a: Xét tứ giác AEMF có 

\(\widehat{MEA}=\widehat{MFA}=\widehat{FME}=90^0\)

Do đó: AEMF là hình chữ nhật

a)Tứ giác AEMF có :

\(\widehat{MEA}=\widehat{MFA}=\widehat{FME}=90^0\)

=>AEMF là hình chữ nhật

\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\Leftrightarrow x^2-4x+4-x^2+9=6\)

\(\Leftrightarrow-4x=-7\)

\(\Leftrightarrow x=\dfrac{7}{4}\)

\(b,9x^2-4-\left(3x-2\right)\left(4x-5\right)=0\)

\(\Leftrightarrow9x^2-4-12x^2+23x-10=0\)

\(\Leftrightarrow-3x^2+23x-14=0\)

\(\Leftrightarrow-3x^2+21x+2x-14=0\)

\(\Leftrightarrow-3x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(2-3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,4x^2\left(x-1\right)-x+1=0\)

\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(4x^2-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x=-1\\x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\) \(d,x^2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow x^2\left(x+3\right)-x\left(x+3\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-3\end{matrix}\right.\)

Bài 2:

\(\frac{n^3-3n^2-3n-1}{n^2+n+1}\)

\(=\frac{n^3+n^2+n-4n^2-4n-4+3}{n^2+n+1}=n-4+\frac{3}{n^2+n+1}\)

Để \(n^3-3n^2-3n-1\) ⋮\(n^2+n+1\) thì 3⋮\(n^2+n+1\)

=>\(n^2+n+1\in\left\lbrace1;3\right\rbrace\)

TH1: \(n^2+n+1=1\)

=>\(n^2+n=0\)

=>n(n+1)=0

=>n=0 hoặc n=-1

TH2: \(n^2+n+1=3\)

=>\(n^2+n-2=0\)

=>(n+2)(n-1)=0

=>n=-2 hoặc n=1

Bài 2:

a: \(f\left(x\right)=x^3-9x^2+27x-27\)

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

\(=x^2\left(x-3\right)-6x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3\)

b: \(g\left(x\right)=x^2-6x+9\)

\(=x^2-2\cdot x\cdot3+3^2\)

\(=\left(x-3\right)^2\)

=>f(x)⋮g(x)

=>f(x) chia g(x) thì dư 0

\(\frac{f\left(x\right)}{k\left(x\right)}=\frac{x^3-9x^2+27x-27}{x^2-6x+10}\)

\(=\frac{x^3-6x^2+10x-3x^2+18x-30-x+3}{x^2-6x+10}=x-3+\frac{-x+3}{x^2-6x+10}\)

=>f(x) chia k(x) dư -x+3

mình là minh chúng ta kết bạn đi !

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tui có thi hsg toán nhưng thấy vẫn hơi ngu 

\(a,a^2+b^2=\left(a+b\right)^2-2ab=3^2-2\left(-10\right)=29\\ b,a^2+b^2=\left(a-b\right)^2+2ab=2^2+2\cdot24=52\)

\(1;\)Từ \(\left(a+b\right)=-7\Rightarrow\left(a+b\right)^3=-343\)

\(\Rightarrow a^3+3a^2b+3ab^2+b^3=-343\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-343\)

\(\Rightarrow a^3+b^3=-343-3.6.\left(-7\right)=-217\)

\(x^2+y^2=\left(x+y\right)^2-2xy=7^2-2.10=29\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=7^3-3.10.7=133\)

\(P=\left(x+y\right)\left(x^2+y^2\right)\left(x^3+y^3\right)\)

\(=7.29.133=26999\)

ĐKXĐ: \(\begin{cases}x<>0\\ x<>-1\\ -x^2+3x+1<>0\end{cases}\)

=>\(\begin{cases}x\notin\left\lbrace0;-1\right\rbrace\\ x^2-3x-1<>0\end{cases}\Rightarrow\begin{cases}x\notin\left\lbrace0;-1\right\rbrace\\ x^2-3x+\frac94<>\frac{13}{4}\end{cases}\)

=>\(\begin{cases}x\notin\left\lbrace0;-1\right\rbrace\\ \left(x-\frac32\right)^2<>\frac{13}{4}\end{cases}\Rightarrow\begin{cases}x\notin\left\lbrace0;-1\right\rbrace\\ x-\frac32<>\pm\frac{\sqrt{13}}{2}\end{cases}\)

=>\(x\notin\left\lbrace0;-1;\frac{3-\sqrt{13}}{2};\frac{3+\sqrt{13}}{2}\right\rbrace\)

a: \(D=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x-x^2+1}{3x}\)

\(=\frac{\left(x+2\right)\left(x+1\right)+6x-3\cdot3x\cdot\left(x+1\right)}{3x\left(x+1\right)}\cdot\frac{x+1}{2-4x}+\frac{x^2-3x-1}{3x}\)

\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(2-4x\right)}+\frac{x^2-3x-1}{3x}\)

\(=\frac{-8x^2+2}{3x\left(-4x+2\right)}+\frac{x^2-3x-1}{3x}=\frac{2\left(4x^2-1\right)}{3x\cdot2\cdot\left(2x-1\right)}+\frac{x^2-3x-1}{3x}\)

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

b: Khi x=2010 thì \(D=\frac{2010-1}{3}=\frac{2009}{3}\)

c: D<0

=>x-1<0

=>x<1

=>x<1 và x∉{0;-1;\(\frac{3-\sqrt{13}}{2}\) }