Đáp án:
 
Giải thích các bước giải:

`1)2x^2-3(x-2)-1=10`

`<=>2x^2-3x+6-1-10=0`

`<=>2x^2-3x-5=0`

`<=>(2x-5)(x+1)=0`

`<=>` \(\left[ \begin{array}{l}2x-5=0\\x+1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=-1\end{array} \right.\)

Vậy `S={-1;5/2}`

`2)4x^2+2x=-3(x-2)`

`<=>4x^2+2x=-3x+6`

`<=>4x^2+2x+3x-6=0`

`<=>4x^2+5x-6=0`

`<=>(4x-3)(x+2)=0`

`<=>` \(\left[ \begin{array}{l}4x-3=0\\x+2=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{3}{4}\\x=-2\end{array} \right.\)

Vậy `S={-2;3/4}`

`3)2x(x-3)=3x-7`

`<=>2x^2-6x-3x+7=0`

`<=>2x^2-9x+7=0`

`<=>(2x-7)(x-1)=0`

`<=>` \(\left[ \begin{array}{l}2x-7=0\\x-1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=1\end{array} \right.\)

Vậy `S={7/2;1}`

`4)2x(x+1)=x(x-2)+5`

`<=>2x^2+2x=x^2-2x+5`

`<=>2x^2-x^2+2x+2x-5=0`

`<=>x^2+4x-5=0`

`<=>(x-1)(x+5)=0`

`<=>` \(\left[ \begin{array}{l}x-1=0\\x+5=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=1\\x=-5\end{array} \right.\)

Vậy `S={-5;1}`

`5)(x-2)(x+3)-6=0`

`<=>x^2+x-12=0`

`<=>(x-3)(x+4)=0`
`<=>` \(\left[ \begin{array}{l}x-3=0\\x+4=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.\)

Vậy `S={-4;3}`

`6)(x+2)(x-5)+6=0`

`<=>x^2-3x-4=0`

`<=>(x+1)(x-4)=0`

`<=>` \(\left[ \begin{array}{l}x+1=0\\x-4=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-1\\x=4\end{array} \right.\)

Vậy `S={-1;4}`

`7)(x-1)(3x+2)=x^2`

`<=>3x^2-x-2=x^2`

`<=>2x^2-x-2=0`

`<=>(\sqrt{2}x)^2-2.\sqrt{2}x. 1/(2\sqrt{2})+1/8-17/8=0`

`<=>(\sqrt{2}x-1/(2\sqrt{2}))^2=17/8`

`<=>` \(\left[ \begin{array}{l}\sqrt{2}x-\dfrac{1}{2\sqrt{2}}=\dfrac{\sqrt{17}}{2\sqrt{2}}\\\sqrt{2}x-\dfrac{1}{2\sqrt{2}}=\dfrac{-\sqrt{17}}{2\sqrt{2}}\end{array} \right.\)

`<=>` \(\left[ \begin{array}{l}\sqrt{2}x=\dfrac{\sqrt{17}+1}{2\sqrt{2}}\\\sqrt{2}x=\dfrac{-\sqrt{17}+1}{2\sqrt{2}}\end{array} \right.\)

`<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{17}+1}{4}\\x=\dfrac{-\sqrt{17}+1}{14}\end{array} \right.\)

`8)(3x+2)(x-5)=x(2-x)-5`

`<=>3x^2-13x-10=2x-x^2-5`

`<=>4x^2-15x-5=0`

`\Delta=(-15)^2-4.4.(-15)=305`
`=>x=(-(-15)+-\sqrt{305})/(2.4)=(15+-\sqrt{305})/8`

`9)(3x-1)(x+5)=12`

`<=>3x^2+14x-17=0`

`<=>(3x+17)(x-1)=0`

`<=>` \(\left[ \begin{array}{l}3x+17=0\\x-1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-\dfrac{17}{3}\\x=1\end{array} \right.\)

Vậy `S={-17/3;1}`

`10)x^2+(3x-1)(x-2)+1=0`

`<=>4x^2-7x+3=0`

`<=>(4x-3)(x-1)=0`

`<=>` \(\left[ \begin{array}{l}4x-3=0\\x-1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{3}{4}\\x=1\end{array} \right.\)

Vậy `S={3/4;1}`

Đáp án:

$1$. $x= -1$; `5/2`

$2$. $x=$; `3/4`

$3$. $ x=1$; `7/2`

$4$. $x=-5; 1$

$5$. $x=-4; 3$

$6$.$ x=-1; 4$

$7$. $ x=$ $\frac{1-\sqrt{17}}{4}$ ; $\frac{1+\sqrt{17}}{4}$ 

$8$. $x=$ $\frac{15-\sqrt{305}}{8}$ ; $\frac{15+\sqrt{305}}{8}$ 

$9$. $x=1$ ; `-17/3`

$10$. $x=1$ ; `3/4`

Giải thích các bước giải:

$1$. $2x^{2}$ $-3(x-2)-1$ $=10$

⇔ $2x^{2}$ $-3x+6-1$ $=10$ 

⇔ $2x^{2}$ $-3x+5$ $=10$

⇔ $2x^{2}$ $-3x+5-10$ $=0$

⇔ $2x^{2}$ $-3x-5$ $=0$

⇔ $x=$ $\frac{-(-3)±\sqrt{(-3)^2 - 4.2.-5}}{2.2}$ 

⇔ $x=$ $\frac{3±\sqrt{9+40}}{4}$ 

⇔ $x=$ $\frac{3±\sqrt{49}}{4}$ 

⇔ $x=$ $\frac{3±7}{4}$ 

⇔ \(\left[ \begin{array}{l}x=\frac{3+7}{4}\\x=\frac{3-7}{4}\end{array} \right.\) 

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