Chọn D.

(x − 5)(x + 3) = x(x + 3) – 5( x + 3) =  x 2  + 3x - 5x - 15 =  x 2 − 2x − 15

Lời giải:

a)

$A=5-8x-x^2=21-(x^2+8x+16)=21-(x+4)^2$Vì $(x+4)^2\geq 0$ nên $A=21-(x+4)^2\leq 21$

Vậy GTLN của $A$ là $21$. Giá trị này đạt tại $x+4=0\Leftrightarrow x=-4$

b) 

$B=5-x^2+2x-4y^2-4y=5-(x^2-2x)-(4y^2+4y)$

$=7-(x^2-2x+1)-(4y^2+4y+1)$

$=7-(x-1)^2-(2y+1)^2$

Vì $(x-1)^2\geq 0; (2y+1)^2\geq 0$ với mọi $x,y$ nên $B=7-(x-1)^2-(2y+1)^2\leq 7$Vậy GTLN của $B$ là $7$ tại $x=1; y=\frac{-1}{2}$

\(a.x^2-7x-3x+21=0\Leftrightarrow\left(x^2-7x\right)-\left(3x-21\right)=0\)

\(\Leftrightarrow x\left(x-7\right)-3\left(x-7\right)=0\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)

\(b.x^2+6x+2x+12=0\Leftrightarrow\left(x^2+6x\right)+\left(2x+12\right)=0\)

\(\Leftrightarrow x\left(x+6\right)+2\left(x+6\right)=0\Leftrightarrow\left(x+2\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-6\end{matrix}\right.\)

\(c.x^2+4x+5x+20=0\Leftrightarrow\left(x^2+4x\right)+\left(5x+20\right)=0\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\Leftrightarrow\left(x+5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(\left(3x-5\right)\left(x+8\right)+8x\left(3x-5\right)=0\)

=>(3x-5)(9x+8)=0

=>x=5/3 hoặc x=-9/8

\(x_1-x_2=\dfrac{5}{3}+\dfrac{9}{8}=\dfrac{40}{24}+\dfrac{27}{24}=\dfrac{67}{24}\)

a: \(=\dfrac{x^3-3x^2-7x+x^2-3x-7}{x^2-3x-7}=x+1\)

b:\(=\dfrac{x^3+x^2+3x^2+3x+5x+5}{x+1}=x^2+3x+5\)

c:\(=\dfrac{x^3-3x^2-7x+2x^2-6x-14}{x^2-3x-7}=x+2\)

d: \(=\dfrac{x^2\left(x+5\right)+5x+25-25}{x+5}=x^2+5-\dfrac{25}{x+5}\)

Ta có 3 x 2 + 8x + 5 = 0

ó 3 x 2 + 3x + 5x + 5 = 0 ó 3x(x + 1) + 5(x + 1) = 0

ó (3x + 5)(x + 1) = 0 

Vậy  x = - 5 3 ;   x = - 1

Đáp án cần chọn là: A

\(a,\dfrac{5}{2x+6}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)};\dfrac{3}{x^2-9}=\dfrac{6}{2\left(x-3\right)\left(x+3\right)}\\ b,\dfrac{2x}{x^2-8x+16}=\dfrac{6x}{3\left(x-4\right)^2};\dfrac{x}{3x^2-12x}=\dfrac{1}{3x-12}=\dfrac{x-4}{3\left(x-4\right)^2}\)

a)\(\dfrac{5}{2x+6}=\dfrac{5}{2\left(x+3\right)}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}=\dfrac{5x-15}{2\left(x+3\right)\left(x-3\right)}\\ \dfrac{3}{x^2-9}=\dfrac{3}{\left(x-3\right)\left(x+3\right)}=\dfrac{6}{2\left(x-3\right)\left(x+3\right)}\)