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x^2 - x - y^2 - y
= x^2 - y^2 - x - y
= ( x - y ) ( x + y ) - ( x + y )
= ( x + y ) ( x - y - 1 )
x^2 - 2xy + y^2 - z^2
= ( x- y ) ^2 - z^2
= ( x - y - z ) ( x - y + z )

Xét tứ giác ABEC có
AB//EC
AC//BE
Do đó: ABEC là hình bình hành
Suy ra: AC=BE
mà AC=BD
nên BE=BD
hay ΔBED cân tại B

a) x\(^2\) - 10x + 9 =0
x\(^2\) - 2x . 5 + 25 = 16
(x - 5)\(^2\) = 4\(^2\)
=> x - 5 = 4
x = 9
Vậy x = 9
b) x\(^2\) - 7x + 6 = 0
x\(^2\) - 2x . 3,5 + 12,25 = 6,25
(x - 3,5)\(^2\) = 2,5\(^2\)
=> x - 3,5 = 2,5
x = 6
Vậy x = 6
c) x\(^2\) + 13x + 12 = 0
x\(^2\) + 2x . 6,5 + 42,25 = 30,25
(x + 6,5)\(^2\) = 5,5\(^2\)
=> x + 6,5 = 5,5
x = -1
Vậy x = -1
d) x\(^2\) - 24x + 23 = 0
x\(^2\) - 2x . 12 + 244 = 121
(x - 12)\(^2\) = 11\(^2\)
=> x - 12 = 11
x = 23
Vậy x = 23
e) 3x\(^2\) + 14x + 8 = 0
3x\(^2\) + 2 . \(\sqrt{3}\)x . \(\frac{7}{\sqrt{3}}\) + \(\frac{49}{3}\) = \(\frac{25}{3}\)
(\(\sqrt{3}\)x + \(\frac{7}{\sqrt{3}}\))\(^2\) = \(\left(\frac{5}{\sqrt{3}}\right)^2\)
=> \(\sqrt{3}\)x + \(\frac{7}{\sqrt{3}}\) = \(\frac{5}{\sqrt{3}}\)
=> \(\sqrt{3}\)x = \(\frac{-2}{\sqrt{3}}\)
=> x = \(\frac{-2}{3}\)

18, \(\frac{x}{2}+\frac{x^2}{8}=0\Leftrightarrow4x+x^2=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow x=-4;x=0\)
19, \(4-x=2\left(x-4\right)^2\Leftrightarrow\left(4-x\right)-2\left(4-x\right)^2=0\)
\(\Leftrightarrow\left(4-x\right)\left[1-2\left(4-x\right)\right]=0\Leftrightarrow\left(4-x\right)\left(-7+2x\right)=0\Leftrightarrow x=4;x=\frac{7}{2}\)
20, \(\left(x^2+1\right)\left(x-2\right)+2x-4=0\Leftrightarrow\left(x^2+1\right)\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3>0\right)=0\Leftrightarrow x=2\)
21, \(x^4-16x^2=0\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\Leftrightarrow x=0;x=\pm4\)
22, \(\left(x-5\right)^3-x+5=0\Leftrightarrow\left(x-5\right)^3-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\Leftrightarrow\left(x-5\right)\left(x-6\right)\left(x-4\right)=0\Leftrightarrow x=4;x=5;x=6\)
23, \(5\left(x-2\right)-x^2+4=0\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\Leftrightarrow x=2;x=3\)

ta có : CK vuông góc DB (1)
AH vuông góc DB (2)
từ (1),(2) suy ra AH//CK (*)
xét tam giác vuông AHD và tam giác vuông CBK:ta có
góc H=góc K=90
góc ADH=góc CBK(slt)
suy ra 2 tam giác đó bằng nhau
suy ra AH=CK (*')
từ (*),(*') ta có tứ giác AHCK là hình bình hình

Câu 4:
a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)
b: \(A=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2}{2x\left(x+5\right)}+\dfrac{2\left(x^2-25\right)}{2x\left(x+5\right)}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
c: Để A=-3 thì x-1=-6
hay x=-5(loại)

\(1)15-3x=0.\\ \Leftrightarrow3x=15.\\ \Leftrightarrow x=5.\)
\(2)4x+12=2x\left(x+3\right).\\ \Leftrightarrow4\left(x+3\right)-2x\left(x+3\right)=0.\\ \Leftrightarrow\left(4-2x\right)\left(x+3\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}4-2x=0.\\x+3=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=-3.\end{matrix}\right.\)
\(3)\dfrac{1}{x-1}+\dfrac{2x^2-5}{x^3-1}=\dfrac{4}{x^2+x+1}.\left(x\ne1\right).\\ \Leftrightarrow\dfrac{x^2+x+1+2x^2-5-4x+4}{\left(x-1\right)\left(x^2+x+1\right)}=0.\\ \Rightarrow3x^2-3x=0.\\ \Leftrightarrow3x\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right).\\x=1\left(koTM\right).\end{matrix}\right.\)
1A
2B =)