\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)

\(\Rightarrow\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)\(\Rightarrow\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)

\(\Rightarrow\left(x-2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=\left(x-2004\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)

Với \(x-2004\ne0\)

\(\Rightarrow\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\left(KTM\right)\)

Với \(x-2004=0\)

\(\Rightarrow x=2004\)

\(\frac{x+10}{2000}+\frac{x+20}{1990}+\frac{x+30}{1980}+\frac{x+40}{1970}=-4\)

\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)

\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)

Vì  \(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}>0\)

\(\Rightarrow x+2010=0\)

\(\Leftrightarrow x=-2010\)

\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)

\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)

mà\(\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)\ne0\Rightarrow\left(x+2010\right)=0\\ \Rightarrow x=-2010\)

44

tuổi ủng hộ mk nha

xin loi minh moi hok lop 6

1. là -2005

2 làm tròn là 49,8

Ta có x+1/99 + x+2/98 + x+3/97 = x+4/96 + x+5/95 + x+10/90

=> x+1/99 + x+2/98 + x+3/97 - x+4/96 - x+5/95 - x+10/90=0

=> (x+1/99 + 1) + (x+2/98 + 1) + (x+3/97 +1) - (x+4/96 + 1) - (x+5/95 + 1) - (x+10/90 + 1) = 0

=> x+100/99 + x+100/98 + x+100/97 - x+100/96 - x+100/95 - x+100/90 =0

=> (x+100)(1/99+1/98+1/97-1/96-1/95-1/90) = 0

Mà 1/99+1/98+1/97-1/96-1/95-1/90 khác 0

=> x+100=0 => x=-100

Vậy phương trình có nghiệm là x=-100

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}+\frac{x+10}{99}\)

\(\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1-\left(\frac{x+4}{96}+1+\frac{x+5}{95}+1+\frac{x+10}{99}+1\right)=0\)

\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}-\left(\frac{x+100}{96}+\frac{x+100}{95}+\frac{x+100}{90}\right)=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{90}\right)=0\)

Mà\(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{90}\ne0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

a) Ta có: \(\frac{x+a}{x+2}+\frac{x-2}{x-a}=2\left(1\right)\)

Với a = 4

Thay vào phương trình (t) ta được:

  \(\frac{x+2}{x+2}+\frac{x-2}{x-2}=2\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow x^2-4+x^2-4=2\left(x^2-4\right)\)

\(\Leftrightarrow2x^2=2x^2-8\)

\(\Leftrightarrow0x=-8\)

Vậy phương trình vô nghiệm

b) Nếu x = -1

\(\Rightarrow\frac{-1+a}{-1+2}+\frac{-1-2}{-1-a}=2\)

\(\Leftrightarrow\frac{-1+a}{1}+\frac{-3}{-1-a}=2\)

\(\Leftrightarrow\frac{\left(-1+a\right)\left(-1-a\right)}{-1-a}+\frac{-3}{-1-a}=\frac{2\left(-1-a\right)}{-1-a}\)

\(\Leftrightarrow1+a-a-a^2-3=-2-2a\)

\(\Leftrightarrow-a^2+2a=-2-1+3\)

\(\Leftrightarrow a\left(2-a\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=0\\2-a=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)

Vậy a = {0;2}

NĂM MỚI VUI VẺ

\(a,\frac{x+4}{x+2}+\frac{x-2}{x-4}=2\)

\(\frac{x+2+2}{x+2}+\frac{x-4+2}{x-4}=2\)

=> \(1+\frac{2}{x+2}+1+\frac{2}{x-4}=2\)

=>\(2\left(\frac{x-4+x+2}{\left(x+2\right)\left(x-4\right)}\right)=0\)

=> x=1 (t/m \(x\ne-2\) và \(x\ne4\))

có: x(x+3)(x^2+3x+4)=-4

\(\Leftrightarrow\)(x^2+3x)(x^2+3x+4)+4=0

\(\Leftrightarrow\)(x^2+3x)\(^2\)+4(x^2+3x)+4=0

\(\Leftrightarrow\)(x^2+3x+2)\(^2\)=0

\(\Leftrightarrow\)x\(^2\)+3x+2=0

\(\Leftrightarrow\)(x+1)(x+2)=0

\(\Leftrightarrow\)x+1=0 hoặc x+2=0

*) Nếu x+2=0\(\Leftrightarrow\)x=-2

*) Nếu x+1=0\(\Leftrightarrow\)x=-1

Vậy S={ 2;-1}

<=> x(x³+3³)=-4 <=> x⁴+27x+4=0 <=> 

x² - 5x + 4 + x² - 5x + 6 = 2

<=> 2x² - 10x + 8 = 0

<=> x² - 5x + 4 = 0

<=> x = 1 hoặc x = 4

X^2-4x-x+4+x^2-2x-3x+6=2                                                                                                                                                               rút gọn và chuyển vế  : 2x^2-10x+8=0                                                                                                                                                bấm máy tính ; x=4 và x=1        

\(\left(x-1\right)\left(x-4\right)+\left(x-3\right)\left(x-2\right)=2\)

\(< =>x^2-4x-x+4+x^2-2x-3x+6=2\)

\(< =>2x^2-10x+10-2=0\)

\(< =>2x^2-10x+8=0\)

\(< =>x^2-5x+4=0\)

\(< =>x^2-x-4\left(x-1\right)=0\)

\(< =>\left(x-4\right)\left(x-1\right)=0\)

\(< =>\orbr{\begin{cases}x=1\\x=4\end{cases}}\)