Giải phương trình :
\(\dfrac{1}{4x-2006}\)+\(\dfrac{1}{5x+2004}\)=\(\dfrac{1}{15x-2007}\)-\(\dfrac{1}{6x-2005}\)
Tìm x:
\(\dfrac{1}{4x-2006}+\dfrac{1}{5x+2004}=\dfrac{1}{15x-2007}-\dfrac{1}{6x-2005}\)
SORRY MÌNH SỬA LẠI ĐỀ XÍU 1/15X - 2007 - 1/6X -2005 NHÉ
C= \(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+.....+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+.....+\dfrac{1}{2006}}\)
GIÚP mình nha
Lèm ơn đấy !!!!!
Ta có: \(C=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{1+\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}}\)
\(=\dfrac{2006}{2007}\)
TÍNH NHANH:
C= \(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4} +....+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+....+\dfrac{1}{2006}}\)
\(C=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)+1}\)
\(=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2007}}=\dfrac{2006}{2007}\)
Tính theo cách hợp lí:
A=3+6+9+12+...+2007
B=\(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
số số hạng của A là :
( 2007 - 3 ) : 3 + 1 = 669 ( số )
tổng A là :
( 2007 + 3 ) . 669 : 2 = 672345
B = \(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(\dfrac{2005}{2}+1\right)+\left(\dfrac{2004}{3}+1\right)+...+\left(\dfrac{1}{2006}+1\right)+1}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}+\dfrac{2007}{2007}}\)
B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{2007.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2006}+\dfrac{1}{2007}\right)}\)
B = \(\dfrac{2006}{2007}\)
\(A=3+6+9+12+...+2007\)
\(A=\dfrac{\left(2007+3\right)\left(\dfrac{2007-3}{3}+1\right)}{2}\)
\(A=\dfrac{2010.669}{2}=672235\)
\(B=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
\(B=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{2006+\left(\dfrac{2005}{2}+1\right)+\left(\dfrac{2004}{3}+1\right)+...\left(\dfrac{1}{2006}+1\right)-2005}\)
\(B=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2005}{3}+...+\dfrac{2007}{2006}}\)
\(B=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{2007\left(\dfrac{1}{2007}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}\right)}=\dfrac{2006}{2007}\)
tik mik nha !!!
giải phương trình :
a, \(\dfrac{4x-1}{\sqrt{4x-3}}+\dfrac{11-2x}{\sqrt{5-x}}=\dfrac{15}{2}\)
b, \(\left(\sqrt{5x-1}+\sqrt{x-1}\right)\left(3x-1-\sqrt{5x^2-6x+1}\right)=4x\)
Giải phương trình: \(\dfrac{1}{4-2014}+\dfrac{1}{5x+2012}+\dfrac{1}{6x-2013}=\dfrac{1}{15x-2015}\)
A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}\)
B = \(\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)
Tính \(\dfrac{B}{A}\)
Đặt: \(L_2=\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)
\(L_2=1+\left(\dfrac{2006}{2}+1\right)+\left(\dfrac{2005}{3}+1\right)+...+\left(\dfrac{2}{2006}+1\right)+\left(\dfrac{1}{2007}+1\right)\)
\(L_2=\dfrac{2008}{2008}+\dfrac{2008}{2}+\dfrac{2008}{3}+...+\dfrac{2008}{2006}+\dfrac{2008}{2007}\)
\(L_2=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+..+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)
\(\dfrac{L_1}{L_2}=\dfrac{1}{2008}\)
So Sánh:
A=\(\dfrac{10^{11}-1}{10^{12}-1}\) và B=\(\dfrac{10^{10}+1}{10^{11}+1}\)
C=\(\dfrac{2005^{2005}+1}{2005^{2006}+1}\) và D=\(\dfrac{2005^{2004}+1}{2005^{2005}+1}\)
giải các phương trình sau
1, \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
2, \(\dfrac{3}{2+x}-\dfrac{x-1}{x^2-4}=\dfrac{2}{x-2}\)
3, \(\dfrac{x-5}{2x-3}-\dfrac{x}{2x+3}=\dfrac{1-6x}{4x^2-9}\)
1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)
\(\Leftrightarrow-3x-12-3+5x-x+4=0\)
\(\Leftrightarrow x=11\left(nhận\right)\)
2. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Leftrightarrow \frac{3(x-2)-(x-1)}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)
\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)
Vậy pt vô nghiệm
3. ĐKXĐ: $x\neq \pm \frac{3}{2}$
PT \(\Leftrightarrow \frac{(x-5)(2x+3)-x(2x-3)}{(2x-3)(2x+3)}=\frac{1-6x}{(2x-3)(2x+3)}\)
\(\Rightarrow (x-5)(2x+3)-x(2x-3)=1-6x\)
\(\Leftrightarrow 2x^2-7x-15-2x^2+3x+6x-1=0\)
\(\Leftrightarrow 2x-16=0\Leftrightarrow x=8\) (thỏa mãn)