a,2x-(-17)=15

2x+17=15

2x=15-17

2x=-2

=>x=-1

a, 2x -(-17) = 15

2x + 17 = 15 

2x        = 15 - 17 

2x        =(-2)

 x         = (-2) : 2 

x          = (-1)

a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\left(2x+1\right)^2-\left[2\left(x+2\right)\right]^2=9\)

\(\left[2x+1-2\left(x+2\right)\right]\left[2x+1+2\left(x+2\right)\right]=9\)

\(\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\)

\(-3\left(4x+5\right)=9\)

\(4x+5=-3\)

\(4x=-8\)

\(x=-2\)

b) \(x^2-2x-15=0\)

\(x^2-5x+3x-15=0\)

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

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

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

c) \(2x^2+3x-5=0\)

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

\(2x\left(x-1\right)+5\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}}\)

1) x (x-2016) + 2015 (2016-x) = 0

 x (x-2016) - 2015 (x- 2016) = 0

(x-2015)(x-2016) =0

\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)

Vậy x= 2015; 2016

2) -5x (x-15) + (15-x) = 0

-5x (x-15) - (x-15) =0

(-5x -1) (x-15) =0

\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)

Vậy x= -1/5; 15

3) 3x (3x-7) - (7-3x) =0

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

(3x+1) (3x-7) =0

\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)

Vậy x= -1/3 ; 7/3

a,x.(x+7)=0

suy ra x=o hoặc x+7=0

vs x+7=0

         x=0+7

         x=7

vậy x=0 hoặc x=7                                

b(2+2x)(7-x)=0

suy ra 2+2x=0 hoặc 7-x=0

vs2+2x=0        vs7-x=0

2x        =0-2            x=0+7

2x        =(-2)            x=7

           x=(-2);2

           x=-1

vậy x=-1 hoặc x=7

d(x^2-9)(3x+15)=0

suy ra x^2-9=0 hoặc 3x+15=0

vsx^2-9=0         vs 3x+15=0

    x^2   =0+9          3x     =0-15

    x^2   =9               3x    =-15

    x^2   =3^2                 x=(-15):3

suy ra x=3 hoặc x=-3    x=-5

vậy x=3 x=-3 hoặc x=-5

e,(4x-8)(x^2+1)=0

suy ra4x-8=0 hoặc x^2+1=0

vs  4x-8=0         vs x^2+1=0

     4x    =0+8          x^2    =0-1

     4x    =8              x^2    =-1

        x   =8:4           x^2     =-1^2 hoặc 1^2

        x    =2                suy ra x=-1 hoặc x=1

vậy x=2, x=-1 hoặc x=1        

Trả lời:

a, \(\left(3x+1\right)\left(x-3\right)-x\left(3x-14\right)=15\)

\(\Leftrightarrow3x^2-9x+x-3-3x^2+14x=15\)

\(\Leftrightarrow6x-3=15\)

\(\Leftrightarrow6x=18\)

\(\Leftrightarrow x=3\)

Vậy x = 3 là nghiệm của pt.

b, \(\left(x-3\right)^2=9-x^2\)

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

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

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

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

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

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

Vậy x = 3; x = 0 là nghiệm của pt.

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

\(\Leftrightarrow4x^2-2x+\frac{1}{4}-\left(1-4x+4x^2\right)=2\)

\(\Leftrightarrow4x^2-2x+\frac{1}{4}-1+4x-4x^2=2\)

\(\Leftrightarrow2x-\frac{3}{4}=2\)

\(\Leftrightarrow2x=\frac{11}{4}\)

\(\Leftrightarrow x=\frac{11}{8}\)

Vậy x = 11/8 là nghiệm của pt.

d, \(4x^2+4x-3=0\)

\(\Leftrightarrow4x^2-2x+6x-3=0\)

\(\Leftrightarrow2x\left(2x-1\right)+3\left(2x-1\right)=0\)

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

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

Vậy x = 1/2; x = - 3/2 là nghiệm của pt.

1) \(x^2-6x+9=\left(5-3x\right)^2\)

\(\left(x-3\right)^2=\left(5-3x\right)^2\)

\(\Rightarrow x-3=5-3x\)

\(\Rightarrow x+3x=5+3\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=2\)

\(3x\left(2x-3\right)=5\left(3-2x\right)\)

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

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

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

3) \(x^2-2x-15=0\)

\(x^2-2x+1-16=0\)

\(\left(x-1\right)^2-4^2=0\)

\(\left(x-1-4\right)\left(x-1+4\right)=0\)

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

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

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+2x=15\)

\(\Leftrightarrow2x=-12\)

\(\Leftrightarrow x=-6\)

a) 5x^2 + 15x = 0
=> 5x(x + 3) = 0
=> x = 0 hoặc x + 3 = 0
=> x = 0 hoặc x = -3

@Hoàng

g: \(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4=0\)

\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+36=0\)

\(\Leftrightarrow\left(x+3\right)^2\left(x^2+6x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\sqrt{5}-3\\x=-\sqrt{5}-3\end{matrix}\right.\)