\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{5;-2\right\}\)

\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)

\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)

Câu d xem lại đề

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)

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

\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)

\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)

\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

3.15:

a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)

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

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

c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3.16

\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)

\(\Leftrightarrow-14m+35-2m^2+8=0\)

\(\Leftrightarrow-14m-2m^2+43=0\)

\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)

\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)

\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)

\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)

pt vô nghiệm

a: =>9x^2+6x+1-6(2x^2-13x+21)=0

=>9x^2+6x+1-12x^2+78x-126=0

=>-3x^2+84x-125=0

=>\(x\in\left\{26.42;1.58\right\}\)

b: =>(3x+1)[(2x-5)^2-(x-3)^2]=0

=>(3x+1)(2x-5-x+3)(2x-5+x-3)=0

=>(3x+1)(x-2)(3x-8)=0

=>\(x\in\left\{-\dfrac{1}{3};2;\dfrac{8}{3}\right\}\)

c; =>(x+5)(0,75x-3+1,25x)=0

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

=>x=3/2 hoặc x=-5

Vậy phương trình đã cho có tập nghiệm là S ={-3; 2}

Ta có: ( x - 2 )( 3x + 5 ) = ( 2x - 4 )( x + 1 )

⇔ ( x - 2 )( 3x + 5 ) - 2( x - 2 )( x + 1 ) = 0

⇔ ( x - 2 )[ ( 3x + 5 ) - 2( x + 1 ) ] = 0

⇔ ( x - 2 )( x + 3 ) = 0

Vậy phương trình đã cho có tập nghiệm là S = { - 3;2 }.

`5-(x-6)=4(3-2x)`

`<=>5-x+6-4(3-2x)=0`

`<=> 5-x+6-12 +8x=0`

`<=> 7x -1=0`

`<=> 7x=1`

`<=>x=1/7`

Vậy pt đã cho có nghiệm `x=1/7`

__

`3-x(1-3x) =5(1-2x)`

`<=> 3-x+3x^2=5-10x`

`<=> 3-x+3x^2-5+10x=0`

`<=> 3x^2 +9x-2=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{105}}{6}\\x=\dfrac{-9-\sqrt{105}}{6}\end{matrix}\right.\)

Vậy pt đã cho có tập nghiệm \(S=\left\{\dfrac{-9+\sqrt{105}}{6};\dfrac{-9-\sqrt{106}}{5}\right\}\)

__

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

`<=>x^2+4x-3x-12- 6x +4 =x^2 -8x+16`

`<=>x^2-5x-8=x^2-8x+16`

`<=> x^2 -5x-8-x^2+8x-16=0`

`<=> 3x-24=0`

`<=>3x=24`

`<=>x=8`

Vậy pt đã cho có nghiệm `x=8`

a) 5-(x-6)=4(3-2x)

=> 5 – x + 6 = 12 – 8x

=> -x + 8x = 12 – 5 – 6

=> 7x = 1

=> x=1/7

Vậy phương trình có nghiệm x=1/7

 b) 3 - x ( 1 - 3x)=5(1-2x)

=> 3-x+3x^2=5-10x

=> 3x^2+9x-2= 0

0=105

=> x =\(\dfrac{-9-\sqrt{105}}{6}\)

a: Ta có: \(3x-5\ge2\left(x-6\right)-12\)

\(\Leftrightarrow3x-5\ge2x-24\)

hay \(x\ge-19\)

b: Ta có: \(2\left(5-2x\right)\ge3-x\)

\(\Leftrightarrow10-4x-3+x\ge0\)

\(\Leftrightarrow-3x\ge-7\)

hay \(x\le\dfrac{7}{3}\)