a: =>6x-3x^2-5=4-3x^2-2

=>6x-5=2

=>6x=7

=>x=7/6

b: =>20x+5-12x^2-3x=6x^2-10x+3x-5

=>-12x^2+17x+5-6x^2+7x+5=0

=>-18x^2+24x+10=0

=>x=5/3 hoặc x=-1/3

a, ĐKXĐ:\(x\ne-3\)

\(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+5}{x+3}-\dfrac{2}{x+3}\\ \Leftrightarrow x+1=\dfrac{x+3}{x+3}\\ \Leftrightarrow x+1=1\\ \Leftrightarrow x=0\left(tm\right)\)

b, ĐKXĐ:\(x>2\)

\(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\\ \Leftrightarrow x^2-4x-2=x-2\\ \Leftrightarrow x^2-5x=0\\ \Leftrightarrow x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

\(P=\dfrac{4x}{x-2}=\dfrac{4\left(x-2\right)+8}{x-2}=4+\dfrac{8}{x-2}\in Z\)

\(\Rightarrow\left(x-2\right)\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

\(\Rightarrow x\in\left\{-6;-2;0;1;3;4;6;10\right\}\)

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

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\)

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

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

\(\Leftrightarrow3x^2-17x+24=0\)

\(\Leftrightarrow3x^2-9x-8x+24=0\)

\(\Leftrightarrow\left(3x-8\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=3\end{matrix}\right.\left(\text{thỏa}\right)\)

\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\left(x\ne\left\{2;4\right\}\right)\\ =>\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1\\ =>x^2-3x-4x+12+x^2-4x+4=-\left(x-2\right)\left(x-4\right)\\ =>2x^2-11x+16=-x^2+6x-8\\ =>3x^2-17x+24=0\\ =>\left(x-3\right)\left(3x-8\right)=0\\ =>\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\left(TMDK\right)\)

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

\(\dfrac{x^2-7x+12}{x^2-6x+8}+\dfrac{x^2-4x+2}{x^2-6x+8}=\dfrac{-x^2+6x-8}{x^2-6x+8}\)

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

\(\dfrac{3x^2-17x+22}{x^2-6x+8}=0\)

\(\dfrac{\left(3x-11\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=0\)

\(\dfrac{3x-11}{x-4}=0\)

Với x khác 4 

=> 3x -11 = 0 

=> \(x=\dfrac{11}{3}\)

\(x^2\) + (\(x\) + 2)(11\(x\) - 7) = 4

        (\(x\) + 2)(11\(x\) - 7) + \(x^2\) - 4 = 0

        (\(x+2\))(11\(x-7\)) + (\(x-2\))(\(x+2\)) = 0

        (\(x+2\))(11\(x\) - 7 + \(x-2\)) = 0

        (\(x\) + 2)(12\(x\) - 9) = 0

         \(\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=-2\\x=\dfrac{9}{12}\end{matrix}\right.\)

            \(\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)

.

a: \(\dfrac{3x+2}{4}-\dfrac{3x+1}{3}=\dfrac{5}{6}\)

=>3(3x+2)-4(3x+1)=10

=>9x+6-12x-4=10

=>-3x+2=10

=>-3x=8

=>x=-8/3

b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)

=>(x-1)(x-2)-x(x+2)=-9x+10

=>x^2-3x+2-x^2-2x=-9x+10

=>-5x+2=-9x+10

=>x=2(loại)

ĐKXĐ: \(x\ge3\)

\(pt\Leftrightarrow5\sqrt{x-3}+3\sqrt{x-3}-\sqrt{x-3}=7\)

\(\Leftrightarrow7\sqrt{x-3}=7\Leftrightarrow\sqrt{x-3}=1\)

\(\Leftrightarrow x-3=1\Leftrightarrow x=4\left(tm\right)\)