Tìm x biết :
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+....+\frac{4}{\left(3x-1\right).\left(3x+3\right)}=\frac{3}{10}\)
Cần cách làm vì mk bt kết quả rồi !
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...
tìm x biết\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}......\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)
cho mình công thức luôn nha
\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-....-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)
Nên 3x + 3 = 30
3x = 30 - 3 = 27
x = 27 : 3 = 9
4( 1/3 - 1/7 + 1/7 - 1/11 +...+ 1/(3x-1) - 1/(3x+3) =3/10
4(1 - 1/(3x+3))= 3/10
phần còn lại tự giải nghe x=-71/111
Tìm x biết \(\frac{4}{3.7}+\frac{4}{7.11}+.....+\frac{4}{\left(3x-1\right).\left(3x+3\right)}=\frac{3}{10}\)
Tìm x, biết:
\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{x\left(x+4\right)}=\frac{43}{552}\)
\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{x\left(x+4\right)}=\frac{43}{552}\)
\(\Leftrightarrow\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{x}-\frac{1}{x+4}\right)=\frac{43}{552}\)
\(\Leftrightarrow\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{x+4}\right)=\frac{43}{552}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+4}=\frac{43}{552}\div\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+4}=\frac{43}{138}\Leftrightarrow\frac{1}{x+4}=\frac{1}{3}-\frac{43}{138}\)
\(\Leftrightarrow\frac{1}{x+4}=\frac{1}{46}\Leftrightarrow x+4=46\Rightarrow x=46-4=42\)
Vậy x = 42
\(s=\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{x\left(x+4\right)}=\)\(\frac{43}{552}\)
\(\Rightarrow S=\frac{4}{4}\left(\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{x\left(x+4\right)}\right)=\frac{43}{552}\)
\(\Rightarrow S=\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{x\left(x+4\right)}\right)=\frac{43}{552}\)
\(\Rightarrow S=\frac{1}{4}\left(\frac{4}{3}-\frac{4}{7}+\frac{4}{7}-\frac{4}{11}+...+\frac{4}{x}-\frac{4}{x+4}\right)=\frac{43}{552}\)
\(\Rightarrow S=\frac{1}{4}\left(\frac{4}{3}-\frac{4}{x+4}\right)=\frac{43}{552}\)
\(\Rightarrow\frac{4}{3}-\frac{4}{x+4}=\frac{43}{552}:\frac{1}{4}\)
\(\frac{\Rightarrow4}{3}-\frac{4}{x+4}=\frac{43}{138}\)
\(\frac{\Rightarrow4}{x+4}=\frac{4}{3}-\frac{43}{138}=\frac{47}{46}\)
\(\Rightarrow x+4=4:\frac{47}{46}=\frac{184}{47}\)
\(\Rightarrow x=\frac{184}{47}-4=\frac{-4}{47}\)
Đặt A=đã cho
=>\(4A=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{x\cdot\left(x+4\right)}\)
=>\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{n}-\frac{1}{n+4}=\frac{1}{3}-\frac{1}{n+4}\)
=>\(4\cdot\frac{43}{552}=\frac{1}{3}-\frac{1}{x+4}\)
=>\(\frac{43}{138}=\frac{46}{138}-\frac{1}{x+4}\left(1\right)\)
Từ đt (1),ta có thể suy ra 1/x+4=3/138
=>3*(x+4)=138
=>x+4=46
=>x=42
Vậy x=42
1) Tìm số tự nhiên x nhỏ nhất đê \(3^{2014}+3^x\)chia het cho 10
2) tìm \(\frac{x}{y}biet\frac{4x}{6y}=\frac{2x+8}{3y+11}\)
3) tìm X biết\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.........+\frac{4}{\left(3X-1\right)\left(3X+3\right)}=\frac{3}{10}\)
4) Tam giácABC có goca=90 độ BC=30cm AB :AC= 3 :4 tim AB= .......cm
5) Tam giác ABC góc a=90 đường caoAH ;BH=9 cm ;CH=16cm Tim AH
6) Tìm số tự nhiên x ;y biết \(2^{x+1}.3^y=36^x\)
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,\(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}< \frac{1}{4}\)
Chứng minh \(\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}=\frac{5n}{4n+3}\)
tìm \(E=\frac{\frac{4}{3.7}-\frac{4}{11.15}}{1-\frac{3}{7}-\frac{3}{11}+\frac{1}{5}}-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2006.2007}\right)\)
\(E=\frac{\frac{4}{3\cdot7}-\frac{4}{11.15}}{1-\frac{3}{7}-\frac{3}{11}+\frac{1}{5}}-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2006.2007}\right)\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{11}+\frac{1}{15}}{\frac{192}{385}}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2006}-\frac{1}{2007}\right)\)
\(=\frac{\frac{64}{385}}{\frac{192}{385}}-\left(\frac{1}{3}-\frac{1}{2007}\right)\)
\(=\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{2007}\right)=\frac{1}{2007}\)
Vậy : \(E=\frac{1}{2007}\)
tìm x\(\in\) Q,biết:
a,\(\frac{x}{2}\left(\frac{3x}{5}\right)-\frac{13}{5}=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
b,\(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
c,\(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
d,\(\frac{13}{x-1}+\frac{5}{2x+2}-\frac{6}{3x-3}\)
e,\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-2\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
Ai bt câu nào làm câu đấy nhé ! Làm hết cũng đc nhé!