K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
M
10 tháng 1 2016
Ta có: \(\frac{x-5}{1990}+\frac{x-15}{1980}=\frac{x-1980}{15}+\frac{x-1990}{5}\)
=> \(\left(\frac{x-5}{1990}-1\right)+\left(\frac{x-15}{1980}-1\right)=\left(\frac{x-1980}{15}-1\right)+\left(\frac{x-1990}{5}-1\right)\)
=> \(\frac{x-5-1990}{1990}+\frac{x-15-1980}{1980}=\frac{x-1980-15}{15}+\frac{x-1990-5}{5}\)
=> \(\frac{x-1995}{1990}+\frac{x-1995}{1980}=\frac{x-1995}{15}+\frac{x-1995}{5}\)
=> \(\frac{x-1995}{1990}+\frac{x-1995}{1980}-\frac{x-1995}{15}-\frac{x-1995}{5}=0\)
=> \(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\right)=0\)
Vì \(\frac{1}{1990}+\frac{1}{1980}\ne\frac{1}{15}+\frac{1}{5}\) => \(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\ne0\)
=> x - 1995 = 0
=> x = 1995
MT
10 tháng 1 2016
\(\frac{x-5}{1990}+\frac{x-15}{1980}=\frac{x-1980}{15}+\frac{x-1990}{5}\)
\(\Leftrightarrow\frac{x-5}{1990}-1+\frac{x-15}{1980}-1-\frac{x-1980}{15}+1-\frac{x-1990}{5}+1=0\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}-\frac{x-1995}{15}-\frac{x-1995}{5}=0\)
\(\Leftrightarrow\left(x-1995\right).\left(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\right)=0\)
<=>x=1995
17 tháng 1 2018
\(\frac{x-5}{1990}-1+\frac{x-15}{1980}-1=\frac{x-1980}{15}-1+\frac{x-1990}{5}-1\)
\(\frac{x-1995}{1990}+\frac{x-1995}{1980}-\frac{x-1995}{15}-\frac{x-1995}{5}=0\)
\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\right)=0\)
Mà \(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\ne0\)
Nên \(x-1995=0\Leftrightarrow x=1995\)
E
29 tháng 3 2017
BPT \(\Leftrightarrow\dfrac{x+1987}{2002}+\dfrac{x+1988}{2003}-\dfrac{x+1989}{2004}+\dfrac{x+1990}{2005}>0\)
\(\Leftrightarrow\left(\dfrac{x+1987}{2002}-1\right)+\left(\dfrac{x+1988}{2003}-1\right)-\left(\dfrac{x+1989}{2004}-1\right)-\left(\dfrac{x+1990}{2005}-1\right)>0\)
\(\Leftrightarrow\dfrac{x-15}{2002}+\dfrac{x-15}{2003}-\dfrac{x-15}{2004}-\dfrac{x-15}{2005}>0\)
\(\Leftrightarrow\left(x-15\right)\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\right)>0\)
Vì \(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}>0\)
\(\Rightarrow x-15>0\)
\(\Leftrightarrow x>15\)
Vậy bpt có nghiệm x > 15
TK
29 tháng 3 2017
\(\dfrac{x+1987}{2002}+\dfrac{x+1988}{2003}-2>\dfrac{x+1989}{2004}+\dfrac{x+1990}{2005}-2\)
\(\Leftrightarrow\left(\dfrac{x+1987}{2002}-1\right)+\left(\dfrac{x+1988}{2003}-1\right)\)
\(-\left(\dfrac{x+1989}{2004}-1\right)-\left(\dfrac{x+1990}{2005}-1\right)\)
quy đồng lên ta được:
\(\left(\dfrac{x+1987-2002}{2002}\right)+\left(\dfrac{x-1998-2003}{2003}\right)\)
\(-\left(\dfrac{x+1989-2004}{2004}\right)-\left(\dfrac{x+1990-2005}{2005}\right)>0\)
\(\Leftrightarrow\left(\dfrac{x-15}{2002}\right)+\left(\dfrac{x-15}{2003}\right)-\left(\dfrac{x-15}{2004}\right)-\left(\dfrac{x-15}{2005}\right)>0\)
đặt nhân tử chung ta được:
\(\Leftrightarrow\left(x-15\right)\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\right)>0\)
Vì:
\(\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\in Z\right)\) nên ta xét \(x-15>0\Rightarrow x>15\)
TL
3
TC
11 tháng 3 2023
`(x-2003)/16 +(x-1997)/11 +(x-1992)/9 +(x-1991)/7=10`
`<=>((x-2003)/16-1)+((x-1997)/11-2)+((x-1992)/9-3)+((x-1991)/7-4)=0`
`<=>(x-2019)/16+ (x-2019)/11 +(x-2019)/9+(x-2019)/7 =0`
`<=> (x-2019)(1/16+1/11+1/9+1/7)=0`
<=> x-2019=0`
`<=> x=2019`
5 tháng 3 2015
pạn -1 vào mỗi phân số là xong. Rùi ra x\(\frac{x-2015}{1986}\)+\(\frac{x-2015}{1988}\)+ \(\frac{x-2015}{1990}\)+...+\(\frac{x-2015}{x1996}\)-\(\frac{x-2015}{29}\)-\(\frac{x-2015}{27}\)-...\(\frac{x-2015}{19}\)=0
<=>(x-2015)(\(\frac{1}{1986}\)+\(\frac{1}{1988}\)+... -\(\frac{1}{19}\))=0...(mà \(\frac{1}{1986}\)+...- \(\frac{1}{19}\) khác 0)
=>x-2015=0
<=> x=2015
23 tháng 6 2020
a)
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\\ \Leftrightarrow\frac{201-x}{99}+\frac{99}{99}+\frac{203-x}{97}+\frac{97}{97}+\frac{205-x}{95}+\frac{95}{95}+4=4\\ \Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\) (*)
Do \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)\ne0\)
nên (*) \(\Leftrightarrow300-x=0\\ \Leftrightarrow x=300\)
b)
\(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\\ \Leftrightarrow\frac{2-x}{2002}+\frac{2002}{2002}-1+1=\frac{1-x}{2003}+\frac{2003}{2003}-\frac{x}{2004}+\frac{2004}{2004}\\ \Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\\ \Leftrightarrow\frac{2004-x}{2002}-\frac{2004-x}{2003}+\frac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\) (*)
Do \(\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)\ne0\)
nên (*) \(\Leftrightarrow2004-x=0\)
\(\Leftrightarrow x=2004\)
c) \(\left|2x-3\right|=2x-3\) (1)
ĐKXĐ: \(\\ 2x-3\ge0\)
\(\Leftrightarrow x\ge\frac{3}{2}\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}2x-3=2x-3\\2x-3=-2x+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}0x=0\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\forall x\in R\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\frac{3}{2}\right\}\)
16 tháng 2 2022
\(\Leftrightarrow\left(\dfrac{x-5}{1990}-1\right)+\left(\dfrac{x-15}{1980}-1\right)+\left(\dfrac{x-25}{1970}-1\right)\\ +\left(\dfrac{x-1990}{5}-1\right)+\left(\dfrac{x-1980}{15}-1\right)+\left(\dfrac{x-1970}{25}-1\right)=0\\ \Leftrightarrow\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}+\dfrac{x-1995}{1970}+\dfrac{x-1995}{5}\\ +\dfrac{n-1995}{15}+\dfrac{n-1995}{25}=0\\ \Rightarrow\left(x-1995\right)\left(\dfrac{1}{1990}+\dfrac{1}{1980}+\dfrac{1}{1970}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}\right)=0\)
\(\Rightarrow x-1995=0\\ \Rightarrow x=1995\)
BG
2
PT
6 tháng 2 2019
ta có : x^5+2x^4+3x^3+3x^2+2x+1=0
\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0
\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0
\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0
\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0
\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0
VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)
\(\Rightarrow\)x+1=0
\(\Rightarrow\)x=-1
CÒN CÂU B TỰ LÀM (02042006)
14 tháng 2 2023
b: x^4+3x^3-2x^2+x-3=0
=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0
=>(x-1)(x^3+4x^2+2x+3)=0
=>x-1=0
=>x=1
15 tháng 4 2017
\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2004}-\frac{x+2005}{2003}-\frac{x+2005}{2003}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\Leftrightarrow x=-2005\)
15 tháng 4 2017
=> (x+1)/2004+1+(x+2)/2003+1=(x+3)/2002+1+(x+4)/2001+1
=> (x+2005)/2004+(x+2005)/2003=(x+2005)/2002+(x+2005)/2001
=> (x+2005)(1/2004+1/2003-1/2002-1/2001)=0
=> x+2005=0
=> x=-2005
Để \(\left(x-1990\right)\left(2003-x\right)>0\)
Suy ra x-1990 và 2003-x cùng dấu
Vậy