a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)

b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)

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

a: A>0

=>\(x^2-3x>0\)

=>x(x-3)>0

TH1: \(\left\{{}\begin{matrix}x>0\\x-3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>0\\x>3\end{matrix}\right.\)

=>x>3

TH2: \(\left\{{}\begin{matrix}x< 0\\x-3< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 0\\x< 3\end{matrix}\right.\)

=>x<0

d: Để D<0 thì \(x^2+\dfrac{5}{2}x< 0\)

=>\(x\left(x+\dfrac{5}{2}\right)< 0\)

TH1: \(\left\{{}\begin{matrix}x>0\\x+\dfrac{5}{2}< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>0\\x< -\dfrac{5}{2}\end{matrix}\right.\)

=>Loại

Th2: \(\left\{{}\begin{matrix}x< 0\\x+\dfrac{5}{2}>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 0\\x>-\dfrac{5}{2}\end{matrix}\right.\)

=>\(-\dfrac{5}{2}< x< 0\)

e: ĐKXĐ: x<>2

Để E<0 thì \(\dfrac{x-3}{x-2}< 0\)

TH1: \(\left\{{}\begin{matrix}x-3>=0\\x-2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=3\\x< 2\end{matrix}\right.\)

=>Loại

TH2: \(\left\{{}\begin{matrix}x-3< =0\\x-2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =3\\x>2\end{matrix}\right.\)

=>2<x<=3

g: Để G<0 thì \(\left(2x-1\right)\left(3-2x\right)< 0\)

=>\(\left(2x-1\right)\left(2x-3\right)>0\)

TH1: \(\left\{{}\begin{matrix}2x-1>0\\2x-3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x>\dfrac{3}{2}\end{matrix}\right.\)

=>\(x>\dfrac{3}{2}\)

TH2: \(\left\{{}\begin{matrix}2x-1< 0\\2x-3< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x< \dfrac{3}{2}\end{matrix}\right.\)

=>\(x< \dfrac{1}{2}\)

a: ĐKXĐ: x<>1; x<>-1

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

\(=\dfrac{x+1}{x-1}-\dfrac{1}{x-1}=\dfrac{x}{x-1}\)

b: x^2+3x+2=0

=>x=-1(loại) hoặc x=-2(nhận)

Khi x=-2 thì A=-2/(-3)=2/3

\(A=\left(\frac{2X-1}{x^2-4}+\frac{x+2}{x^2-x-2}\right):\frac{x-2}{x^2+3x+2}ĐK:x\ne\left\{2,-2,-1\right\}\)

a)  \(A=\left[\frac{\left(2x-1\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x+1\right)\left(x-2\right)}\right]:\frac{x-2}{\left(x+2\right)\left(x+1\right)}\)

\(A=\left[\frac{\left(2x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}\frac{\left(x+2\right)\left(x+2\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\right].\frac{\left(x+2\right)\left(x+1\right)}{x-2}\)

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

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

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

Ta có :\(3x^2+5x+3\)

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

\(=3\left[x^2+2.\frac{5}{6}x+\frac{25}{36}+\frac{9}{36}\right]\)

\(=3\left[\left(x+\frac{5}{6}\right)^2+\frac{9}{36}\right]>0\)

Mà \(\left(x-2\right)^2>0\)

\(\Rightarrow A>0\left(dpcm\right)\)

\(b,A=11\Leftrightarrow\frac{3x^2+5x+3}{\left(x-2\right)^2}=11\)

\(\Rightarrow3x^2+5x+3=11.\left(x-2\right)^2\)

\(\Rightarrow3x^2+5x+3=11.\left(x^2-4x+4\right)\)

\(\Rightarrow8x^2-49x+41=0\)

\(\Rightarrow8x^2-8x-41x+41=0\)

\(\Rightarrow8x\left(x-1\right)-41\left(x-1\right)=0\)

\(\Rightarrow\left(8x-41\right)\left(x-1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}8x-41=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{41}{8}\\x=1\end{cases}}}\)(Thỏa mãn)

5.\(C\text{ó}x^2-12=0\Rightarrow x^2=12\Rightarrow x=\sqrt{12}ho\text{ặc}x=-\sqrt{12}\)

Mà x>0\(\Rightarrow x=\sqrt{12}\)

6.Vì x-y=4\(\Rightarrow\left(x-y\right)^2=x^2-2xy+y^2=x^2-10+y^2=4^2=16\Rightarrow x^2+y^2=26\)

Có \(\left(x+y\right)^2=x^2+2xy+y^2=26+10=36=6^2=\left(-6\right)^2\)

Vì xy>0 và x>0 =>y>0=>x+y>0=>x+y=6

7. \(3x^2+7=\left(x+2\right)\left(3x+1\right)\)

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

\(3x^2+7-3x^2-7x-2=0\)

-7x+5=0

-7x=-5

\(x=\frac{5}{7}\)

8.\(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\left(2x+1\right)^2-\left(2x+4\right)^2=9\)

(2x+1-2x-4)(2x+1+2x+4)=9

-3(4x+5)=9

4x+5=-3

4x=-8

x=-2

Còn câu 9 và 10 để mình nghiên cứu đã

biet x+y =2 tinh min 3x^2 + y^2

Ta có: \(\hept{\begin{cases}\left(2x-y-4\right)^{2016}\ge0\\\left(3x+2y-13\right)^{2016}\ge0\end{cases}}\)

\(\Rightarrow\left(2x-y-4\right)^{2016}+\left(3x+2y-13\right)^{2016}\ge0\)

Dấu bằng xảy ra khi

\(\hept{\begin{cases}2x-y-4=0\\3x+2y-13=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}\)

\(\Rightarrow A=\left(x-y\right)^{2016}+2016=\left(3-2\right)^{2016}+2016=2017\)

\(2\left(3x-5\right)-4\left(2+3\left(x-1\right)\right)=3\left(x-5\right)\)

           \(6x-10-4\left(2+3x-3\right)=3x-15\)

              \(6x-10-8-12x+12=3x-15\)

                                  \(6x-12x-3x=-15+10+8-12\)

                                                      \(-9x=-9\)

                                                            \(x=1\)