a: Đặt M=0

=>2x-12=0

hay x=12

b: Đặt N=0

=>x+5-4x-1=0

=>-3x+4=0

hay x=4/3

à quên ghi đề nx:))

thu gọn về hằng đẳng thức nha mng<3

a,(2x+7)+135=0                     b, 1/2x-2/5=1/5

   2x+7=0-135                             1/2x=1/5+2/5

   2x+7=-135                               1/2x=3/5

   2x=-135-7                                     x=3/5:1/2

   2x=-142                                         x=6/5

     x=-142:2                                Vậy x=6/5

     x=-71

Vậy x=-71

c, 10-|x+1|=5                                      d, 1/2x+150%x=2014

         |x+1|=10-5                                         2x=2014

         |x+1|=5                                                x=2014:2

*TH1:x+1=5      *TH2:x+1=-5                       x=1007

         x=5-1                x=-5-1                   Vậy x=1007

         x=4                   x=-6

  Vậy x=4 hoặc x=-6

Cám ơn ak!

a: =>(x-5)(x+5)+(x-5)(3x-15)=0

=>(x-5)(x+5+3x-15)=0

=>(x-5)(4x-10)=0

=>x=5 hoặc x=5/2

c: =>x^3-3x^2+2x^2-6x-8x+24=0

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

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

=>\(x\in\left\{3;-4;2\right\}\)

a) \(\left(2x-1\right)^2-25=0\)

\(\left(2x-1\right)^2=0+25=25\)

\(\left(2x-1\right)^2=5^2=\left(-5\right)^2\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-1=5\\2x-1=-5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}2x=6\\2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\end{array}\right.\)

b) \(8x^3-50x=0\)

\(2x\left(4x^2-25\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=0\\4x^2-25=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-\frac{5}{2}\end{array}\right.\end{array}\right.\)

h)\(\dfrac{x+2}{x^2+2x+1}\ge0\)

⇔\(\dfrac{x+2}{\left(x+1\right)^2}\ge0\)

⇔\(\dfrac{x+2}{\left(x+1\right)\left(x+1\right)}\ge\dfrac{0.\left(x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x+1\right)}\)

⇒\(x+2\ge0\)

⇔\(x+2-2\ge0-2\)

⇔\(x\ge-2\)

i)\(\dfrac{x-1}{x-3}>1\)

⇔\(\dfrac{x-1}{x-3}>\dfrac{1.\left(x-3\right)}{1.x-3}\)

⇒\(x-1>x-3\)

⇔\(x-x>-3+1\)

⇔\(0x>-2\)

25 – 2x = 16 – 3x

-2x + 3x = 16 - 25

x = -9

Vậy x = -9

x - (47 - 22 ) = 5 + (10-4x)

x - 47 + 22    = 5 + 10 - 4x

x + 4x = 5 + 10 - 22 + 47

5x       = 40

x         = 40 : 5

x         = 8

Vậy x = 8.

~ HOK TỐT ~

\(\text{25 – 2x = 16 – 3x}\)

\(2x+3x=16-25\)

\(5x=-9\)

\(\Rightarrow x=\frac{-9}{5}\)

\(\text{d) x – (47 – 22) = 5+ (10 – 4x)}\)

\(x-47+22=5+10-4x\)

\(x+4x=5+10-22+47\)

\(5x=40\)

\(\Rightarrow x=8\)

học tốt

\(1,\)

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

\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)

\(2,\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

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

\(\Leftrightarrow x^2\left(x^2-1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

Tìm tập nghiệm :)))??

\(a,\left(x+1\right)\left(4x-11\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{11}{4}\end{matrix}\right.\\ b,\left(x+1\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x+\dfrac{5}{3}\end{matrix}\right.\\ c,\left(x+1\right)\left(4x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{9}{4}\end{matrix}\right.\)

ít thôi, từng câu 1 mới có động lực lm