Giải các pt sau:
a.\(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
b.\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}=\dfrac{x^2-10x-1973}{27}\)
Giaỉ các phương trình sau
\(a,\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)\(a,\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
nên \(x^2-10x-2000=0\)
\(\Leftrightarrow x^2+40x-50x-2000=0\)
\(\Leftrightarrow x\left(x+40\right)-50\left(x+40\right)=0\)
\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Vậy: S={-40;50}
\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-100-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
`(x^2-10x-29)/1971+(x^2-10x-27)/1973=(x^2-10x-1971)/1929+(x^2-10x-1973)/1927`
`<=>(x^2-10x-29)/1971-1+(x^2-10x-27)/1973-1=(x^2-10x-1971)/1929-1+(x^2-10x-1973)/1927-1`
`<=>(x^2-10x-200)/1971+(x^2-10x-200)/1973=(x^2-10x-200)/1971+(x^2-10x-200)/1927`
`<=>(x^2-10x-200)(1/1971+1/1973-1/1929-1/1927)=0`
`<=>x^2-10x-200=0` do `1/1971+1/1973-1/1929-1/1927<0`
`<=>x^2-20x+10x-200=0`
`<=>x(x-20)+10(x-20)=0`
`<=>(x-20)(x+10)=0`
`<=>` \(\left[ \begin{array}{l}x=20\\x=-10\end{array} \right.\)
Vậy `S={20,-10}`
Giải phương trình sau :
\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\left(\dfrac{x^2-10x-27}{1973}-1\right)+\left(\dfrac{x^2-10x-29}{1971}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1973}+\dfrac{x^2-10x-2000}{1971}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1973}+\dfrac{1}{1971}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)=0\) vì \(\left(\dfrac{1}{1973}+\dfrac{1}{1971}-\dfrac{1}{29}-\dfrac{1}{27}\right)\ne0\)
\(\Leftrightarrow x^2-50x+40x-2000=0\)
\(\Leftrightarrow x\left(x-50\right)+40\left(x-50\right)=0\)
\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-50=0\\x+40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)
Vậy: Giá trị x thỏa mãn là: \(x=-40;50\)
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
vì \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
Nên \(x^2-10x-2000=0\)
<=> \(x^2-50x+40x-2000=0\)
<=> \(x\left(x-50\right)+40\left(x-50\right)=0\)
<=> \(\left(x-50\right)\left(x+40\right)=0\)
<=> \(x=50\) hoặc \(x=-40\)
Vậy tập nghiệm của phương trình là \(S=\left\{50;-40\right\}\)
1.\(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
2.\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
I : Giải các phương trình sau :
a) \(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
b) \(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
c) \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
d) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
e) \(\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)
f) \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
g) \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
các bạn ơi giúp mình với
mình cảm ơn
b) x-45/55 + x-47/53 = x-55/45 + x-53/47
<=>x-45/55 -1 + x-47/53 -1= x-55/45 -1 + x-53/47 - 1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55+1/53-1/45-1/47)=0
<=>x-100=0
<=>x=100
Vậy x = 100
\(a,\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+1+2004}{2004}+\dfrac{x+2+2003}{2003}-\dfrac{x+3+2002}{2002}-\dfrac{x+4+2001}{2001}=0\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}-\dfrac{x+2005}{2002}-\dfrac{x+2005}{2001}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
Vậy pt có tập nghiệm S = { 2005 }
Giải các phương trình sau :
1, 2x (x + 2)\(^2\) - 8x\(^2\) = 2 (x - 2) (x\(^2\) + 2x +4)
2, 8x (x +3 )\(^2\) - 3x = ( x +2)\(^3\) +1
3, \(\dfrac{10x+3}{12}\) = 1 + \(\dfrac{6+8x}{9}\)
4,\(\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)
5,\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
Giúp mk nha mấy bạn mk đg cần gấp nên mong mấy bạn giúp đỡ
Giải phương trình :
a, x-45/55 +x-47/53 =x-55/45 + x-53,47
b, x^2-10x-29/1971 +x^2-10x-27/1973 = x^2-10x-1971/29 + x^2-10x-1973/27
c, 2-x/2002-1= 1-x/2003 -x/2004
Các bạn giúp mik vs nhé !
a) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\Leftrightarrow\left(\frac{x-45}{55}-1\right)+\left(\frac{x-47}{53}-1\right)=\left(\frac{x-55}{45}-1\right)+\left(\frac{x-53}{47}-1\right)\)
\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
Vì \(\hept{\begin{cases}\frac{1}{55}< \frac{1}{45}\\\frac{1}{53}< \frac{1}{47}\end{cases}}\Rightarrow\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}< 0\)
\(\Rightarrow x-100=0\Rightarrow x=100\)
Vậy x = 100
Các phần sau tương tự nhé bạn
Giải phương trình sau:
a)\(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3=0\)
b)\(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
c)\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)
a)\(\frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{95}+1=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
Mà \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\Rightarrow300-x=0\Rightarrow x=300\)
b)\(\frac{2-x}{2002}+1=\frac{1-x}{2003}+2-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{1-x}{2003}+1+1-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}+\frac{2004-x}{2004}\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
Mà \(\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\ne0\Rightarrow2004-x=0\Rightarrow x=2004\)
c)\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}-2=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}-2\)
\(\Leftrightarrow\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}=\frac{x^2-10x-2000}{29}+\frac{x^2-10x-2000}{27}\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\right)=0\)
Mà\(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\ne0\)
\(\Rightarrow x^2-10x-2000=0\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Giải các phương trình sau:
\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)
em moi hoc lop 7 thoi a doi xong ki 2 nha
\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\frac{x^2-10x-29}{1971}-1+\frac{x^2-10x-27}{1973}-1=\frac{x^2-10x-1971}{29}-1+\frac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\frac{x^2-10x-29-1971}{1971}+\frac{x^2-10x-27-1973}{1973}=\frac{x^2-10x-1971-29}{29}+\frac{x^2-10x-193-27}{27}\)
\(\Leftrightarrow\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}-\frac{x^2-10x-2000}{29}-\frac{x^2-10x-2000}{27}=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\right)=0\)
\(\Leftrightarrow x^2-10x-2000=0\)
\(\Leftrightarrow x^2-50x+40x-2000=0\)
\(\Leftrightarrow x\left(x-50\right)+40\left(x-50\right)=0\)
\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)
Th1 \(x-50=0\Leftrightarrow x=50\)
Th2 \(x+40=0\Leftrightarrow x=-40\)
Vậy tập nghiệm của phương trình là \(S=\left\{50;-40\right\}\)