\(\Leftrightarrow\dfrac{x^2+2x+1-1}{x+1}+\dfrac{x^2+8x+16+4}{x+4}=\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+6x+9+3}{x+3}\)

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

\(\Leftrightarrow2x+5-\dfrac{1}{x+1}+\dfrac{4}{x+4}=2x+5+\dfrac{2}{x+2}+\dfrac{3}{x+3}\)

=>-x-4+4x+4=2x+6+3x+6

=>3x=5x+12

=>-2x=12

hay x=-6(nhận)

\(\dfrac{x^2+2x+2}{x+1}+\dfrac{x^2+8x+20}{x+4}=\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+6x+12}{x+3}\)\(\Leftrightarrow\)\(\dfrac{x^2+2x+1+1}{x+1}+\dfrac{x^2+8x+16+4}{x+4}=\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+6x+9+3}{x+3}\)

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

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

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

\(\Leftrightarrow\) \(\dfrac{1}{x+1}\) + \(\dfrac{4}{x+4}\) - \(\dfrac{2}{x+2}\) - \(\dfrac{3}{x+3}\) = 0

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

\(\Leftrightarrow\) \(\dfrac{x+4}{\left(x+1\right)\left(x+4\right)}\) + \(\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x+4\right)}\) = \(\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x+2\right)}\) + \(\dfrac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}\)

\(\Leftrightarrow\) \(\dfrac{x+4+4x+4}{x^2+5x+4}\) = \(\dfrac{2x+6+3x+6}{x^2+5x+6}\)

\(\Leftrightarrow\) \(\dfrac{5x+8}{x^2+5x+4}\) = \(\dfrac{5x+12}{x^2+5x+6}\)

Đặt 5x + 8 = y; x² + 5x + 4 = t, ta có:

\(\dfrac{y}{t}\) = \(\dfrac{y+4}{t+2}\)

\(\Leftrightarrow\) \(\dfrac{y\left(t+2\right)}{t\left(t+2\right)}\) = \(\dfrac{t\left(y+4\right)}{t\left(t+2\right)}\)

\(\Leftrightarrow\) yt + 2y = yt + 4t

\(\Leftrightarrow\) 2y = 4t

\(\Leftrightarrow\) 2(5x + 8) = 4(x² + 5x + 4)

\(\Leftrightarrow\) 10x + 16 = 4x² + 20x + 16

\(\Leftrightarrow\) 16 - 16 = 4x² + 20x - 10x

\(\Leftrightarrow\) 0 = 4x² + 10x

\(\Leftrightarrow\) 2x(2x + 5) = 0

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

CHÚC BN HOK TỐT...

a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1

=>1,7x=6,7

hay x=67/17

b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)

=>150x+120-45x-75=96x+216-40x+360

=>105x+45=56x+576

=>49x=531

hay x=531/49

e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

\(\Leftrightarrow x=-1\left(TM\right)\)

a.

\(\Leftrightarrow4x^2-6x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(4x^2-2x+1\right)\left(4x^2+2x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2-2x+1}=a>0\\\sqrt{4x^2+2x+1}=b>0\end{matrix}\right.\) ta được:

\(2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

\(\Leftrightarrow\left(a-\dfrac{b}{\sqrt{3}}\right)\left(2a+\sqrt{3}b\right)=0\)

\(\Leftrightarrow a=\dfrac{b}{\sqrt{3}}\)

\(\Leftrightarrow3a^2=b^2\)

\(\Leftrightarrow3\left(4x^2-2x+1\right)=4x^2+2x+1\)

\(\Leftrightarrow...\)

b.

\(x^2-3x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

Lặp lại cách làm câu a

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

\(\dfrac{1}{x+2}+\dfrac{6x+12}{x^3+8}-\dfrac{7}{x^2-2x+4}=0\) \(\left(đk:x\ne-2\right)\)

\(\Leftrightarrow\dfrac{x^2-2x+4+6x+12-7\left(x+2\right)}{x^3+8}=0\)

\(\Leftrightarrow\dfrac{x^2-3x+2}{x^3+8}=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)(TM)

Vậy ...

dk : x khac -2 

\(\Rightarrow x^2-2x+4+6x+12-7\left(x+2\right)=0\)

\(\Leftrightarrow x^2+4x+16-7x-14=0\Leftrightarrow x^2-3x+2=0\)

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

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

\(\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}=\dfrac{2}{-x^2+6x-8}\left(đk:x\ne2,x\ne4\right)\Leftrightarrow\dfrac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{-2}{x^2-6x+8}\Leftrightarrow\dfrac{2x^2-4x-2}{x^2-6x+8}=\dfrac{-2}{x^2-6x+8}\Leftrightarrow2x^2-4x-2=-2\Leftrightarrow2x^2-4x=0\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)\(\Leftrightarrow x=0\)( do x≠2)

2)Biện luận PT

`m(mx-1)=x+1`

`<=>m^2x-m=x+1`

`<=>x(m^2-1)=m+1`

PT vô nghiệm `<=>{(m^2-1=0),(m+1\ne0):}<=>m=1`

PT vô số nghiệm `<=>{(m^2-1=0),(m+1=0):}<=>m=-1`

PT có nghiệm duy nhất `m^2-1\ne0<=>m^2\ne1<=>m\ne+-1=>x=(m+1)/(m^2-1)=1/(m-1)`

\(m\left(mx-1\right)=x+1\Leftrightarrow m^2x-x-m-1=0\Leftrightarrow x\left(m-1\right)\left(m+1\right)-\left(m+1\right)=0\Leftrightarrow\left(m+1\right)\left[x\left(m-1\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m-1=0\\m^2x-m=x+1\end{matrix}\right.\\\left\{{}\begin{matrix}x\left(m-1\right)-1=0\\m^2x-m=x+1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m=1\\x-1=x+1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\m=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\m=2\end{matrix}\right.\)(do x-1≠x+1)