Tìm x, biết a. (3x + 4) ² - (3x-1)(3x+1) = 49 b. x ² - 4x + 4 = 9.(x-2) c. x ² - 25 = 3x - 15 d. (x-1) ³ + 3(x+1) ² = (x ² - 2x + 4)(x+2)

Đáp án: a. x=$\frac{4}{3}$ 

 b. x∈{2; 11}  

c. x=5 hoặc x=-2

d. x=$\frac{2}{3}$ 

Giải thích các bước giải:

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

     ($9x^{2}$+24x+16)-(9$x^{2}$-1)=49

      9$x^{2}$+24x+16-9$x^{2}$+1=49

       24x+17=49

        24x=32

         x=$\frac{4}{3}$ 

b) $x^{2}$-4x+4=9(x-2)

     ($x^{2}$-4x+4)-9(x-2)=0

      $(x-2)^{2}$-9(x-2)=0

       (x-2)[(x-2)-9]=0

        (x-2)(x-11)=0

        ⇒$\left \{ {{x-2=0} \atop {x-11=0}} \right.$ 

        ⇒$\left \{ {{x=2} \atop {x=11}} \right.$ 

c. $x^{2}$-25=3x-15

⇔($x^{2}$-25)-(3x-15)=0

⇔(x-5)(x+5)-3(x-5)=0

⇔(x-5)(x+5-3)=0

⇔(x-5)(x+2)=0

⇔\(\left[ \begin{array}{l}x-5=0\\x+2=0\end{array} \right.\)

⇒\(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\) 

d. $(x-1)^{3}$+3($(x+1)^{2}$=($x^{2}$-2x+4)(x+2)

⇔$x^{3}$-3$x^{2}$+3x-1+3($x^{2}$+2x+1)=$x^{3}$+8

⇔-3$x^{2}$+3x+3$x^{2}$+6x=8+1-3

⇔9x=6

⇔x=$\frac{2}{3}$