a: Để C vô nghĩa thì x+2=0

hay x=-2

Để C có nghĩa thì x+2<>0

hay x<>-2

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

Để C=0 thì \(x\in\varnothing\)

Để C>0 thì x+2>0

hay x>-2

Để C<0 thì x+2<0

hay x<-2

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

\(a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}\)

\(b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}\)

\(c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}\)

\(d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

\(e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)

\(h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)\)

Đặt \(x+\dfrac{1}{x}=t\), pt trở thành:

\(t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\)

Tick plzz

a: Ta có: \(\left(x-2\right)\left(x+3\right)-3\left(4x-2\right)=\left(x-4\right)^2\)

\(\Leftrightarrow x^2+x-6-12x+6-x^2+8x-16=0\)

\(\Leftrightarrow-3x=16\)

hay \(x=-\dfrac{16}{3}\)

b: Ta có: \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)

\(\Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\)

\(\Leftrightarrow-14x+7+4x-6=0\)

\(\Leftrightarrow10x=1\)

hay \(x=\dfrac{1}{10}\)

c: Ta có: \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)

\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)

\(\Leftrightarrow23x+70=10x+200\)

\(\Leftrightarrow x=10\)

a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)

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

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

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

c: Thay x=2017 vào C, ta được:

\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)

a)Tử: \(x^5-2x^4+2x^3-4x^2-3x+6\)

\(=x^5+2x^3-3x-2x^4-4x^2+6\)

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

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

\(=\left(x-2\right)\left[x^4-x^2+3x^2-3\right]\)

\(=\left(x-2\right)\left[x^2\left(x^2-1\right)+3\left(x^2-1\right)\right]\)

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

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

Mẫu: \(x^2+2x-8=x^2-2x+4x-8\)

\(=x\left(x-2\right)+4\left(x-2\right)\)

\(=\left(x-2\right)\left(x+4\right)\)

Suy ra \(A=\dfrac{\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x^2+3\right)}{\left(x-2\right)\left(x+4\right)}=\dfrac{\left(x-1\right)\left(x+1\right)\left(x^2+3\right)}{x+4}\)

b)\(A=0\Rightarrow\dfrac{\left(x-1\right)\left(x+1\right)\left(x^2+3\right)}{x+4}=0\)

\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(x^2+3\right)=0\)

Dễ thấy: \(x^2+3\ge3>0\forall x\) (vô nghiệm)

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

A có nghĩa khi \(x+4\ne0\Rightarrow x\ne-4\)

A vô nghĩa khi \(x+4=0\Rightarrow x=-4\)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

\(=\left[\dfrac{2x-3}{\left(2x-5\right)\left(2x-1\right)}-\dfrac{3}{2x-1}-\dfrac{2\left(x-4\right)}{\left(x-4\right)\left(2x-5\right)}\right].\dfrac{2x\left(2x+3\right)-\left(2x+3\right)}{-2x\left(4x-7\right)-3\left(4x-7\right)}+1\)

\(=\left[\dfrac{2x-3-6x+15-4x+2}{\left(2x-5\right)}\right].\dfrac{2\left(x+\dfrac{3}{2}\right)}{\left(-2x-3\right)\left(4x-7\right)}+1\)

\(=\dfrac{-2\left(4x-7\right)}{2x-5}.\dfrac{2\left(x+\dfrac{3}{2}\right)}{\left(-2x-3\right)\left(4x-7\right)}+1\)

\(=\dfrac{1}{2x-5}.2+1\)

\(=\dfrac{2+2x-5}{2x-5}\)

\(=\dfrac{-3+2x}{2x-5}\)

a: ĐKXĐ: x<>2; x<>0

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

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

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

\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)

c: M>=-3

=>(x+1+6x)/2x>=0

=>(7x+1)/x>=0

=>x>0 hoặc x<=-1/7

a) ĐKXĐ: x khác 0

\(x+\dfrac{5}{x}>0\)

\(\Leftrightarrow x^2+5>0\) ( luôn đúng)

Vậy bất pt vô số nghiệm ( loại x = 0)

d)

\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)

\(\Leftrightarrow2x+2-4x+4>-15\)

\(\Leftrightarrow-2x>-21\)

\(\Leftrightarrow x< \dfrac{21}{2}\)

Vậy....................

a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)

\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)

Mà \(x^2+5>0\)

\(\Rightarrow x>0\)

d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)

\(\Leftrightarrow-x>-\dfrac{21}{2}\)

\(\Leftrightarrow x< \dfrac{21}{2}\)

c)

\(\left(x^2-2x\right)\left(x^3-3x^2-18x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^3-6x^2+3x^2-18x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left[\left(x^3-6x^2\right)+\left(3x^2-18x\right)\right]=0\)

\(\Leftrightarrow x\left(x-2\right)\left[x^2\left(x-6\right)+3x\left(x-6\right)\right]=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x-6\right)\left(x^2+3x\right)=0\)

\(\Leftrightarrow x^2\left(x-6\right)\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\\x=2\\x=-3\end{matrix}\right.\)

a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)

\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)

hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)

b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)

nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)

hay \(x=\dfrac{8}{41}\)

c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|2x-1\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)