx phải nhân luôn con cuối chứ
\(\frac{1}{15}+\frac{1}{21}+...+\frac{2}{x.\left(x+1\right)}=\frac{806}{2015}\)
\(\Rightarrow2.\left(\frac{1}{30}+\frac{1}{42}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{806}{2015}\)
\(\Rightarrow2.\left(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{806}{2015}\)
\(\Rightarrow2.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{806}{2015}\)
\(\Rightarrow2.\left(\frac{1}{5}-\frac{1}{x}\right)=\frac{806}{2015}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x}=\frac{806}{2015}:2\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x}=\frac{403}{2015}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{5}-\frac{403}{2015}\)
\(\Rightarrow\frac{1}{x}=\frac{403}{2015}-\frac{403}{2015}\)
\(\Rightarrow\frac{1}{x}=0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
Chúc bạn học tốt !!!!
\(\Rightarrow\frac{1}{2}\left(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{806}{2015}.\frac{1}{2}\)
\(\Rightarrow\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{x\left(x+1\right)}=\frac{403}{2015}\)
\(\Rightarrow\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}=\frac{403}{2015}\)
\(\Rightarrow\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+...+\frac{x+1-x}{x\left(x+1\right)}=\frac{403}{2015}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{403}{2015}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{403}{2015}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{403}{2015}\)
rồi bạn tự giải nốt nhé
\(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+....+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)
\(\Leftrightarrow\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+....+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)
\(\Leftrightarrow\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+....+\frac{2}{x\left(x+1\right)}=\frac{11}{40}\)
\(\Leftrightarrow2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{11}{40}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{11}{40}\div2\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{11}{80}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{16}\Rightarrow x=16-1=15\)
a) 3/4 + -1/8 = 5/8
b)-5/12 + -7/24 = -9/8
c) 4/21 - -5/28 = 31/84
d) 1 + -7/28 = 3/4
e) -4/3 - 17/6= -25/6
f) 1/3 - ( 1/2 +1/8 )= -7/24
g)1/21 - ( 1/7 - 1/3 ) = 5/21
h)1/2 - 1/4 + 1/13 + 1/8= 47/104
a) x - 1/10 = 1/15
x=1/15+1/10
x=1/6 Vay x=1/6b) -4/21 - x = -3/7
x=-4/21+3/7 x=5/21 Vay x=5/21c) x + 1/2 = 3/4 - (-1/2)
x+1/2= 5/4
x= 5/4-1/2
x=3/4
Vay x=3/4
d) 4/7 - x = 1/3 - (-2/3)
x= 4/7-1/3-2/3 x= -3/7 Vay x=-3/7
\(\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\times\left(x+1\right)}=\dfrac{1}{3}\\ \dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\times\left(x+1\right)}=\dfrac{1}{3}\\ 2\times\left(\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+...+\dfrac{1}{x\times\left(x+1\right)}\right)=\dfrac{1}{3}\\ \dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{3}:2\\ \dfrac{1}{5}-\dfrac{1}{x+1}=\dfrac{1}{6}\\ \dfrac{1}{x+1}=\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{1}{30}\\ x+1=30\\ x=29\)
Lời giải:
** Sửa đề:
$\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{2}{x(x+1)}=\frac{1}{3}$
$\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x(x+1)}=\frac{1}{3}$
$\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{x(x+1)}=\frac{1}{3}$
$2(\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+...+\frac{x+1-x}{x(x+1)})=\frac{1}{3}$
$2(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1})=\frac{1}{3}$
$2(\frac{1}{5}-\frac{1}{x+1})=\frac{1}{3}$
$\frac{1}{5}-\frac{1}{x+1}=\frac{1}{6}$
$\frac{1}{x+1}=\frac{1}{5}-\frac{1}{6}=\frac{1}{30}$
$\Rightarrow x+1=30$
$\Rightarrow x=29$.