a, ABD đồng dạng ACE (g.g) (có chung góc A và cùng có 1 góc vuông)

b, từ câu a => AD/AB = AE/AC

2 tam giác có chung góc A => đồng dạng  (c.g.c)

\(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}\right)=\left(x-23\right)\left(\frac{1}{26}+\frac{1}{27}\right)\text{ nhận thấy:}\frac{1}{24}+\frac{1}{25}>\frac{1}{26}+\frac{1}{27}\)

\(\Rightarrow x-23=0\Leftrightarrow x=23\)

\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\Rightarrow\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)=\left(\frac{x+3}{2002}+1\right)+\left(\frac{x+4}{2001}+1\right)\)

\(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\text{dạng giống câu a rồi nha}\)

\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\Leftrightarrow300-x=0\)

Vậy: x=300

1) \(\frac{x^2-1}{A}=\frac{x+1}{x^2+y-2}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)}{A}=\frac{x+1}{x^2+y-2}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)}{A}=\frac{\left(x+1\right)\left(x-1\right)}{\left(x^2+y-2\right)\left(x-1\right)}\)

\(\Leftrightarrow A=\left(x^2+y-2\right)\left(x-1\right)\)

\(\Leftrightarrow A=x^3+xy-x^2-y-2x+2\)

Vậy A = x³ + xy - x² - y - 2x + 2

2) \(\frac{x^3+8}{x^2-xy+2x-2y}=\frac{x^2-2x+4}{A}\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x-y\right)\left(x+2\right)}=\frac{x^2-2x+4}{A}\)

\(\Leftrightarrow\frac{x^2-2x+4}{x-y}=\frac{x^2-2x+4}{A}\)

\(\Leftrightarrow A=x-y\)

Vậy A = x - y

Tìm 2 số x và  y hay tìm hiệu x - y ?

\(\left|3x-2y\right|=20x-50-2x^2\)

\(\Leftrightarrow\left|3x-2y\right|=-2\left(x^2-10x+25\right)\)

\(\Leftrightarrow\left|3x-2y\right|=-2\left(x-5\right)^2\)

Ta thấy : \(\left|3x-2y\right|\ge0\)

               \(-2\left(x-5\right)^2\le0\)

Dấu " = " xảy ra khi và chỉ khi :

\(\Leftrightarrow\hept{\begin{cases}3x-2y=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=5\\y=7,5\end{cases}}\)

\(\Leftrightarrow x-y=5-7,5=-2,5\)

Vậy x - y = -2,5

Gọi O là giao điểm của AD và FC CMR. S OCD = 1/4 S OCA 

1) Với x = 0; y = 2

A = (x^2 - y^2 + x + y)/(2 + x^2 + y^2)

A  = (0^2 - 2^2 + 0 + 2)/(2 + 0^2 + 2^2)

A = -1/3

3) Với x = 2; y = -2

A = (x^2 - y^2 + x + y)/(2 + x^2 + y^2)

A = [2^2 - (-2)^2 + 2 + (-2)]/[2 + 2^2 + (-2)^2]

A = 0