Bài 3:
a: \(\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)
=>\(\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}\right)-\left(x-23\right)\left(\dfrac{1}{26}+\dfrac{1}{27}\right)=0\)
=>\(\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)
=>x-23=0
=>x=23
b: \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
=>\(\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}=\dfrac{x+4+96}{96}+\dfrac{x+5+95}{95}\)
=>\(\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
=>x+100=0
=>x=-100
c: \(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
=>\(\left(\dfrac{x+1}{2004}+1\right)+\left(\dfrac{x+2}{2003}+1\right)=\left(\dfrac{x+3}{2002}+1\right)+\left(\dfrac{x+4}{2001}+1\right)\)
=>\(\dfrac{x+1+2004}{2004}+\dfrac{x+2+2003}{2003}=\dfrac{x+3+2002}{2002}+\dfrac{x+4+2001}{2001}\)
=>\(\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)
=>x+2005=0
=>x=-2005
d: Sửa đề:\(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}=-3\)
=>\(\left(\dfrac{201-x}{99}+1\right)+\left(\dfrac{203-x}{97}+1\right)+\left(\dfrac{205-x}{95}+1\right)=0\)
=>\(\dfrac{201-x+99}{99}+\dfrac{203-x+97}{97}+\dfrac{205-x+95}{95}=0\)
=>\(\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
=>300-x=0
=>x=300
Bài 1:
1: a: 15-8x=9-5x
=>-8x+5x=9-15
=>-3x=-6
=>x=2
b: 3+2x=5x+2
=>2x-5x=2-3
=>-3x=-1
=>\(x=\dfrac{1}{3}\)
c: \(5-\left(x-6\right)=4\left(3-2x\right)\)
=>\(5-x+6=12-8x\)
=>\(11-x=12-8x\)
=>\(-x+8x=12-11\)
=>7x=1
=>\(x=\dfrac{1}{7}\)
d: \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
=>\(2x^3+8x^2+8x-8x^2=2x^3-16\)
=>8x=-16
=>x=-2
2:
a: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
=>\(\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(8x+6\right)}{36}\)
=>\(30x+9=36+32x+24\)
=>32x+60=30x+9
=>2x=-51
=>\(x=-\dfrac{51}{2}\)
b: \(\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)
=>\(\dfrac{5\left(7x-1\right)}{30}+\dfrac{60x}{30}=\dfrac{6\left(16-x\right)}{30}\)
=>5(7x-1)+60x=6(16-x)
=>35x-5+60x=96-6x
=>101x=101
=>x=1
c: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
=>\(\dfrac{3\left(3x+2\right)-3x-1}{6}=\dfrac{10}{6}+\dfrac{12x}{6}\)
=>9x+6-3x-1=10+12x
=>12x+10=6x+5
=>6x=-5
=>\(x=-\dfrac{5}{6}\)
d: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
=>\(\dfrac{6\left(x+4\right)}{30}+\dfrac{30\left(-x+4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
=>6x+24-30x+120=10x-15x+30
=>-24x+144=-5x+30
=>-19x=-114
=>x=6
