tìm x
\(\frac{1}{2.4}+\frac{1}{4.6}+.........+\frac{1}{\left(2x-2\right)2x}=\frac{1}{8}\)
giúp mình với nha
Tìm x biết:
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\left(x\in N,x\ge2\right)\)
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}\right)=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{8}:\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
TL:
\(\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{\left(2x-2\right)2x}\right)=\frac{1}{8}\)
\(\frac{1}{2}-\frac{1}{4x}=\frac{1}{8}\)
\(\frac{1}{4x}=\frac{3}{8}\)
=>x=2/3
hc tốt
i dont know
tìm x ạ
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{\left(2x-2\right)2x}=\frac{1}{8}\)
có 1 bài này hoy ạ mọi người giúp em giải đi
Ai nhanh và đúng nhất em tích cho
Nha nha mơn nhìu
\(\Rightarrow2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2x-2\right).2x}\right)=\frac{1}{8}.2\).2
\(\Rightarrow\frac{2}{2.4}+\frac{2}{4.6}+...\frac{2}{\left(2x-2\right).2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\Rightarrow\frac{1}{2x}=\frac{1}{2.2}\)
\(\Rightarrow x=2\)
=>1/2 - 1/4 + 1/4 -1/6 +1/6- 1/8 +1/2x-2 - 1/2x =1/8
=>1/2 - 1/8 + 1/2x-2 - 1/2x= 1/8
=>3/8 + 1/2x-2 - 1/2x =1/8
=>1/2x-2 - 1/2x = 1/2
=>1/(2x-2) . 2x =1/2
=> (2x-2) . 2x =2
=> 4x2 -4x=2
=> (x+1). 4x =2=-2.-1=2.1
=>x+1=-2 =>x=-3
Hay x+1=2 =>x=1
=>4x=1 => x=1/4
Hay 4x=-1 =>x=-1/4
K MK NHA. CHÚC BẠN HỌC GIỎI
Tìm x, biết:
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)(\(x\inℕ,\)\(x\ge2\))
\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)\cdot2x}=\frac{1}{8}\left(x\inℕ;x\ge2\right)\)
Đặt \(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)2x}\)
\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{\left(2x-2\right)2x}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2x-2}-\frac{1}{2x}\)
\(2A=\frac{1}{2}-\frac{1}{2x}=\frac{x-1}{2x}\)
\(\Rightarrow A=\frac{x-1}{2x}:2=\frac{x-1}{2x}\cdot\frac{1}{2}=\frac{x-1}{4x}\)
Mà \(A=\frac{1}{8}\Rightarrow\frac{x-1}{4}=\frac{1}{8}\)
\(\Leftrightarrow8x-8=4\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\frac{12}{8}=\frac{3}{2}\left(ktm\right)\)
Vậy không có x thỏa mãn yêu cầu đề bài
Tìm x, biết :
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).}2x=\frac{11}{84}\)
(x là số tự nhiên, x > hoặc = 2
I. Tìm x,biết :
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right)2x}=\frac{11}{48}\)
( \(x\in N,x\ge2)\)
*Bài làm:
~I) Tìm x:
➤Ta có: \(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + ... + \(\frac{1}{\left(2x-2\right)2x}\) = \(\frac{11}{48}\)
⇒ \(2\) . (\(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + ... + \(\frac{1}{\left(2x-2\right)2x}\)) = \(2\) . \(\frac{11}{48}\)
⇒ \(\frac{2}{2.4}\) + \(\frac{2}{4.6}\) + ... + \(\frac{2}{\left(2x-2\right)2x}\) = \(\frac{22}{48}\)
⇒ (\(\frac{1}{2}\) - \(\frac{1}{4}\)) + (\(\frac{1}{4}\) - \(\frac{1}{6}\)) + ... + (\(\frac{1}{2x-2}\) - \(\frac{1}{2x}\)) = \(\frac{22}{48}\)
⇒ \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{6}\) + \(\frac{1}{6}\) - ... - \(\frac{1}{2x-2}\) + \(\frac{1}{2x-2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{1}{2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x}{x}\) . \(\frac{1}{2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x}{2x}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x-1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x-1}{2x}\) = \(\frac{22}{48}\)
⇒ \(x-1\) = \(\frac{22}{48}\) . \(2x\)
⇒ \(x-1\) = \(\frac{44x}{48}\)
⇒ \(x\) = \(\frac{44x}{48}\) + \(1\)
⇒ \(x\) = \(\frac{44x}{48}\) + \(\frac{48}{48}\)
⇒ \(x\) = \(\frac{44x+48}{48}\)
⇒ \(x\) = \(12\) (Chỗ này mình bấm máy tính nên hơi tắt;Bạn thông cảm)
*Vậy \(x\) = \(12\) .
I.
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right)2x}=\frac{11}{48}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right)2x}\right)=\frac{11}{48}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)=\frac{11}{48}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2x}=\frac{11}{24}\)
\(\Leftrightarrow\frac{1}{2x}=\frac{1}{2}-\frac{11}{24}\)
\(\Leftrightarrow\frac{1}{2x}=\frac{1}{24}\)
\(\Leftrightarrow2x=24\)
\(\Leftrightarrow x=24:2\)
\(\Rightarrow x=12\left(TM\right)\)
Vậy \(x=12.\)
Chúc bạn học tốt!
Bài 1:Tìm x, biết
\(\frac{1}{2.3}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{11}{48}\left(x\in N,x\ge2\right)\)
Bài 2:Chứng tỏ rằng với mọi \(n\in Nsao\),ta có
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right).\left(3n+2\right)}\)
Bài 1:
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}\)\(=\frac{11}{48}\)
\(\frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(x-1\right).x}\right)\)\(=\frac{11}{48}\)
\(\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x-1}-\frac{1}{x}\right)\)\(=\frac{11}{48}\)
\(\frac{1}{4.}.\left(1-\frac{1}{x}\right)=\frac{11}{48}\)
\(1-\frac{1}{x}=\frac{11}{48}:\frac{1}{4}\)
\(1-\frac{1}{x}=\frac{11}{12}\)
\(\frac{1}{x}=1-\frac{11}{12}\)
\(\frac{1}{x}=\frac{1}{12}\)
Vậy x= 12
Bài 2 :
Xét vế trái ta có :
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{\left(3n-1\right).\left(3n+2\right)}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\frac{1}{2\left(3n+2\right)}=\frac{n}{2\left(3n+2\right)}\)
VẾ TRÁI ĐÚNG BẰNG VẾ PHẢI .ĐẲNG THỨC ĐÃ CHỨNG TỎ LÀ ĐÚNG
cHÚC BẠN HỌC TỐT ( -_- )
1.Tìm x biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
2. Tìm x biết:
\(\left(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}\right)+\frac{58}{57}+2.\left(x-1\right)=2x+\frac{7}{3}+5x-\frac{8}{4}\)
3. Chứng minh với mọi n>1 thì:
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)....\left(1+\frac{1}{n.\left(n+2\right)}\right)<2\)
1/
\(1+\frac{2014}{2}+...+\frac{4024}{2012}=1+\left(1+\frac{2012}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{2012}{2012}\right)\)
\(=2012+2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)
Phương trình đã cho tương đương:
\(\left(1+\frac{1}{2}+...+\frac{1}{2012}\right).503x=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)
\(\Leftrightarrow503x=2012\)
\(\Leftrightarrow x=4\)
2/
\(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}+\frac{58}{57}+2x-2=2x+\frac{7}{3}+5x-\frac{8}{4}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{49}-\frac{1}{57}+\left(1+\frac{1}{57}\right)-2-\frac{7}{3}+\frac{8}{4}=5x\)
\(\Leftrightarrow\)\(5x=\frac{17}{3}\Leftrightarrow x=\frac{17}{15}\)
3/
Ta có: \(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{n\left(n+2\right)}\right)\)\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}.......\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
\(=2.\frac{n+1}{n+2}
Tìm x:
a) \(\frac{1}{2.4}\)+\(\frac{1}{4.6}\)+...+\(\frac{1}{\left(2x-2\right).2x}\)=\(\frac{1}{8}\)( x\(\in\)\(ℕ\), x\(\ge\)2)
b) \(\frac{2x+1}{3}\)=\(\frac{x-5}{2}\)
c) \(\frac{x+3}{-4}\)=\(\frac{-9}{x+3}\)
\(a,\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left[2x-2\right]\cdot2x}=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left[\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{\left[2x-2\right]\cdot2x}\right]=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right]=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left[\frac{1}{2}-\frac{1}{2x}\right]=\frac{1}{8}\)
\(\Rightarrow\left[\frac{1}{2}-\frac{1}{2x}\right]=\frac{1}{8}:\frac{1}{2}\)
\(\Rightarrow\left[\frac{1}{2}-\frac{1}{2x}\right]=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow2x=4\Leftrightarrow x=2\)
Vậy x = 2
Mun ảnh đại diện cute
<3
À tk mk nhé. giờ mk tk bn trước
a) \(\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}\right)=\frac{1}{8}\)
\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2x}\right)=\frac{1}{8}\)
\(\frac{1}{4}-\frac{1}{4x}=\frac{1}{8}\)
\(\frac{1}{4x}=\frac{1}{8}\)
\(4x=8\)
\(x=2\)
Vậy ...
b) \(\frac{2x+1}{3}=\frac{x-5}{2}\)
\(\Leftrightarrow\left(2x+1\right).2=\left(x-5\right).3\)
\(\Leftrightarrow4x+2=3x-15\)
\(\Leftrightarrow4x-3x=-15-2\)
\(\Leftrightarrow x=-17\)
c) Tương tự phần b
Rút gọn biểu thức sau:
\(A=\left(\frac{x^2-2x}{2x^2+8}-\frac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\)
^ Giúp tui nhanh zới nha! ^
=[x(x-2)/2(x2+4)-2x2/(4+x2)(2-x)][x(x-2)(x+1)/x3]
={[x(x-2)(2-x)-4x2 ]/2(2-x)(4+x2)} .[x(x-2)(x+1)/x3 ]
=[-x(x2+4)/2(2-x)(4+x2)].[x(x-2)(x+1)/x3 ]
=-x.x(x-2)(x+1)/2(2-x)x3
=(x+1)/2x