a) \(\frac{4-3x}{5}-\frac{4-x}{10}=\frac{x+2}{2}\)

 \(\frac{8-6x-4+x}{10}=\frac{5x+10}{10}\) 

\(4-5x=5x+10\) 

\(4-5x-5x-10=0\) 

\(-6-10x=0\) 

\(\Rightarrow x=\frac{-3}{5}\) 

Vậy....

\(\frac{4-3x}{5}-\frac{4-x}{10}=\frac{x+2}{2}\)

\(\Leftrightarrow\)\(\frac{2.\left(4-3x\right)}{10}-\frac{4-x}{10}=\frac{5.\left(x+2\right)}{10}\)

\(\Rightarrow\) 2.( 4 - 3x ) - 4 + x = 5.( x + 2 ) 

\(\Leftrightarrow\)8 - 6x  - 4+ x = 5x + `10

\(\Leftrightarrow\)-6x + x - 5x = -8 + 4 + 10

\(\Leftrightarrow\) -10x = 6

\(\Leftrightarrow\)\(x=\frac{-3}{5}\)

Vậy phương trình có nghiệm là: \(x=\frac{-3}{5}\)

b ) \(\frac{x+1}{2009}+\frac{x+2}{2008}=\frac{x+2007}{3}+\frac{x+2006}{4}\)

\(\Leftrightarrow\) \(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1\)\(=\frac{x+2007}{3}+1+\frac{x+2006}{4}+1\)

\(\Leftrightarrow\)\(\frac{x+1}{2009}+\frac{2009}{2009}+\frac{x+2}{2008}+\frac{2008}{2008}\)\(=\frac{x+2007}{3}+\frac{3}{3}+\frac{x+2006}{4}+\frac{4}{4}\)

\(\Leftrightarrow\)\(\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{3}+\frac{x+2006}{4}\)

\(\Leftrightarrow\)\(\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{3}-\frac{x+2010}{4}=0\)

\(\Leftrightarrow\)\(\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}\right)=0\)

\(\Leftrightarrow\)\(x+2010=0\)    ( Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{3}-\frac{1}{4}\ne0\))

\(\Leftrightarrow\) \(x=-2010\)

Vậy phương trình có nghiệm là: x = -2010

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x²
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :

\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)

\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)

Đến đây ta đặt  \(x+\frac{60}{x}+16=t\left(1\right)\)

Ta được :

\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)

Từ đó ta lắp vào ( 1 ) tính được x 

\(\frac{5x-3}{6}-\frac{7x-1}{4}-\frac{4x+2}{7}+5=0\)

<=> \(\frac{14\left(5x-3\right)-21\left(7x-1\right)-12\left(4x+2\right)+420}{84}=0\)

<=> 70x - 42 - 147x + 21 - 48x -24 + 420 = 0

<=> -125x + 375 = 0

<=> -125x = -375

<=> x = 3

Vậy S = {3}

\(\frac{3\left(2x+1\right)}{4}-5-\frac{3x+2}{10}=\frac{2\left(3x-1\right)}{5}\)

<=> \(\frac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\frac{8\left(3x-1\right)}{20}\)

<=> 30x + 15 - 100 - 6x - 4 = 24x - 8

<=> 24x - 24x = -8 + 89

<=> 0x = 81

=> pt vô nghiệm

\(a,\frac{x}{x-3}-\frac{6}{x^2-9}=\frac{1}{x+3}\) (đkxđ: x khác 3, -3)

\(\frac{x\left(x+3\right)-6}{\left(x-3\right)\left(x+3\right)}=\frac{1}{x+3}\)

\(x\left(x+3\right)-6=x-3\)

\(x^2+2x-3=0\)

\(\left(x+3\right)\left(x-1\right)=0\)

\(\Longrightarrow\left[\begin{array}{l}x=-3\left(L\right)\\ x=1\left(N\right)\end{array}\right.\)

\(b,\frac{x^2}{x-2}+\frac{x}{1-x}=\frac{4}{x^2-3x+2}\) (đkxđ: \(x\ne1,x\ne2)\)

\(\frac{x^2}{x-2}-\frac{x}{x-1}=\frac{4}{\left(x-1\right)\left(x-2\right)}\)

\(\frac{x^2\left(x-1\right)-x\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\frac{4}{\left(x-1\right)\left(x-2\right)}\)

\(x^2\left(x-1\right)-x\left(x-2\right)=4\)

\(x^3-x^2-x^2+2x=4\)

\(x^3-2x^2+2x-4=0\)

\(\left(x^3-2x^2\right)+\left(2x-4\right)=0\)

\(x^2\left(x-2\right)+2\left(x-2\right)=0\)

\(\left(x-2\right)\left(x^2+2\right)=0\)

vì \(x^2+2>0\forall x\) ⇒ x - 2 = 0

⇒ x = 2 (ko thoả mãn)

vậy phương trình vô nghiệm

\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)

Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)

A . 3x + 2(x + 1) = 6x - 7

<=> 3x + 2x + 2 = 6x -7

<=> 5x - 6x = -7 - 2

<=> -x = -9

<=> x =9

B . \(\frac{x+3}{5}\).< \(\frac{5-x}{3}\)

=> 3(x +3) < 5(5 -x)

<=> 3x+9 < 25 - 5x

<=> 3x + 5x < 25 - 9

<=> 8x < 16

<=> x < 2

C . \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2-3x-4}\)=\(\frac{2}{x-4}\)

<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2+x-4x-4_{ }}\)= \(\frac{2}{x-4}\)

<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{\left(x+1\right)\left(x-4\right)}\)= \(\frac{2}{x-4}\)

<=> 5(x - 4) + 2x = 2(x +1)

<=> 5x - 20 + 2x = 2x + 2

<=>7x - 2x = 2 + 20

<=> 5x = 22

<=> x =\(\frac{22}{5}\)