câu f là 9+3x hay 9-3x vậy???

a) 

$2x+6=0$

$2x=-6$

$x=-3$

b) $4x+20=0$

$4x=-20$

$x=-5$

c) 

$2(x-1)=5x-7$

$2x-2=5x-7$

$3x=5$

$x=\frac{5}{3}$

d) $2x-3=0$

$2x=3$

$x=\frac{3}{2}$

e) 

$3x-1=x+3$

$2x=4$

$x=2$

f) 

$15-7x=9-3x$

$6=4x$

$x=\frac{3}{2}$

g) $x-3=18$

$x=18+3=21$

h) 

$2x+1=15-5x$

$7x=14$

$x=2$

2) \(\left(x+3\right)\left(5x-15\right)+2x-6=0\)

\(\Leftrightarrow\left(x+3\right).5.\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[5\left(x+3\right)+2\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x+17\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x+17=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{17}{5}\end{matrix}\right.\)

câu 1 phải là \(\left(x-5\right)\left(x+9\right)+6x=30\)

\(\Leftrightarrow\left(x-5\right)\left(x+9\right)+6x-30=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+9\right)+6\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+9+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+15\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-15\end{matrix}\right.\)

tìm hả bn

a) \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-15=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\3x-9=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=3\end{matrix}\right.\)

a. \(\left[{}\begin{matrix}2x+3=0\\5x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=3\end{matrix}\right.\)

b. \(\left[{}\begin{matrix}3x+1=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=3\end{matrix}\right.\)

a)\(\left(2x+3\right).\left(5x-15\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-3\\5x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1,5\\x=3\end{matrix}\right.\)

b)\(\left(3x+1\right).\left(3x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\3x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=3\end{matrix}\right.\)

a) 7x+4=3x+16\(\Leftrightarrow\)4x=12\(\Leftrightarrow\)x=3

b)(x+9)(3x-15)=0\(\Leftrightarrow\)x+9=0 hoặc 3x-15=0

\(\Rightarrow\)x\(\in\){-9;5}

c) |-5x|=2x+21

Nếu x\(\le\)0 thì -5x=2x+21\(\Leftrightarrow\)x=-3 (t/m)

Nếu x>0 thì -5x=-2x-21\(\Leftrightarrow\)x=7 (t/m)

Vậy x\(\in\){-3;7}

d) 3x-5>15-x\(\Leftrightarrow\)4x>20\(\Leftrightarrow\)x>5

e) \(\dfrac{x+1}{2001}+\dfrac{x+5}{2005}< \dfrac{x+9}{2009}+\dfrac{x+13}{2013}\)

\(\Leftrightarrow\dfrac{x+1}{2001}-1+\dfrac{x+5}{2005}-1< \dfrac{x+9}{2009}-1+\dfrac{x+13}{2013}-1\)

\(\Leftrightarrow\)\(\dfrac{x-2000}{2001}+\dfrac{x-2000}{2005}-\dfrac{x-2000}{2009}-\dfrac{x-2000}{2013}< 0\)

\(\Leftrightarrow\)(x-2000)(\(\dfrac{1}{2001}+\dfrac{1}{2005}-\dfrac{1}{2009}-\dfrac{1}{2013}\))<0

Vì \(\dfrac{1}{2001}+\dfrac{1}{2005}-\dfrac{1}{2009}-\dfrac{1}{2013}>0\) nên x-2000<0

\(\Leftrightarrow\)x<2000

\(2x-8x^2=0\Rightarrow2x\left(1-4x\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\1-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}}\)

\(x-x^2=0\Rightarrow x\left(1-x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Cn lại lm tương tự nha e!

=.= hok tốt!!

\(a,-2x+\left(5x-9\right)=0\)

\(-2x+5x-9=0\)

\(3x-9=0\)

\(3x=9\)

\(x=3\)

\(b,11+3.\left(15-x\right)=-16\)

\(3.\left(15-x\right)=-16-11\)

\(3.\left(15-x\right)=-27\)

\(15-x=-27:3\)

\(15-x=-9\)

\(x=15+9\)

\(x=24\)

\(-2x+\left(5x-9\right)=0\)

\(\Leftrightarrow-2x+5x-9=0\)

\(\Leftrightarrow-2x+5x=9\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\)

b, \(11+3\cdot\left(15-x\right)=-16\)

\(\Leftrightarrow11+45-3x=-16\)

\(\Leftrightarrow-3x=-72\)

\(\Leftrightarrow x=24\)

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

a,5x(4x-3)+6-8x=0

=>5x(4x-3)-8x+6=0

=>5x(4x-3)-2(4x-3)=0

=>(5x-2)(4x-3)=0

=>\(\orbr{\begin{cases}5x-2=0\\4x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=\frac{3}{4}\end{cases}}}\)

b, 5x(x-9)-x+9=0

=>5x(x-9)-(x-9)=0

=>(5x-1)(x-9)=0

=>\(\orbr{\begin{cases}5x-1=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=9\end{cases}}}\)