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Bài 15:
Ta có: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)
\(\Leftrightarrow\frac{x+2+2008}{2008}+\frac{x+3+2007}{2007}+\frac{x+4+2006}{2006}+\frac{x+2028-18}{6}=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}>0\)
nên x+2010=0
hay x=-2010
Vậy: x=-2010
Bài 17:
Ta có: \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)
\(\Leftrightarrow\frac{x+1+65}{65}+\frac{x+3+63}{63}=\frac{x+5+61}{61}+\frac{x+7+59}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
Vì \(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)
nên x+66=0
hay x=-66
Vậy: x=-66
Ta có :
\(\frac{x+1}{65}+\frac{x+3}{63}< \frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)< \left(\frac{x+5}{61}+1\right)+\left(\frac{x+7}{59}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+5}{61}-\frac{x+7}{59}< 0\)
\(\Leftrightarrow\)\(\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)< 0\)
Vì \(\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)< 0\)
\(\Rightarrow\)\(x+66>0\)
\(\Rightarrow\)\(x>-66\)
Vậy \(x>-66\)
Bài 1:
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
\(\Leftrightarrow x+66=0\)
\(\Leftrightarrow x=-66\)
b) \(\frac{m^2\left(\left(x+2\right)^2-\left(x-2\right)^2\right)}{8}-4x=\left(m-1\right)^2+3\left(2m+1\right)\)
\(\Leftrightarrow m^2x-4x=m^2+4m+4\)
\(\Leftrightarrow\left(m^2-4\right)x=m^2+4m+4\)
Để phương trình vô nghiệm thì \(\hept{\begin{cases}m^2-4=0\\m^2+4m+4\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}m=2\vee m=-2\\\left(m+2\right)^2\ne0\end{cases}}\Leftrightarrow m=2\)
\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)
\(-18x^3+51x^2+9x-60=0\)
\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)
a) \(\frac{x+2}{2002}\)+\(\frac{x+5}{1999}\)+\(\frac{x+201}{1803}\)=-3
⇔\(\frac{x+2}{2002}\)+\(\frac{x+5}{1999}\)+\(\frac{x+201}{1803}\)+3=0
⇔\(\frac{x+2}{2002}\)+1+\(\frac{x+5}{1999}\)+1+\(\frac{x+201}{1803}\)+1=0
⇔\(\frac{x+2004}{2002}\)+\(\frac{x+2004}{1999}\)+\(\frac{x+2004}{1803}\)=0
⇔(x+2004)(\(\frac{1}{2002}\)+\(\frac{1}{1999}\)+\(\frac{1}{1803}\))=0
Mà (\(\frac{1}{2002}\)+\(\frac{1}{1999}\)+\(\frac{1}{1803}\))≠0
⇒x+2004=0
⇔x=-2004
Vậy tập nghiệm của phương trình đã cho là:S={-2004}
a) 3x - 2(5 + 2x) =45 - 2x
=> 3x - 10 - 4x = 45 - 2x
=> 3x - 4x + 2x = 45 + 10
=> x = 55
b) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
=> \(\frac{x-3}{5}=\frac{2x+17}{3}\)
=> 5(2x + 17) = 3(x - 3)
=> 10x + 85 = 3x - 9
=> 7x = -94
=> x = -94/7
c) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x-33}{7}\)
=> \(\frac{10x-6}{12}-\frac{21x-3}{12}=\frac{4x-33}{7}\)
=> \(\frac{-11x-3}{12}=\frac{4x-33}{7}\)
=> (-11x - 3).7 = (4x - 33).12
= -77x - 21 = 48x - 396
=> x = 3
d) (x - 1)(5x + 3) = (3x - 8)(x - 1)
=> (x - 1)(5x + 3) - (3x - 8)(x -1) = 0
=> (x - 1)(2x + 11) = 0
=> \(\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5,5\end{cases}}\)
e) (x - 1)(x2 + 5x - 2) - (x3 - 1) = 0
=> (x - 1)(x2 + 5x - 2) - (x - 1)(x2 + x + 1) = 0
=> (x - 1)(4x - 3) = 0
=> \(\orbr{\begin{cases}x-1=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0,75\end{cases}}\)
f) \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
=> \(\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)
=> \(\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
=> \(\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
=> x - 50 = 0 (Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\))
=> x = 50
b, \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
\(\Leftrightarrow\frac{x-3}{5}=\frac{17+2x}{3}\Leftrightarrow3x-9=85+10x\)
\(\Leftrightarrow-7x=94\Leftrightarrow x=-\frac{94}{7}\)
f, sửa : \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)=0\)
\(\Leftrightarrow x=-66\)
sê đài
Sửa đề:
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\\\Leftrightarrow \frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\\ \Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\\\Leftrightarrow \left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\\\Leftrightarrow x+66=0\left(Vi\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)\\ \Leftrightarrow x=-66\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-66\right\}\)