Tìm x biết
a) ( 3x + 1 ) ( 3x – 1 ) – (x^2 + 2x + 4 ) = x ( 6 – x^2 )
b) 5x ( x – 3 )^2 + 5 ( x – 4 ) ( x + 4 ) – 5x^3 = -5x ( 6x – 1 )
c) x^2 ( 3x + 1 ) – ( x – 3 )^2 = -9
d) ( 2x + 5 )( 4x^2 – 10x + 25 ) – ( 2x + 1 )^2 = 52
Lm đầy đủ cho mik nha
Giải thích các bước giải:
$a)(3x+1)(3x-1)-(x-2)(x^2+2x+4)=x(6-x^2)$
$⇔(9x^2-1)-(x^3-8)=6x-x^3$
$⇔9x^2-1-x^3+8-6x+x^3=0$
$⇔9x^2-6x+7=0$
$⇒x∈∅$
$b)5x(x-3)^2+5(x-4)(x+4)-5x^3=-5x(6x-1)$
$⇔5x(x^2-6x+9)+5(x^2-16)-5x^3=-30x^2+5x$
$⇔5x^3-30x^2+45x+5x^2-80-5x^3=-30x^2+5x$
$⇔-25x^2+45x-80=-30x^2+5x$
$⇔-25x^2+45x-80+30x^2-5x=0$
$⇔5x^2+40x-80=0$
$⇔5(x^2+8x-16)=0$
$⇔x^2+8x-16=0$
$\text{ Kết quả này bạn chưa học nên $x∈∅$ nha (Nếu chưa học)}$
$\text{ Kết quả: $x∈\{4-4\sqrt{2};-4+4\sqrt{2}\}$ (Nếu đã học)}$
$\text{Vậy $x∈\{4-4\sqrt{2};-4+4\sqrt{2}\}$}$
$c)x^2(3x+1)-(x-3)^2=-9$
$⇔3x^3+x^2-x^2+6x-9=-9$
$⇔3x^3+6x=0$
$⇔3x(x^2+2)=0$
$⇔$ \(\left[ \begin{array}{l}3x=0\\x^2+2=0 \text{(Vô lý)}\end{array} \right.\) $⇔$ $x=0$
$\text{Vậy $x=0$}$
$d)(2x+5)(4x^2-10x+25)-(2x+1)^2=52$
$⇔8x^3+125-4x^2-4x-1=52$
$⇔8x^3-4x^2-4x+124=52$
$⇔8x^3-4x^2-4x+124-52=0$
$⇔8x^3-4x^2-4x+72=0$
$⇔4(2x^3-x^2-x+18)=0$
$⇔4(2x^2+4x^2-5x^2-10x+9x+18)=0$
$⇔4[2x^2(x+2)-5x(x+2)+9(x+2)]=0$
$⇔4(x+2)(2x^2-5x+9)=0$
$⇔$ \(\left[ \begin{array}{l}x+2=0\\2x^2-5x+9=0 \text{(Vô lý)}\end{array} \right.\) $⇔$ $x=-2$
$\text{Vậy $x=2$}$
Học tốt!!!
Đáp án:
$a)(3x+1)(3x-1)-(x-2)(x^2+2x+4)=x(6-x^2)$
$⇔(9x^2-1)-(x^3-8)=6x-x^3$
$⇔9x^2-1-x^3+8-6x+x^3=0$
$⇔9x^2-6x+7=0$
$⇒x∈∅$
$b)5x(x-3)^2+5(x-4)(x+4)-5x^3=-5x(6x-1)$
$⇔5x(x^2-6x+9)+5(x^2-16)-5x^3=-30x^2+5x$
$⇔5x^3-30x^2+45x+5x^2-80-5x^3=-30x^2+5x$
$⇔-25x^2+45x-80=-30x^2+5x$
$⇔-25x^2+45x-80+30x^2-5x=0$
$⇔5x^2+40x-80=0$
$⇔5(x^2+8x-16)=0$
$⇔x^2+8x-16=0$
$\text{ $x∈\{4-4\sqrt{2};-4+4\sqrt{2}\}$ }$
$c)x^2(3x+1)-(x-3)^2=-9$
$⇔3x^3+x^2-x^2+6x-9=-9$
$⇔3x^3+6x=0$
$⇔3x(x^2+2)=0$
$⇔$ $\left \{ {{x^2+2=0 } \atop {3x=0}} \right. $
Vậy $x=0$
$d)(2x+5)(4x^2-10x+25)-(2x+1)^2=52$
$⇔8x^3+125-4x^2-4x-1=52$
$⇔8x^3-4x^2-4x+124=52$
$⇔8x^3-4x^2-4x+124-52=0$
$⇔8x^3-4x^2-4x+72=0$
@Yết gửi bạn nhé~~
HỌC TỐT~~
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