\(\frac{1}{3.7}+\frac{1}{7.11}+.....+\frac{1}{\left(4x+3\right)\left(4x+7\right)}=\frac{5}{12}\)
Những câu hỏi liên quan
Chứng minh \(\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}=\frac{5n}{4n+3}\)
\(CMR:\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}=\frac{5n}{4n+3}\)
Chứng minh rằng:
a,\(\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}=\frac{5n}{3.\left(4n+3\right)}\)
b,\(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}< \frac{1}{4}\)
Tìm x biết :
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{3\left(x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{3\left(x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)(Vì 3x + 3 lớn hơn 3x - 1 là 4 đơn vị)
\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{x+1-1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{x}{3x+3}=\frac{3}{10}\)
\(\Rightarrow10x=3.\left(3x+3\right)\)
\(\Rightarrow10x=9x+9\)
\(\Rightarrow x=9\)
Vậy...
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Chứng minh:
\(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)\(=\frac{5n}{4n+3}\)
=\(\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+........+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{7-3}{7.3}+\frac{11-7}{7.11}+........+\frac{\left(4n+3\right)-\left(4n-1\right)}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{7}{7.3}-\frac{3}{7.3}+\frac{11}{7.11}-\frac{7}{7.11}+......+\frac{4n+3}{\left(4n-1\right)\left(4n+3\right)}-\frac{4n-1}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7.}-\frac{1}{11}+......+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{3\left(4n+3\right)}-\frac{3}{3\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3-3}{3\left(4n+3\right)}\right)\)
\(=\frac{5}{4}.\frac{4n}{3\left(4n+3\right)}=\frac{4.n.5}{3\left(4n+3\right).4}=\frac{5n}{3\left(4n+3\right)}\)
ban nen xem lai dau bai di minh giai dung 100% do
ma neu dau bai ra nhu ket qua cua to thi tick cho minh nha
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Có ai giúp tôi với
Tôi sắp hi sinh rồi
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Có ai đó giúp tôi với
Tôi sắp tử vong rồi
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Xem thêm câu trả lời
1)2x(25x-4)-(5x-2)(5x+1)8 / 5)2left(x-2right)-3left(3x-1right)left(x-3right)
2)x(4x-3)-(2x-2)(2x-1)5 / 6)frac{2}{x+1}-frac{1}{x-2}frac{3x-11}{left(x+1right)left(x-2right)}
3)frac{5}{2x+3}+frac{3}{9-x^2}frac{8}{7left(x3right)} / 7)frac{5x-2}{6}+frac{3-4x}{2}2-frac{x+7}{3}
4)frac{2}{3left(x-2right)}+frac{5}{12-3x^2}frac{3}{4left(x+2right)} /...
Đọc tiếp
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
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Tìm x biết \(\frac{4}{3.7}+\frac{4}{7.11}+.....+\frac{4}{\left(3x-1\right).\left(3x+3\right)}=\frac{3}{10}\)
,\(a,\left(\frac{4}{3}-2x\right).\left(\frac{1}{2}.x^2-\frac{1}{8}\right)=0\)
\(b.1\frac{1}{2}-\frac{3}{2}\div\left(1-\frac{1}{2^{ }}\right)^2\)
\(c.\frac{5}{3.7}.\frac{12}{5.9}+\frac{12}{3.7}.\frac{54}{3.9}-\frac{1}{3-7}\)
\(d.\frac{1}{4.8}+\frac{1}{8.12}+\frac{1}{12.16}+......+\frac{1}{96.100}\)
\(a.\frac{4x-7}{12}-x=\frac{3x}{8}\\ b.\frac{5x-8}{3}=\frac{1-3x}{2}\\ c.\left(\frac{x-1}{\frac{2}{5}}-3\right)-\left(\frac{3x-2}{\frac{5}{4}}-2\right)=1\)
ai quen thì kb cả 2 tk nhé
a) \(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Rightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Rightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Rightarrow-56-64x=36x\)
\(\Leftrightarrow100x=-56\Leftrightarrow x=\frac{-14}{25}\)
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