Đáp án:

$1)
a)  {\left[\begin{aligned}y=0\\y=-1\\ y=1\end{aligned}\right.}\\
b)  {\left[\begin{aligned}y=0\\y=\dfrac{1}{2}\\ y=\dfrac{-5}{2}\end{aligned}\right.}\\
c)  {\left[\begin{aligned}x=\dfrac{2}{3}\\x=\dfrac{-5}{4}\end{aligned}\right.}\\
d){\left[\begin{aligned}x=\dfrac{6,9}{2,3}=3\\x=\dfrac{-2}{0,1}=-20\end{aligned}\right.}\\
e) x=\dfrac{-1}{2}
f)  {\left[\begin{aligned}x=\dfrac{-7}{2}\\x=5\\ x=\dfrac{-1}{5}\end{aligned}\right.}\\
2)
a)  {\left[\begin{aligned}y=0\\y=\dfrac{11}{2}\end{aligned}\right.}\\
b)  {\left[\begin{aligned}x=3\\x=\dfrac{-5}{2}\end{aligned}\right.}\\
c)  {\left[\begin{aligned}x=\dfrac{7}{2}\\x=2\end{aligned}\right.}\\
d) {\left[\begin{aligned}x=5\\x=\dfrac{3}{2}\end{aligned}\right.}\\$

Giải thích các bước giải:

 $1)
a) y(y^2-1)=0\\
\Leftrightarrow y(y-1)(y+1)=0\\
\Leftrightarrow {\left[\begin{aligned}y=0\\y=-1\\ y=1\end{aligned}\right.}\\
b) y(y-\dfrac{1}{2})(2y+5)=0\\
\Leftrightarrow {\left[\begin{aligned}y=0\\y-\dfrac{1}{2}=0\\ 2y+5=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}y=0\\y=\dfrac{1}{2}\\ y=\dfrac{-5}{2}\end{aligned}\right.}\\
c) (3x-2)(4x+5)=0\\
\Leftrightarrow {\left[\begin{aligned}3x-2=0\\4x+5=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{2}{3}\\x=\dfrac{-5}{4}\end{aligned}\right.}\\
d) (2,3x-6,9)(0,1x+2)=0\\
\Leftrightarrow {\left[\begin{aligned}2,3x-6,9=0\\0,1x+2=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{6,9}{2,3}=3\\x=\dfrac{-2}{0,1}=-20\end{aligned}\right.}\\
e) (4x+2)(x^2+1)=0\\
\Leftrightarrow {\left[\begin{aligned}4x+2=0\\x^2+1=0 (VL)\end{aligned}\right.}\\
\Leftrightarrow x=\dfrac{-2}{4}=\dfrac{-1}{2}$
Vì $x^2>0 \Rightarrow x^2+1>0$
$f) (2x+7)(x-5)(5x+1)=0\\
\Leftrightarrow {\left[\begin{aligned}2x+7=0\\x-5=0\\ 5x+1=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{-7}{2}\\x=5\\ x=\dfrac{-1}{5}\end{aligned}\right.}\\
2)
a) 2y^2-11y=0\\
\Leftrightarrow y(2y-11)=0\\
\Leftrightarrow {\left[\begin{aligned}y=0\\2y-11=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}y=0\\y=\dfrac{11}{2}\end{aligned}\right.}\\
b) 2x(x-3)+5(x-3)=0\\
\Leftrightarrow (x-3)(2x+5)=0\\
\Leftrightarrow {\left[\begin{aligned}x-3=0\\2x+5=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=3\\x=\dfrac{-5}{2}\end{aligned}\right.}\\
c) x(2x-7)-4x+14=0\\
\Leftrightarrow x(2x-7)-2(2x-7)=0\\
\Leftrightarrow (2x-7)(x-2)=0\\
\Leftrightarrow {\left[\begin{aligned}2x-7=0\\x-2=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{7}{2}\\x=2\end{aligned}\right.}\\
d) 3x-15=2x(x-5)\\
\Leftrightarrow 3(x-5)-2x(x-5)=0\\
\Leftrightarrow (x-5)(3-2x)=0\\
\Leftrightarrow {\left[\begin{aligned}x-5=0\\3-2x=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=5\\x=\dfrac{3}{2}\end{aligned}\right.}\\$

Đáp án:

 Bài 1 :

$a) y(y² -1) =0$

$⇔ y = 0$

$⇔ y² -1 = 0⇔y =±1$

Vậy....

$b) y(y-\dfrac{1}{2})(2y+5)=0$

$⇔y =0$

$⇔ y - \dfrac{1}{2} =0 ⇔ y = \dfrac{1}{2}$

$⇔ 2y +5 = 0⇔y = -\dfrac{5}{2}$

Vậy.....

$c)(3x-2)(4x+5)=0#

⇔\(\left[ \begin{array}{l}3x-2=0\\4x+5=0\end{array} \right.\)  

⇔\(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{array} \right.\)

Vậy....

$d) (2,3x-6,9)(0,1x+2)=0$

⇔\(\left[ \begin{array}{l}2,3x-6,9=0\\0,1x+2=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=3\\x=-20\end{array} \right.\) 

Vậy....

$e) (4x+2)(x^2+1)=0$

Vì $x^2 ≥ 0 ⇔ x^2 +1 > 0$ (loại)

$⇔4x+2 = 0 $

$⇔ 4x = -2$

$⇔ x= -2 : 4$

$⇔ x = -\dfrac{1}{2}$

Vậy...

$f) (2x+7)(x-5)(5x+1)=0$

⇔\(\left[ \begin{array}{l}2x+7=0\\x-5=0\\5x+1=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{array} \right.\) 

Vậy....

Bài 2 :

$a) 2y^2 -11y =0$

$⇔ y(2y-11)=0$

⇔\(\left[ \begin{array}{l}y=0\\2y-11=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}y=0\\y=\dfrac{11}{2}\end{array} \right.\) 

Vậy....

$b) 2x(x-3)+5(x-3)=0$

⇔\(\left[ \begin{array}{l}x-3=0\\2x+5=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=3\\x=-\dfrac{5}{2}\end{array} \right.\) 

Vậy...

$c) x(2x-7)-4x+14=0$

$⇔x(2x-7) -2(2x-7)=0$

$⇔(2x-7)(x-2)=0$

⇔\(\left[ \begin{array}{l}2x-7=0\\x-2=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=2\end{array} \right.\) 

Vậy...

$d) 3x-15 = 2x(x-5)$

$⇔ 3x-15 -2x(x-5)=0$

$⇔ 3(x-5) -2x(x-5)=0$

$⇔(x-5)(3-2x)=0$

⇔\(\left[ \begin{array}{l}x-5=0\\3-2x=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=5\\x=\dfrac{3}{2}\end{array} \right.\) 

Vậy...