17) (3x – 2)(x + 9) – (3x – 2)^2 = 0
18) (2x – 3)^2 – 4(x – 3)(x + 3) = -11
19) x^2 (3x + 1) – (x – 3)^2 = -9
20) (2x + 5) (4x^2 – 10x + 25) – (2x + 1)^2 = 52
17) (3x – 2)(x + 9) – (3x – 2)^2 = 0
18) (2x – 3)^2 – 4(x – 3)(x + 3) = -11
19) x^2 (3x + 1) – (x – 3)^2 = -9
20) (2x + 5) (4x^2 – 10x + 25) – (2x + 1)^2 = 52
Đáp án:
$17)
{\left[\begin{aligned}x=\dfrac{2}{3}\\x=\dfrac{11}{2}\end{aligned}\right.}\\
18)
x=\dfrac{14}{3}\\
19)
x=0\\
20)
x=-2$
Giải thích các bước giải:
$17)
(3x-2)(x+9)-(3x-2)^2=0\\
\Leftrightarrow (3x-2)(x+9-3x+2)=0\\
\Leftrightarrow (3x-2)(11-2x)=0\\
\Leftrightarrow {\left[\begin{aligned}3x-2=0\\11-2x=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{2}{3}\\x=\dfrac{11}{2}\end{aligned}\right.}\\
18)
(2x-3)^2-4(x-3)(x+3)=-11\\
\Leftrightarrow 4x^2-12x+9-4(x^2-9)+11=0\\
\Leftrightarrow 4x^2-12x+9-4x^2+36+11=0\\
\Leftrightarrow -12x=-56\\
\Leftrightarrow x=\dfrac{14}{3}\\
19)
x^2(3x+1)-(x-3)^2=-9\\
\Leftrightarrow 3x^3+x^2-(x^2-6x+9)+9=0\\
\Leftrightarrow 3x^3+x^2-x^2+6x-9+9=0\\
\Leftrightarrow 3x^3+6x=0\\
\Leftrightarrow 3x(x^2+6)=0\\
\Leftrightarrow {\left[\begin{aligned}3x=0\\x^2+6=0(VL)\end{aligned}\right.}\\
\Leftrightarrow x=0\\
20)
(2x+5)(4x^2-10x+25)-(2x+1)^2=52\\
\Leftrightarrow (2x)^3+5^3-(2x+1)^2=52\\
\Leftrightarrow 8x^3+125-(4x^2+4x+1)-52=0\\
\Leftrightarrow 8x^3+125-4x^2-4x-1-52=0\\
\Leftrightarrow 8x^3-4x^2-4x+72=0\\
\Leftrightarrow 8x^3+16x^2-20x^2-40x+36x+72=0\\
\Leftrightarrow 8x^2(x+2)-20x(x+2)+36(x+2)=0\\
\Leftrightarrow (x+2)(8x^2-20x+36)=0\\
\Leftrightarrow (x+2).4.\left [2x^2-5x+9 \right ]=0\\
\Leftrightarrow (x+2).4.\left [(\sqrt{2}x-\dfrac{5}{2\sqrt{2}})^2+\dfrac{47}{8} \right ]=0\\
\Leftrightarrow x+2=0\\
\Leftrightarrow x=-2$
vì $(\sqrt{2}x-\dfrac{5}{2\sqrt{2}})^2+\dfrac{47}{8}>0$