Question

b. x( x – 4) - 2x + 8 = 0 

c. x^2-25 –( x+5 ) = 0

d.(2x -1)^2- (4x2 – 1) = 0 

e. ( 3x – 1)^2 – ( x +5)^2 = 0 

f. x^3 – 8 – (x -2)(x -12) =0

Answer 1

b) x(x-4) - 2x+8 = 0
    x(x-4) - 2(x-4) = 0
    (x-2) (x-4) = 0
TH1: x-2=0              TH2: x-4=0
            x=2                          x=4
Vậy x\(\in\){2;4}

Answer 2

\(b,\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ c,\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\\ d,\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ e,\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ f,\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow x=2\)

Answer 3

b) x(x-4)-2x+8=0

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

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

th1: x-4=0

x=4

th2: x-2=0

x=2

Vậy x thuộc tập hợp 4;-2

Answer 4

c) x²-25-(x+5)=0
    x²-5²-(x+5)=0
    (x-5)(x+5)-(x+5)=0
    (x+5)(x-5-1)=0
TH1: x+5=0            Th2: x-5-1=0
             x=-5                         x= 6
Vậy x\(\in\){-5;6}

Answer 5

d) (2x-1)²-(4x²-1)=0
    (4x²-4x+1)-(4x²-1)=0
    4x²-4x+1-4x²+1=0
    (4x²-4x²)+(1+1)-4x=0
        0+2-4x=0
                4x=2
                  x=1/2

Answer 6

e) (3x-1)²-(x+5)²=0
    (3x-1-x-5)(3x-1+x+5)=0
    (2x-6)(4x+4)=0
    TH1: 2x-6=0            TH2: 4x+4=0
                  x=3                           x=-1
Vậy x\(\in\){-1;3}

Answer 7

f) x³-8-(x-2)(x-12)=0
   x³-2³-(x-2)(x-12)=0
   (x-2)(x²+2x+4)-(x-2)(x-12)=0
   (x-2)(x²+2x+4-x+12)=0
   (x-2)(x²+x+16)=0
TH1: x-2=0      TH2: x²+x+16=0
            x=2                     x²+x=-16
                                            x= vô lí (x²>2\(\Rightarrow\)x\(\notin\)số âm)
Vậy x=2