Sửa đề: \(x^3-a^3b^3=x^3-\left(ab\right)^3\\ =\left(x^3-ab\right)\left(x^2+abx+a^2b^2\right)\)

Sửa đề: \(\left(x-1\right)^3-\left(x-2\right)\left(x^2+2x+4\right)+3x^2-3x\\ =x^3-3x^2+3x-1-\left(x^3-2^3\right)+3x^2-3x\\ =x^3-3x^2+3x-1-x^3+8+3x^2-3x\\ =7\) (k phụ thuộc vào biến x)

Bạn xem lại đề nhé.

\(A\rightarrow-\dfrac{3}{4}\\ B\rightarrow-\dfrac{1}{2}\)

a) \(\left|x-2011\right|=x-2012\left(ĐK:x-2012\ge0\Rightarrow x\ge2012\right)\\ \Rightarrow\left[{}\begin{matrix}x-2011=x-2012\\x-2011=-\left(x-2012\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-x=2011-2012\\x-2011=-x+2012\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0=-1\left(Vô\right)lí\\x+x=2011+2012\end{matrix}\right.\\ \Rightarrow2x=4023\\ \Rightarrow x=\dfrac{4023}{2}\left(KTMDK\right)\)

Vậy không có giá trị x thỏa mãn yc đề bài

Khi thêm vào số bé bao nhiêu đơn vị thì hiệu giảm đi bấy nhiêu đơn vị

Hiệu 2 số ban đầu là:

       155 + 132 = 287

Số bé là:

   (1205 - 287) : 2 = 459

Số lớn là:

   1205 - 459 = 746

   Đáp số: 459 và 746

c) \(\dfrac{x-ab}{a+b}+\dfrac{x-bc}{b+c}+\dfrac{x-ac}{a+c}>a+b+c\\ \Rightarrow\left(\dfrac{x-ab}{a+b}-c\right)+\left(\dfrac{x-bc}{b+c}-a\right)+\left(\dfrac{x-ac}{a+c}-b\right)>0\\ \Leftrightarrow\dfrac{x-ab-ac-bc}{a+b}+\dfrac{x-bc-ab-ac}{b+c}+\dfrac{x-ac-ab-bc}{a+c}>0\\ \Leftrightarrow\left(x-ab-bc-ca\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)>0\left(c\right)\\ \)

Nhận thấy: \(a,b,c>0\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}>0\)

\(\left(c\right)\Rightarrow x-ab-bc-ca>0\\ \Leftrightarrow x>ab+bc+ca\)

b) \(\dfrac{x-2}{2002}+\dfrac{x-4}{2000}< \dfrac{x-3}{2001}+\dfrac{x-5}{1999}\\ \Rightarrow\left(\dfrac{x-2}{2002}-1\right)+\left(\dfrac{x-4}{2000}-1\right)< \left(\dfrac{x-3}{2001}-1\right)+\left(\dfrac{x-5}{1999}-1\right)\\ \Rightarrow\dfrac{x-2004}{2002}+\dfrac{x-2004}{2000}< \dfrac{x-2004}{2001}+\dfrac{x-2004}{1999}\\ \Rightarrow\dfrac{x-2004}{2002}+\dfrac{x-2004}{2000}-\dfrac{x-2004}{2001}-\dfrac{x-2004}{1999}< 0\\ \)

\(\Rightarrow\left(x-2004\right)\left(\dfrac{1}{2002}+\dfrac{1}{2000}-\dfrac{1}{2001}-\dfrac{1}{1999}\right)< 0\left(b\right)\)

Nhận thấy: \(\dfrac{1}{2002}< \dfrac{1}{2001},\dfrac{1}{2000}< \dfrac{1}{1999}\Rightarrow\dfrac{1}{2002}-\dfrac{1}{2001}< 0,\dfrac{1}{2000}-\dfrac{1}{1999}< 0\\ \Rightarrow\dfrac{1}{2002}+\dfrac{1}{2000}-\dfrac{1}{2001}-\dfrac{1}{1999}< 0\)

\(\left(b\right)\Rightarrow x-2004>0\Leftrightarrow x>2004\)

a) \(\dfrac{x+2004}{x+2005}+\dfrac{x+2005}{2006}< \dfrac{x+2006}{2007}+\dfrac{x+2007}{2008}\\ \Rightarrow\left(\dfrac{x+2004}{2005}-1\right)+\left(\dfrac{x+2005}{2006}-1\right)< \left(\dfrac{x+2006}{2007}-1\right)+\left(\dfrac{x+2007}{2008}-1\right)\\ \Rightarrow\dfrac{x-1}{2005}+\dfrac{x-1}{2006}< \dfrac{x-1}{2007}+\dfrac{x-1}{2008}\\ \Rightarrow\dfrac{x-1}{2005}+\dfrac{x-1}{2006}-\dfrac{x-1}{2007}-\dfrac{x-1}{2008}< 0\\ \)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{2005}+\dfrac{1}{2006}-\dfrac{1}{2007}-\dfrac{1}{2008}\right)< 0\left(a\right)\)

Nhận thấy: \(\dfrac{1}{2005}>\dfrac{1}{2007},\dfrac{1}{2006}>\dfrac{1}{2008}\\ \Rightarrow\dfrac{1}{2005}-\dfrac{1}{2007}>0,\dfrac{1}{2006}-\dfrac{1}{2008}>0\\ \Rightarrow\dfrac{1}{2005}+\dfrac{1}{2006}-\dfrac{1}{2007}-\dfrac{1}{2008}>0\)

\(\left(a\right)\Rightarrow x-1< 0\Leftrightarrow x< 1\)

Vậy \(S=\left\{x|x< 1\right\}\)

\(n^2+n-7=\left(n^2-2n\right)+\left(3n-6\right)-1\\ =n\left(n-2\right)+3\left(n-2\right)-1\\ =\left(n-2\right)\left(n+3\right)-1\)

Để: \(\left(n^2+n-7\right)⋮\left(n-2\right)\Rightarrow\left[\left(n-2\right)\left(n+3\right)-1\right]⋮\left(n-2\right)\\ \Rightarrow1⋮\left(n-2\right)\) (Vì: \(\left(n-2\right)\left(n+3\right)⋮\left(n-2\right)\forall n\inℤ\) )

\(\Rightarrow n-2\in\left\{1;-1\right\}\Rightarrow n\in\left\{3;1\right\}\)