Chỉ cần thắng là được rồi

khoảng 4-0

a) \(\frac{x^2+2x-3}{x^2+x-6}\)

\(=\frac{x^2+3x-x-3}{x^2-2x+3x-6}\)

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

\(=\frac{\left(x-1\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x-1}{x-2}\)

b) \(\frac{2x^2+2x+2}{x^2+6x-7}\)

\(=\frac{2\left(x^2+x+1\right)}{x^2+7x-x-7}\)

\(cogidosaisai?\)

a ) Xét ◇AHCE có :

D là trung điểm HE

D là trung điểm AC

\(\Rightarrow\)◇AHCE là hình bình hành

Mà góc AHC = 90°

\(\Rightarrow\)◇AHCE là hình chữ nhật

b ) Xét ◇AEIH có :

AI // HE ( giả thiết )

AE // IH ( do I \(\in\)BC và AE // BC )

\(\Rightarrow\)◇AEIH là hình bình hành

a) \(\frac{x^2}{x-1}-\frac{2x}{x-1}+\frac{1}{x-1}\)

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

\(=\frac{\left(x-1\right)^2}{x-1}=x-1\)

b) \(\left(\frac{1}{1-2x}+\frac{1}{1+2x}\right):\frac{1}{1-2x}\)

\(=\left(\frac{1+2x}{\left(1-2x\right)\left(1+2x\right)}+\frac{1-2x}{\left(1+2x\right)\left(1-2x\right)}\right):\frac{1}{1-2x}\)

\(=\frac{2}{\left(1-2x\right)\left(1+2x\right)}.\left(1-2x\right)\)

\(=\frac{2}{1+2x}\)

\(a.=\frac{4x\left(x^2-2x+1\right)}{x^2-1x-5x+5}\)

\(=\frac{4x\left(x-1\right)^2}{x\left(x-1\right)-5\left(x-1\right)}\)

\(=\frac{4x\left(x-1\right)^2}{\left(x-5\right)\left(x-1\right)}\)

\(=\frac{4x\left(x-1\right)}{x-5}\)

b) \(\frac{4x^3-64x}{x^2-7x+12}\)

\(=\frac{4x\left(x^2-16\right)}{x^2-3x-4x+12}\)

\(=\frac{4x\left(x+4\right)\left(x-4\right)}{x\left(x-3\right)-4\left(x-3\right)}\)

\(=\frac{4x\left(x+4\right)\left(x-4\right)}{\left(x-4\right)\left(x-3\right)}\)

\(=\frac{4x\left(x+4\right)}{x-3}=\frac{4x^2+16x}{x-3}\)

c) \(\frac{x^2-6x+8}{x^3-8}\)

\(=\frac{x^2-2x-4x+8}{\left(x-2\right)\left(x^2+2x+4\right)}\)

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

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

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

\(=xy+3y-2x\left(x+3\right)\)

\(=\left(x+3\right)y-2x\left(x+3\right)\)

\(=\left(x+3\right)\left(y-2x\right)\)

a) Ta có: \(2x^2+2x+3=\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.\frac{1}{\sqrt{2}}+\frac{1}{2}+\frac{5}{2}\)

\(=\left(\sqrt{2}x+\frac{1}{\sqrt{2}}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)

\(\Rightarrow S\le\frac{3}{\frac{5}{2}}=\frac{6}{5}\)

Vậy \(S_{max}=\frac{6}{5}\Leftrightarrow\sqrt{2}x+\frac{1}{\sqrt{2}}=0\Leftrightarrow x=-\frac{1}{2}\)

b) Ta có: \(3x^2+4x+15=\left(\sqrt{3}x\right)^2+2.\sqrt{3}x.\frac{2}{\sqrt{3}}+\frac{4}{3}+\frac{41}{3}\)

\(=\left(\sqrt{3}x+\frac{2}{\sqrt{3}}\right)^2+\frac{41}{3}\ge\frac{41}{3}\)

\(\Rightarrow T\le\frac{5}{\frac{41}{3}}=\frac{15}{41}\)

Vậy \(T_{max}=\frac{15}{41}\Leftrightarrow\sqrt{3}x+\frac{2}{\sqrt{3}}=0\Leftrightarrow x=\frac{-2}{3}\)

c) Ta có: \(-x^2+2x-2=-\left(x^2-2x+1\right)-1\)

\(=-\left(x-1\right)^2-1\le-1\)

\(\Rightarrow V\ge\frac{1}{-1}=-1\)

Vậy \(V_{min}=-1\Leftrightarrow x-1=0\Leftrightarrow x=1\)

d) Ta có: \(-4x^2+8x-5=-\left(4x^2-8x+5\right)\)

\(=-\left(4x^2-8x+4\right)-1\)

\(=-\left(2x-2\right)^2-1\le-1\)

\(\Rightarrow X\ge\frac{2}{-1}=-2\)

Vậy \(X_{min}=-2\Leftrightarrow2x-2=0\Leftrightarrow x=1\)