Bài 3: Rút gọn phân thức

mình không biết kq =mấy

nhứng mình c/m kq =2 là sai

\(A-2=\dfrac{4024.2014-2}{Khongquantam}-2=\dfrac{4024.2014-2-2.2011-2.2012.2010}{Khongquantam}\)

\(A-2=\dfrac{2\left(2012.2014-2011-2012.2010-1\right)}{Khongquantam}=\dfrac{2\left[2012.\left(2014-2010\right)-2011-1\right]}{Khongquantam}\)

\(A-2=\dfrac{2\left[4.2012-2011-1\right]}{Khongquantam}=\dfrac{2\left[3.2011+3\right]}{Khongquantam}\)

\(A-2=\dfrac{2\left[3.\left(2011+1\right)\right]}{Khongquantam}=\dfrac{2.3.2012}{Khongquantam}\ne0\)\(A-2\ne0\)

\(\Rightarrow A\ne2\Rightarrow kq=2=sai\)

a,\(5+3x-\left(2+x\right)< 5x+1\)

\(\Rightarrow5+3x-2-x-5x-1< 0\)

\(\Rightarrow-3x< -5+2+1\Rightarrow x< \dfrac{2}{3}\)

b, \(\left(x-2\right)\left(x+1\right)-x\left(x+4\right)\le x\)

\(\Rightarrow x^2+x-2x-2-\left(x^2+4x\right)-x\le0\)

\(\Rightarrow x^2-x-2-x^2-4x-x\le0\)

\(\Rightarrow-6x\le2\Rightarrow x\le-\dfrac{1}{3}\)

Chúc bạn học tốt!!!

Bài 2: 

Gọi số học sinh nam là x

Số học sinh nữ là 160-x

Theo đề,ta có: 3x+2(160-x)=391

=>x+320=391

hay x=71

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

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

\(=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{1}{x+2}=\dfrac{1}{\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}}=\dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)Ta có:

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\) \(\Rightarrow\dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)

Vậy Max Q = \(\dfrac{4}{3}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

\(\dfrac{2004a}{ab+2004a+2004}+\dfrac{b}{bc+b+2004}+\dfrac{c}{ac+c+1}\)\(=\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)\(=\dfrac{a^2bc}{ab\left(1+ac+c\right)}+\dfrac{b}{b\left(c+1+ac\right)}+\dfrac{c}{ac+c+1}\)\(=\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{ac+c+1}{ac+c+1}=1\)

=> đpcm

2x² + 2xy + y² - 2x + 2y + 2

= x² + x² +2xy + y² - 2x + 2y + 1 + 1

= (x + y)² + (x - 1)² + 2y + 1

Có: \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\forall x,y\\\left(x-1\right)^2\ge0\forall x\end{matrix}\right.\)

(x - 1)² nhỏ nhất khi x = 1

<=> ( x + y)² nhỏ nhất khi x = 1; y = -1

Thế vào D ta có: 0 + 0 + 2. (-1) + 1 = -1

Vậy D_min = -1 khi x = 1 ; y = -1

\(\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)

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

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

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

\(=\dfrac{\left(x-1\right)\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=x-1\)

b) \(C=0\Rightarrow x-1=0\Rightarrow x=1\)

c) C luôn dương => \(C\ge0\Rightarrow x-1\ge0\Rightarrow x\ge1\)