Tìm x,biết rằng:
a)1/5.8+1/8.11+1/11.14+...1/x(x+3)=101/1540
b)1+1/3+1/6+1/10+...+1/x(x+1):2=1 1991/1993
*1 1991/1993 là hỗn số nha
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tìm x biết
1/5.8+1/8.11+1/11.14+...+1/x(x+3)=101/15401+1/3+1/6+1/10+...+1/x(x+1):2=1 1991/1993
tìm x :
a, 1/5.8+1/8.11+11/11.14+...+1/x.(x+3)=101/1540
b, 1+1/3+1/6+1/10+...+1/x.(x+1):2=1 và 1991/1993
tìm x biết
a 1/5.8+1/8.11+1/11.14+...1/x(x+3)=101/1540
b 1+1/3+1/6+1/10+...1/x(x+1)=\(1\frac{1991}{1993}\)
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(=3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x.3}=\frac{303}{1540}\)
\(=\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(=\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(=\frac{1}{x+3}=\frac{1}{308}\)
\(x+3=308\)
\(\Rightarrow x=305\)
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Tìm x, biết:
a) 1/5.8 + 1/8.11 + 1/11.14 + ... + 1/x.(x+3)= 101/1540
b) 1+ 1/3 + 1/6 + 1/10 +...+ 1/x.(x+1):2 = \(1\frac{1991}{1993}\)
a) 1/5.8+1/8.11+1/11.14+......+1/x.(x+3)=101/1540
1/3.3.[1/5.8+1/8,11+1/11.14+......+1/x.(x+3)=101/1540
1/3.[3/5.8+3/8.11+3/11.14+........+3/x.(x+3)]=101/1540
1/3.[1/5-1/8+1/8-1/11+1/11-1/14+....+1/x-1/x+3=101/1540
1/3.[1/5-1/x+3]=101/1540
1/5-1/x+3=101/1540.3
1/5-1/x+3=303/1540
1/x+3=1/3-303/1540=1/308
=>x+3=308 =>x=305
Vậy x=305
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1/3.3(1/5.8+1/8.11+1/11.14+.....1/x(x+1)_101/1540
1/3.(1/5-1/8+1/8-1/11+1/11-1/14+....1/x+1/x+3)=101/1540
1/3.(1/5-1/x+3)=101/1540
1/5-1/x+3=101/1540/1/3=303/1540
1/x+3=1/5-303/1540=1/308
x+3+308
x=305
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sao lại 1/x+3=1/3-303/1540=1/308 phải là 1/x+3=1/5-303/1540=1/108
x+3=108
x=105
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Tìm x, biết:
a) 1/5.8 + 1/8.11 + 1/11.14 + ... + 1/x.(x+3)= 101/1540
b) 1+ 1/3 + 1/6 + 1/10 +...+ 1/x.(x+1):2 = \(1\frac{1991}{1993}\)
tính x biết: 1/5.8 + 1/8.11+1/11.14+ ....... +1/x(x+3)= 1 1991/1993
Tìm x, biết rằng:
a)\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
b)\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
a)Ta có \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
=)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
=)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
Suy ra \(\frac{1}{5}-\frac{1}{x+3}\)= \(\frac{303}{1540}\)=)\(\frac{1}{x+3}=\frac{1}{305}\)=) \(x+3=305\)=) \(x=302\)
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Giup minh voi!
Tim x:
a) 1/5.8 + 1/8.11 + 1/11.14 +........+ 1/x(x+3) = 101/1540
b) 1 + 1/3 + 1/6+ 1/10+......+ 1/x(x+1):2 = hon so 1 1991/1993
Ai nhanh nhat dung nhat tick lun!(Nho ghi cach giai ro rang)
còn chi tiết đây
a)1/5.8+1/8.11+1/11.14+...+1/x.(x+3)=101/1540
1/(5.8)+1/(8.11)+1/(11.14)+...1/x.(… =101/1540
3/(5.8)+3/(8.11)+...+3/x(x+3)=3.(10…
1/5-1/8+1/8-1/11+...+1/x-1/(x+3)=30…
1/5-1/(x+3)=303/1540
1/(x+3)=1/5-303/1540=1/308
=>x=305
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lời giải nè : ấn vô dòng đen đen ở dưới ấy nhé
Tìm x, biết:a) 1/5.8 + 1/8.11 + 1/11.14 + ... + 1/x.(x+3)= 101/1540b) 1+ 1/3 + 1/6 + 1/10 +...+ 1/x.(x+1):2 = $1\frac{1991}{1993}$119911993
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Bài 4:Tìm x biết:
a,15.8+1/8.11+1/11.14+...+1/x.(x+3)=101/1540
b,1+1/3+1/6+1/10+...+1/x.(x+1)+2=1 và 1991/1993