bài 1:

a. \((x+1)(x+3) - x(x+2)=7 \)

    \(x^2+ 3x +x +3 - x^2 -2x =7\)

    \(x^2+4x+3-x^2-2x=7\)

\(=> 2x+3=7\)

    \(2x=4\)

    \(x = 2\)

Bài 2:

a)

\((3x-5)(2x+11) -(2x+3)(3x+7) \)

\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)

\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)

\(=-65\)

\(\)

\(A=x^2+2y^2+2xy+2x-4y+2016\)

\(=\left(x^2+2xy+2x+y^2+2y+1\right)+\left(y^2-6y+9\right)+2006\)

\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2006\ge2006\)

\(\Rightarrow A\ge2006\)

Dấu = khi \(\begin{cases}\left(x+y+1\right)^2=0\\\left(y-3\right)^2=0\end{cases}\)\(\Rightarrow\begin{cases}x+y+1=0\\y-3=0\end{cases}\)

\(\Rightarrow\begin{cases}x+y+1=0\\y=3\end{cases}\)\(\Rightarrow\begin{cases}x+3+1=0\\y=3\end{cases}\)\(\Rightarrow\begin{cases}x=-4\\y=3\end{cases}\)

Vậy Min_A=2006 khi \(\begin{cases}x=-4\\y=3\end{cases}\)

a: \(A=\frac{x^2+2x}{x^2-4x+4}:\left(\frac{x+2}{x}-\frac{1}{2-x}+\frac{6-x^2}{x^2-2x}\right)\)

\(=\frac{x\left(x+2\right)}{\left(x-2\right)^2}:\frac{\left(x+2\right)\left(x-2\right)+x+6-x^2}{x\left(x-2\right)}\)

\(=\frac{x\left(x+2\right)}{\left(x-2\right)^2}\cdot\frac{x\left(x-2\right)}{x^2-4+x+6-x^2}=\frac{x^2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x^2}{x-2}\)

b: |2x+1|=3

=>2x+1=3 hoặc 2x+1=-3

=>2x=2 hoặc 2x=-4

=>x=1(nhận) hoặc x=-2(loại)

Thay x=1 vào A, ta được:

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

c: A<0

=>\(\frac{x^2}{x-2}<0\)

=>x-2<0

=>x<2

Kết hợp ĐKXĐ, ta được: x<2 và x∉{0;-2}

d: Để A nguyên thì \(x^2\vdots x-2\)

=>\(x^2-4+4\vdots x-2\)

=>4⋮x-2

=>x-2∈{1;-1;2;-2;4;-4}

=>x∈{3;1;4;0;6;-2}

Kết hợp ĐKXĐ, ta được: x∈{3;1;4;6}

Violympic toán 8

a: \(=2\left(x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}\right)\)

\(=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}>=-\dfrac{9}{2}\)

Dấu '=' xảy ra khi x=3/2

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

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

Dấu '=' xảy ra khi x=1/2

\(H=x^2+2xy+y^2+2x+2y+x^2+4x+2019=\left(x+y\right)^2+2\left(x+y\right)+\left(x+2\right)^2+2015\)

\(=\left(x+y+1\right)^2+\left(x+2\right)^2+2014\ge2014\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=-2;y=1\)

\(I=\left(1-x\right)^2+\left(-2-y\right)^2+\left(x+y\right)^2\ge\frac{\left(1-x-2-y+x+y\right)^2}{3}=\frac{1}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(1-x=-2-y=x+y\)\(\Leftrightarrow\)\(x=\frac{4}{3};y=\frac{-5}{3}\)