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

=>x=5 hoặc x=-3

2: =>(x-1)(5x-1)=0

=>x=1/5 hoặc x=1

5: =>(x-4)*x=0

=>x=0 hoặc x=4

10: =>(x+5)(x-3)=0

=>x=3 hoặc x=-5

9: =>(x-2)(x-4)=0

=>x=2 hoặc x=4

7: =>(x-6)(2x-1)=0

=>x=1/2 hoặc x=6

8: =>(2x-1)(3x-12)=0

=>x=4 hoặc x=1/2

a) ( 2x + 3 )^2 - 4( x - 1 )( x + 1 ) = 49

=>4x²+12x+9-4x²+4=49

 =>12x+13=49

=>12x=36

=>x=3

b) 16x^2 - ( 4x - 5 )^2 = 15

=>16x²-16x²+40x-25=15

=>40x-25=15

=>40x=40

=>x=1

c) ( 2x + 1 )^2 - ( x - 1)^2 = 0

=>4x²+4x+1-x²+2x-1=0

=>3x²+6x=0

=>3x(x+2)=0

=>3x=0 hoặc x+2=0

=>x=0 hoặc x=-2

a) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ =>4x^2+12x+9-4x^2+4=49\\=>12x+13=49\\ =>12x=36\\ =>x=3\)

b) \(16x^2-\left(4x-5\right)^2=15\\ =>16x^2-16x^2+40x-25=15\\ =>40x-25=15\\ =>40x=40\\ =>x=1\)

c) \(\left(2x+1\right)^2-\left(x-1\right)^2=0\\ =>4x^2+4x+1-x^2+2x-1=0\\ =>3x^2+6x=0\\ =>3x\left(x+2\right)=0\\ =>\left[\frac{3x=0}{x+2=0}\right]=>\left[\frac{x=0}{x=-2}\right]\)

a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)

\(\Rightarrow6x^2+2-6x^2+12x-6=10\)

\(\Rightarrow12x-4=10\)

\(\Rightarrow12x=14\)

\(\Rightarrow x=\dfrac{7}{6}\)

b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)

\(\Rightarrow x^3-25x-x^3-8=42\)

\(\Rightarrow-25x-8=42\)

\(\Rightarrow-25x=50\)

\(\Rightarrow x=\dfrac{50}{-25}=-2\)

c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Rightarrow24x+25=49\)

\(\Rightarrow24x=24\)

\(\Rightarrow x=\dfrac{24}{24}=1\)

a) \(\left(2x-5\right)^2=49\)    

     \(\left(2x-5\right)^2=\left(\pm7\right)^2\)   

      \(=>2x-5=7\)  hoặc \(2x-5=-7\)  

      \(\cdot2x-5=7\)                                        \(\cdot2x-5=-7\)   

       \(2x=5+7\)                                          \(2x=-7+5\)    

       \(2x=12\)                                                \(2x=-2\)   

         \(x=12:2\)                                             \(x=-2:2\)  

         \(x=6\)                                                     \(x=-1\)  

   Vậy x=6 hoặc x=-1

b/ \(\left(2x+5\right)^2-\left(1-2x\right)^2=10\)   

     \(4x^2+20x+25-\left(1-4x+4x^2\right)=10\)   

     \(4x^2+20x+25-1+4x-4x^2=10\)   

                  \(24x+24=10\)   

                   \(24x=10-24\)   

                    \(24x=-14\)    

                         \(x=\frac{-14}{24}\)

                          \(x=\frac{-7}{12}\)   

c/ \(\left(9-2x\right)^3=27\)   

     \(\left(9-2x\right)^3=3^3\)

      \(9-2x=3\)   

       \(2x=9-3\)

        \(2x=6\)

         \(x=6:2\)

         \(x=3\)

a/ (2x-5)^2=49

TH1: 2x-5=7

x=6

TH2: 2x-5=-7

x=-1

\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a) 9x²-49=0

(3x)²-7²=0

<=> (3x-7)(3x+7)=0

th1: 3x-7=0

<=>3x=7

<=>x=\(\dfrac{7}{3}\)

th2: 3x+7=0

<=>3x=-7

<=>x=\(-\dfrac{7}{3}\)

3.|x+1|-2=4

3.|x+1|=4+2

3.|x+1|=6                                    

|x+1|=6:3

|x+1|=2

Trường hợp 1       x+1=2

x=2-1

x=1

trường hợp 2

x+1=-2

x=(-2)-1

x=-3

==> x thuộc {1; -3}

k mk nha chúc học tốt                                                                             

a) 16x^2 - (4x - 5)^2 = 15

<=> 16x^2 - 16x^2 + 40x - 25 = 15

<=> 40x = 40

<=> x = 1

b) (2x + 3)^2 - 4(x - 1)(x + 1) = 49

<=> 4x^2 + 12x + 9 - 4x^2 - 4x + 4x + 4 = 49

<=> 12x + 13 = 49

<=> 12x = 36

<=> x = 3

c) (2x + 1)(1 - 2x) + (1 - 2x)^2 = 18

<=> 1 - 4x^2 + 1 - 4x + 4x^2 = 18

<=> 2 - 4x = 18

<=> -4x = 16

<=> x = -4

d)2(x + 1)^2 - (x - 3)(x + 3) - (x - 4)^2 = 0

<=> 2x^2 + 4x + 2 - x^2 + 3^2 - x^2 + 8x - 16 = 0

<=> 12x - 5 = 0

<=> 12x = 5

<=> x = 5/12

e) (x - 5)^2 - x(x - 4) = 9

<=> x^2 - 10x + 25 - x^2 + 4x = 9

<=> -6x + 25 = 9

<=> -6x = 9 - 25

<=> -6x = -16

<=> x = -16/-6 = 8/3

f) (x - 5)^2 + (x - 4)(1 - x) = 0

<=> x^2 - 10x + 25 + x - x^2 - x - 4 + 4x = 0

<=> -5x + 21 = 0

<=> -5x = -21

<=> x = 21/5

Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)