Ban vao trang Đề thi HSG Toán 8 cấp huyện năm 2016-2017 Phòng GD&ĐT Củ Chi

a) \(x^2+5y^2+2xy-4x-8y+2015\)

\(=x^2+2xy+y^2+4y^2-4x-8y+2015\)

\(=\left(x+y\right)^2-4\left(x+y\right)+4+4y^2-4y+2011\)

\(=\left(x+y\right)^2-2\cdot\left(x+y\right)\cdot2+2^2+\left(2y\right)^2-2\cdot2y\cdot1+1^2+2010\)

\(=\left(x+y-2\right)^2+\left(2y-1\right)^2+2010\ge2010\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y-2=0\\2y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{1}{2}\end{cases}}\)

Vậy.....

b) \(\frac{3\left(x+1\right)}{x^3+x^2+x+1}\)

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

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

\(=\frac{3}{x^2+1}\le\frac{3}{1}=3\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy....

\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)

\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)

\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)

\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)

\(\text{Giải}\)

\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)

a, A xác định

\(\Leftrightarrow3x^3-19x^2+33x-9\ne0\)

\(\Leftrightarrow3x^3-x^2-18x^2+6x+27x-9\ne0\)

\(\Leftrightarrow x^2\left(3x-1\right)-6x\left(3x-1\right)+9\left(3x-1\right)\ne0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)^2\ne0\Leftrightarrow\hept{\begin{cases}x\ne\frac{1}{3}\\x\ne3\end{cases}}\)

b, \(\frac{3x^3-14x^2+3x+36}{3x^2-19x^2+33x-9}=\frac{3x^2\left(x-3\right)-5x\left(x-3\right)-12\left(x-3\right)}{\left(3x-1\right)\left(x-3\right)^2}\)

\(=\frac{\left(3x^2-5x-12\right)\left(x-3\right)}{\left(3x-1\right)\left(x-3\right)^2}=\frac{\left(3x+4\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}=\frac{3x+4}{3x-1}\)

\(A=0\Leftrightarrow\frac{3x+4}{3x-1}=0\Leftrightarrow3x+4=0\Leftrightarrow x=-\frac{4}{3}\) (thỏa mãn ĐKXĐ)

c, \(A=\frac{3x+4}{3x-1}=1+\frac{5}{3x-1}\in Z\Rightarrow5⋮\left(3x-1\right)\)

\(\Rightarrow3x-1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(\Rightarrow x\in\left\{-\frac{4}{3};0;\frac{2}{3};2\right\}\)

Mà \(x\in Z,x\ne\left\{\frac{1}{3};3\right\}\Rightarrow x\in\left\{0;2\right\}\)

Bài của Hùng rất thông minh

Đang định có cách khác mà dài hơn cách Hùng nên thui

^^ 2k5 kết bạn nhé 

bạn đặt mỗi biểu thức = 1 số bình phương ví dụ là x^2

ở câu a bn đặt xong nhân 4 lên sau đó biến đổi về 1 hằng đẳng thức 

câu b thì đưa chữ sang 1 vế số sang 1 vế

câu c làm tương tự câu a

\(\text{Giải}\)

\(A=\left(\frac{x+2}{2x-4}-\frac{2-x}{2x+4}+\frac{32}{4x^2-16}\right):\frac{x-1}{x-2}\)

\(A=\left(\frac{x+2}{2x-4}-\frac{2-x}{2x+4}+\frac{32}{\left(2x-4\right)\left(2x+4\right)}\right):\frac{x-1}{x-2}\)

\(A=\left(\frac{\left(x+2\right)\left(2x+4\right)}{\left(2x-4\right)\left(2x+4\right)}-\frac{\left(2-x\right)\left(2x-4\right)}{\left(2x-4\right)\left(2x+4\right)}+\frac{32}{\left(2x-4\right)\left(2x+4\right)}\right):\frac{x-1}{x-2}\)

\(A=\left(\frac{2x^2+8x+8}{\left(2x-4\right)\left(2x+4\right)}-\frac{4x^2-8+4x}{\left(2x-4\right)\left(2x+4\right)}+\frac{32}{\left(2x-4\right)\left(2x+4\right)}\right):\frac{x-1}{x-2}\)

\(A=\frac{2x^2+8x+8-4x^2+8-4x+32}{\left(2x-4\right)\left(2x+4\right)}:\frac{x-1}{x-2}\)

\(A=\frac{4x-2x^2+48}{\left(2x-4\right)\left(2x+4\right)}:\frac{x-1}{x-2}\)

\(A=\frac{2\left(2x-x^2+24\right)}{\left(2x-4\right)\left(2x+4\right)}:\frac{x-1}{x-2}=\frac{2\left(2x-x^2+24\right)\left(x-2\right)}{\left(2x-4\right)\left(2x+4\right)\left(x-1\right)}\)

\(=\frac{2\left(2x-x^2+24\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)\left(x-1\right)}=\frac{2x-x^2+24}{\left(x-2\right)\left(x-1\right)}\)

c, Bạn tự giải hệ pt nhé :)