giải giúp mình nhé.
1/3+1/6+1/10+...+1/x×(x+1)÷2=399/400 (tìm x).
K
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TD
2
3 tháng 7 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{399}{400}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{399}{400}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{399}{400}:2\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{399}{400}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{399}{800}\)
\(\frac{1}{x+1}=\frac{400}{800}-\frac{399}{800}\)
\(\frac{1}{x+1}=\frac{1}{800}\)
\(=>x+1=800\)
\(=>x=800-1=799\)
Vậy x = 799
Ủng hộ mk nha ^_-
TD
1
4 tháng 7 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{399}{400}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{399}{400}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{399}{400}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{399}{400}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{200}\)
\(\frac{1}{x+1}=\frac{-299}{200}\)
\(x+1=\frac{-200}{299}\)
\(x=\frac{-499}{299}\)
TD
0
15 tháng 9 2019
ko ghi lại đề nha !
a) \(\Leftrightarrow x^3+3x^2+9x-3x^2-9x-27+x\left(2^2-x^2\right)=0\)
\(\Leftrightarrow x^3+3x^2+9x-3x^2-9x-27+4x-x^3=0\)
\(\Leftrightarrow-27+4x=0\)
\(\Leftrightarrow4x=27\)
\(\Leftrightarrow x=6,75\)
b)\(\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)
\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=-0,5\)
NN
0
NH
1
26 tháng 5 2016
theo bai co 1/3+1/6+...+2/x.x+1=499/1000
(=)2/2.3+2/3.4+...+2/x.(x+1)=499/1000
(=)2(1/2-...-1/x+1)=499/1000
(=)1/2-1/x+1=499/2000
(=)1/x+1=501/2000
den day thi minh chiu roi!
AH
Akai Haruma
Giáo viên
17 tháng 4 2023
Lời giải:
a.
$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$
b.
$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$
$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$
c.
$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$
AH
Akai Haruma
Giáo viên
17 tháng 4 2023
d.
$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$
$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$
e.
$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$
$\frac{-19}{15}: x=1$
$x=\frac{-19}{15}:1 =\frac{-19}{15}$
f.
$(-\frac{3}{4}+x).2\frac{2}{3}=1$
$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$
$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$
Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=\frac{399}{400}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\left(\text{Quy đồng nhé !}\right)\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{399}{400}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{399}{400}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{399}{400}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{399}{800}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{800}\)
=> x + 1 = 800
<=> x = 799
1/3+1/6+1/10+...+1/x(x+1):2=399/400
2.[1/3.2+1/6.2+1/10.2+...+1/x(x+1)]=399/400
2.[1/6+1/12+1/20+...+1/x(x+1)]=399/400
2.[1/2.3+1/3.4+1/4.5+...+1/x(x+1)]=399/400
1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=399/800
1/2-1/x+1=399/800
1/x+1=1/800
=> x+1=800
=> x=799