giải các phương trình sau bằng cách đưa về dạng ax+b=0

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bài 1:giải các phương trình sau bằng cách đưa về dạng ax+b=0
a.$3-4y+24+6y=y+27+3y$

b.$2x(x+2)^2-8x^2=2(x-2)(x^2+2x+4)$

c.$(x-2)^3+(3x-1)(3x+1)=(x+1)^3$

bài 2:giải các phương trình sau

a. $\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}$

b.($\dfrac{x+2}{98}+1)+(\dfrac{x+3}{97}+1)=\dfrac{x+4}{96}+1)+(\dfrac{x+5}{95}+1)$

c $\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}$

d.]$\dfrac{201-x}{99}+\dfrac{203-x}{97}=\dfrac{205-x}{95}+3=0$

e.$\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53 }{47}$

f.$\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}$

g.$\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}$

h$\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}$

i.$\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}$

 

[thinhrost1]

bài 1:giải các phương trình sau bằng cách đưa về dạng ax+b=0
a.3-4y+24+6y=y+27+3y

)

$3-4y+24+6y=y+27+3y \Leftrightarrow 6y-4y-y-3y+3+24-27=0 \Leftrightarrow -2y=0 \Leftrightarrow y=0$

 

[nhuquynhdat]

$\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}$

$\leftrightarrow (x-23)(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27})=0$

$\leftrightarrow x-23=0 \leftrightarrow x=23$


$\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2 002}+\dfrac{x+4}{2001}$

$\leftrightarrow \dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2 002}+1+\dfrac{x+4}{2001}+1$

$\leftrightarrow \dfrac{x+2015}{2004}+\dfrac{x+2015}{2003}-\dfrac{x+2015}{2 002}-\dfrac{x+2015}{2001}=0$

$\leftrightarrow (x+2015)(\dfrac{1}{2004}+\dfrac{1}{2003}=\dfrac{1}{2 002}+\dfrac{1}{2001}=0$

$\leftrightarrow x+2015=0 \leftrightarrow x=-2015$

Mấy phần còn lại tương tự

 

[huuthuyenrop2]

$\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0$

$ \Leftrightarrow \frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0$

$ \Leftrightarrow (300-x)=0$

$ \Leftrightarrow x=300$

 

[hodoico]

)

$3-4y+24+6y=y+27+3y \Leftrightarrow 6y-4y-y-3y+3+24-27=0 \Leftrightarrow -2y=0 \Leftrightarrow y=0$

3-4y+24+6y=y+27+3y
\Leftrightarrow 27+2y=27+4y
\Leftrightarrow 27-27+2y-4y=0
\Leftrightarrow -2y=0
\Leftrightarrow y=0
@};-

 

[riverflowsinyou1]

1) b)
$2x(x+2)^2-8x^2=2(x-2)[(x+2)^2-2x]$
\Leftrightarrow $x(x+2)^2-4x^2=(x-2)[(x+2)^2-2x]$
\Leftrightarrow $x.[(x+2)^2-2x-2x]=(x-2)[(x+2)^2-2x]$
\Leftrightarrow $x[(x+2)^2-2x]-2x^2=(x-2)[(x+2)^2-2x]$
\Leftrightarrow $2.[x^2+2x+4]=2x^2$
\Rightarrow $2x+4=0$ \Leftrightarrow $x=-2$

c) $(x-2)^3+9x^2-1=(x+1)^3$
\Leftrightarrow $9x^2-1=(x+1)^3-(x-2)^3=9x^2-9x+9$
\Rightarrow $-1=-9x+9$ \Rightarrow $x=\frac{10}{9}$

 

[huuthuyenrop2]

$\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+ \frac{x+8}{92}$

$\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}$

$ \Leftrightarrow (x+100)[\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}]=0$

$ \Leftrightarrow x+100=0$

$ \Leftrightarrow x=-100$

 

[huuthuyenrop2]

$\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53 }{47}$
$\Leftrightarrow \dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}$
$ \Leftrightarrow x-100=0$
$ \Leftrightarrow x=100$

 

[huuthuyenrop2]

$(\dfrac{x+2}{98}+1)+(\dfrac{x+3}{97}+1)=\dfrac{x+4}{96}+1)+(\dfrac{x+5}{95}+1)$

$ \Leftrightarrow \dfrac{x+100}{98}+\dfrac{x+100}{97} -\dfrac{x+100}{96}-\dfrac{x+100}{95}$

$ \Leftrightarrow (x+100)[\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}]$

$ \Leftrightarrow x+100=0$

$\Leftrightarrow x=-100$

 

[ronaldover7]

b)
$2x(x+2)^2-8x^2=2(x-2)[(x+2)^2-2x]$
\Leftrightarrow $x.[(x+2)^2-2x-2x]=(x-2)[(x+2)^2-2x]$
\Leftrightarrow $2.[x^2+2x+4]=2x^2$
\Rightarrow $2x+4=0$ \Rightarrow $x=-2$

 

[ronaldover7]

\Leftrightarrow $\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}$
\Leftrightarrow $\dfrac{2-x}{2002}+1=\dfrac{1-x}{2003}+1-\dfrac{x}{2004}$+1
\Leftrightarrow $\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}$+1
Đễ rùi!

 

[ronaldover7]

$\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}$

$\leftrightarrow (x-23)(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27})=0$

$\leftrightarrow x-23=0 \leftrightarrow x=23$