Giải pt :
\(a,\frac{1-x}{2013}=1+\frac{2-x}{2014}-\frac{x}{2014},\)
\(b,\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(c,\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-a-c}{b}=3\)
Giải Phương trình :
\(a,\frac{1-x}{2013}=1+\frac{2-x}{2012}-\frac{x}{2014}\)
\(b,\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(c,\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-a-c}{b}=3\)
giai các phương trình sau:
a,\(\frac{1-x}{2013}=1+\frac{2-x}{2012}-\frac{x}{2014}\)
b,\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
c,\(\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-a-c}{b}=3\)
d,(x+3)4 + (x+5)4=16
e,x4+ 3x3 - 7x2- 27x-18=0
f,\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
Giải phương trình sau:
a) \(\frac{1-x}{2013}=1+\frac{2-x}{2012}-\frac{x}{2014}\)
b) \(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
a,\(\Leftrightarrow\left(\frac{1-x}{2013}+1\right)=\left(\frac{2-x}{2012}+1\right)-\left(1-\frac{x}{2014}\right)\)
\(\Leftrightarrow\frac{2014-x}{2013}=\frac{2014-x}{2012}-\frac{2014-x}{2014}\)
\(\Leftrightarrow\frac{2014-x}{2013}-\frac{2014-x}{2012}+\frac{2014-x}{2014}\)=0
\(\Leftrightarrow\left(2014-x\right)\left(\frac{1}{2013}-\frac{1}{2012}+\frac{1}{2014}\right)=0\)
\(\Leftrightarrow x=2014\left(do.cái.còn.lại.\ne0\right)\)
b,tương tự +1 vào cái thứ nhất ,+1 vào cái thứ 2,1- vào cái thứ 3 được x=2013
phần a : \(\frac{x+2}{2010}+\frac{x+2}{2011}+\frac{x+2}{2012}=\frac{x+2}{2013}+\frac{x+2}{2014}\)
phần b :\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(a,\frac{x+2}{2010}+\frac{x+2}{2011}+\frac{x+2}{2012}=\frac{x+2}{2013}+\frac{x+2}{2014}\)
\(\Leftrightarrow\frac{x+2}{2010}+\frac{x+2}{2011}+\frac{x+2}{2012}-\frac{x+2}{2013}-\frac{x+2}{2014}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\text{Mà }\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\text{ nên:}\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
\(b,\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Leftrightarrow \frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(M\text{à}:\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0 n\text{ê}n:\)
\(x+2004=0\)
\(\Leftrightarrow x=-2004\)
Giải phương trình
a, \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}\frac{x+4}{2001}\)
b, \(\frac{201-x}{99}+\frac{205-x}{97}+\frac{205-x}{95}+3=0\)
c, \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
a) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
b) Sửa đề :
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow x=300\)
c) \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2-x}{2002}+1=\frac{1-x}{2003}+1-\frac{x}{2004}+1\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\)
\(\Leftrightarrow x=2004\)
Vậy....
Giải các pt sau:
a, (4x-1)(x+5)=(2x-3)^2
b, x(x+1)(x+2)(x+3)=24
c, x^2-2x+1=3x(x-1)
d,\(\frac{x+1}{2017}+\frac{x+2}{2015}=\frac{x+2014}{3}+\frac{x+2013}{4}\)
b) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow\)\(\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)
Đặt \(x^2+3x=t\) ta có:
\(t\left(t+2\right)-24=0\)
\(\Leftrightarrow\)\(t^2+2t-24=0\)
\(\Leftrightarrow\)\(\left(1-4\right)\left(1+6\right)=0\)
đến đây bn giải tiếp
Bài 1: Tìm số tự nhiên n để phân số \(\frac{7n-8}{2n-3}\)có GTLN.
Bài 2: Tìm x, biết: \(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\).
Bài 3: Cho a+b+c=2010 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{3}\).
Tính S=\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
Tìm x biết:
a)(x-5)x+2014-(x-5)x+2015
b)\(\frac{x-1}{2014}+\frac{x-2}{2013}+\frac{x-3}{2012}=\frac{x-10}{2005}+\frac{x-11}{2004}+\frac{x-12}{2003}\)
a) (x-5)x+2015 - (x-5)x+2014 =0
(x-5)x+2014(x-5 -1) =0
+ x -5 =0 => x =5
+ x -6 =0 => x =6
Vậy x = 5 hoặc x =6
Bài 5: Giải các phương trình sau :
a) \(\frac{3x-1-\frac{x-1}{2}}{3}-\frac{2x+\frac{1-2x}{3}}{2}=\frac{\frac{3x-1}{2}-6}{5}\)
b) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
c) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
d) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
e) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)