\(A=x^2-5x+12\\ A=x^2-5x+\dfrac{25}{4}+\dfrac{23}{4}\\ A=\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{23}{4}\\ A=\left[x^2-2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{23}{4}\\ A=\left(x-\dfrac{5}{2}\right)^2+\dfrac{23}{4}\\ Do\text{ }\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-\dfrac{5}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-\dfrac{5}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{5}{2}=0\\ \Leftrightarrow x=\dfrac{5}{2}\\ \text{Vậy }A_{\left(Min\right)}=\dfrac{23}{4}\text{ }khi\text{ }x=\dfrac{5}{2}\)

\(B=2x^2-14x+5\\ \\ A=2x^2-14x+\dfrac{49}{2}-\dfrac{39}{2}\\ A=\left(2x^2-14x+\dfrac{49}{2}\right)-\dfrac{39}{2}\\ A=2\left(x^2-7x+\dfrac{49}{4}\right)-\dfrac{39}{2}\\ A=\left[x^2-2\cdot x\cdot\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2\right]-\dfrac{39}{2}\\ A=\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\\ Do\text{ }\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\\ \Rightarrow A=\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\ge-\dfrac{39}{2}\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-\dfrac{7}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{7}{2}=0\\ \Leftrightarrow x=\dfrac{7}{2}\\ \text{Vậy }B_{\left(Min\right)}=-\dfrac{39}{2}\text{ }khi\text{ }x=\dfrac{7}{2}\)

\(B=2x^2-14x+5\\ B=2x^2-14x+\dfrac{49}{2}-\dfrac{39}{2}\\ B=\left(2x^2-14x+\dfrac{49}{2}\right)-\dfrac{39}{2}\\ B=2\left(x^2-7x+\dfrac{49}{4}\right)-\dfrac{39}{2}\\ B=2\left[x^2-2\cdot x\cdot\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2\right]-\dfrac{39}{2}\\ B=2\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\\ \)

Do \(\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-\dfrac{7}{2}\right)^2\ge0\forall x\)

\(\Rightarrow B=2\left(x-\dfrac{7}{2}\right)^2-\dfrac{39}{2}\ge-\dfrac{39}{2}\forall x\)

Dấu \("="\) xảy ra khi :

\(\left(x-\dfrac{7}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{7}{2}=0\\ \Leftrightarrow x=\dfrac{7}{2}\)

Vậy \(B_{\left(Min\right)}=-\dfrac{39}{2}\) khi \(x=\dfrac{7}{2}\)

Do máy bị lỗi nên câu B bị trục trặc.

Mk xin lỗi.

a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

$\Leftrightarrow (x-3)^2(2x+1)^2=0$

$\Leftrightarrow (x-3)(2x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$

a)      \(2x\left(5-3x^2\right)-10\left(6+x\right)\)

     \(=10x-6x^3-60-10x\)

     \(=\) \(-6x^3-60\)

a) \(2x\left(5-3x^2\right)-10\left(6+x\right)\\ =2x.5-2x.3x^2-10.6-10.x\\ =10x-6x^3-60-10x\)

b) \(3\left(-x+2\right)-6\left(1-x+5x^{20}\right)\\ =-3.x+3.2-6.1+6.x-5.5x^{20}\\ =-3x+6-6+6x-25x^{20}=25x^{20}+3x\)

c) \(7x\left(2-5x^2+\dfrac{1}{2}x^3\right)-14x\left(1-2x^2\right)\\ =7x.2-7x.5x^2+7x.\dfrac{1}{2}x^3-14x.1+14x.2x^2\\ =14x-25x^3+\dfrac{7}{2}x^4-14x+28x^3=3x^2+\dfrac{7}{2}x^4\) 

b)    \(3\left(-x+2\right)-6\left(1-x+5x^{20}\right)\)

  \(=-3x+6-6+6x-30x^{20}\)

  \(=3x-30x^{20}\)

a: \(=\dfrac{6x^2+9x+8x+12}{2x+3}=\dfrac{3x\left(2x+3\right)+4\left(2x+3\right)}{2x+3}\)

=3x+4

b: \(=\dfrac{5x^2-2x+15x-6}{5x-2}\)

\(=\dfrac{x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=x+3\)

c: \(=\dfrac{-8x^2+20x+2x-5-10}{2x-5}=-4x+1+\dfrac{-10}{2x-5}\)

d: \(=\dfrac{14x^2-35x+2x-5}{2x-5}=\dfrac{7x\left(2x-5\right)+\left(2x-5\right)}{2x-5}\)

=7x+1

e: \(=\dfrac{2x^3+x^2+6x^2+3x+12x+6}{2x+1}\)

\(=\dfrac{x^2\left(2x+1\right)+3x\left(2x+1\right)+6\left(2x+1\right)}{2x+1}=x^2+3x+6\)

f: \(=\dfrac{x^3-2x^2+6x^2-12x+x-2}{x-2}=x^2+6x+1\)

g: \(=\dfrac{12x^3+6x^2-4x^2-2x+6x+3}{2x+1}=6x^2-2x+3\)

1)2x-3=11-5x

2x+5x=11+3

7x =14

x=2.

a: \(=\dfrac{6x^3+13x^2-5x}{2x+5}=\dfrac{6x^3+15x^2-2x^2-5x}{2x+5}=3x^2-x\)

b: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)

\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)

\(=x^2-2x+3\)

d: \(=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)

a: ĐKXD: x<>0

\(\dfrac{14x^3+12x^2-14x}{2x}=\left(x+2\right)\left(3x-4\right)\)

=>\(\dfrac{2x\left(7x^2+6x-7\right)}{2x}=\left(x+2\right)\left(3x-4\right)\)

=>\(7x^2+6x-7=3x^2-4x+6x-8\)

=>\(7x^2+6x-7=3x^2+2x-8\)

=>\(4x^2+4x+1=0\)

=>\(\left(2x+1\right)^2=0\)

=>2x+1=0

=>x=-1/2(nhận)

b: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)

=>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)

=>\(24x^2-26x-5-24x^2+23x+12=15\)

=>-3x+7=15

=>-3x=8

=>\(x=-\dfrac{8}{3}\)

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

b: =x^2+4x-5x-20

=x^2-x-20

c: =-7x^2+4x-2

d: \(=\dfrac{-2x^4+2x^3+3x^3-3+2}{x-1}\)

\(=-2x^3+3x+3+\dfrac{2}{x-1}\)