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\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1998}=3\)
\(\Leftrightarrow\left(\frac{x-1}{2006}-1\right)+\left(\frac{x-10}{1997}-1\right)+\left(\frac{x-19}{1998}-1\right)=0\)
\(\Leftrightarrow\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1998}=0\)
\(\Leftrightarrow\left(x-2007\right)\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\right)=0\)
Dễ thấy cái đằng sau luôn > 0 nên x-2007=0 <=> x=2007
Đặt \(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}\left(1\right)\)
\(\left(1\right)\Leftrightarrow\frac{x-2007}{2006}=\frac{x-2007}{1997}=\frac{x-2007}{1998}=0\)
\(\Rightarrow x=2007\)
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)
\(\Leftrightarrow\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1988}=0\)
\(\Leftrightarrow x=2007\)
✰ ღ๖ۣۜDαɾƙ ๖ۣۜBαηɠ ๖ۣۜSĭℓεηтღ✰
lắm tắt thế này đi thi ko đc điểm đâu nhóc =))
Bạn sửa đề lại nha.
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)
=>\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}-3=0\)
=>\(\left(\frac{x-1}{2006}-1\right)+\left(\frac{x-10}{1997}-1\right)+\left(\frac{x-19}{1988}-1\right)=0\)
=>\(\frac{x-1-2006}{2006}+\frac{x-10-1997}{1997}+\frac{x-19-1988}{1988}=0\)
=>\(\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1988}=0\)
=>\(\left(x-2007\right).\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\right)=0\)
Vì \(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\ne0\)
=>x-2007=0
=>x=2007
a, \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
\(\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)
\(\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)
\(=>x+36=0\)
\(=>x=36\)
Bài 2
\( a)4{\left( {x + 1} \right)^2} + {\left( {2x - 1} \right)^2} - 8\left( {x - 1} \right)\left( {x + 1} \right) = 11\\ \Leftrightarrow 4\left( {{x^2} + 2x + 1} \right) + 4{x^2} - 4x + 1 - 8\left( {{x^2} - 1} \right) = 11\\ \Leftrightarrow 4{x^2} + 8x + 4 + 4{x^2} - 4x + 1 - 8{x^2} + 8 = 11\\ \Leftrightarrow 4x + 13 = 11\\ \Leftrightarrow 4x = 11 - 13\\ \Leftrightarrow 4x = - 2\\ \Leftrightarrow x = - \dfrac{1}{2} \)
Bài 2:
\( b)\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + x\left( {x + 2} \right)\left( {2 - x} \right) = 1\\ \Leftrightarrow {x^3} - 27 + x\left( {2 + x} \right)\left( {2 - x} \right) = 1\\ \Leftrightarrow {x^3} - 27 + x\left( {4 - {x^2}} \right) = 1\\ \Leftrightarrow {x^3} - 27 + 4x - {x^3} = 1\\ \Leftrightarrow 4x = 1 + 27\\ \Leftrightarrow 4x = 28\\ \Leftrightarrow x = 7 \)
Đặt \(a=\frac{1}{4453};b=\frac{1}{1997}\)ta có :
\(5\frac{6}{4453}\cdot\frac{1}{1997}-\frac{2}{1997}\cdot2\frac{3}{4453}\)
\(=\left(5+6a\right)\cdot b-2b\left(2+3a\right)\)
\(=5b+6ab-4b-6ab\)
\(=b=\frac{1}{1997}\)
Sửa đề :
\(\dfrac{x+1}{2006}+\dfrac{x+10}{1997}+\dfrac{x+19}{1988}=-3\)
\(\Leftrightarrow\left(\dfrac{x+1}{2006}+1\right)+\left(\dfrac{x+10}{1997}+1\right)+\left(\dfrac{x+19}{1988}+1\right)=0\)
\(\Leftrightarrow\dfrac{x+2007}{2006}+\dfrac{x+2007}{1997}+\dfrac{x+2007}{1988}=0\)
\(\Leftrightarrow\left(x+1007\right)\left(\dfrac{1}{2006}+\dfrac{1}{1997}+\dfrac{1}{1988}\right)=0\)
Mà \(\dfrac{1}{2006}+\dfrac{1}{1997}+\dfrac{1}{1988}\ne0\)
\(\Leftrightarrow x+2007=0\)
\(\Leftrightarrow x=-2007\)
Vậy..