\(y=\dfrac{x^2+3x+3}{x^2+1}\Rightarrow y'=\dfrac{\left(x^2+3x+3\right)'\left(x^2+1\right)-\left(x^2+3x+3\right)\left(x^2+1\right)'}{\left(x^2+1\right)^2}\)

\(y'=\dfrac{\left(x^2+1\right)\left(2x+3\right)-\left(x^2+3x+3\right).2x}{\left(x^2+1\right)^2}\)

\(y'=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)-2x\left(x^2+3x+3\right)=0\)

\(\Leftrightarrow2x^3+3x^2+2x+3-2x^3-6x^2-6x=0\)

\(\Leftrightarrow3x^2+4x-3=0\)

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

Check lai ho t nhe

\(\left(x-1\right)^2+x\left(5-x\right)=0\)

\(\Leftrightarrow x^2-2x+1+5x-x^2=0\)

\(\Leftrightarrow3x+1=0\)

\(\Leftrightarrow3x=-1\)

\(x=\frac{-1}{3}\)

\(3=x-2\)

\(\Leftrightarrow x=5\)

vậy phương trình có ngiệm là 5

x=3+2 x=5

\(a,\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\left(x\ne2;x\ne3\right)\\ \Leftrightarrow\dfrac{5}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}=0\\\Leftrightarrow\dfrac{5-\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}=0 \\ \Leftrightarrow5-x^2+9=0\\ \Leftrightarrow14-x^2=0\\ \Leftrightarrow x^2=14\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{14}\\x=-\sqrt{14}\end{matrix}\right.\)

\(b,\dfrac{x}{2x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\left(x\ne-1;x\ne3\right)\\ \Leftrightarrow\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{x}{2\left(3-x\right)}\\ \Leftrightarrow\dfrac{x\left(x-3\right)-2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\\ \Leftrightarrow x^2-3x-4x=-x^2-x\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(c,\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\left(x\ne1\right)\\ \Leftrightarrow\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ \Leftrightarrow-2x^2+x+1=2x^2-2x\\ \Leftrightarrow4x^2-3x-1=0\\ \Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)

\(d,\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}=\dfrac{5-x}{2x^2+10x}\left(x\ne5;x\ne-5\right)\\ \Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\\ \Leftrightarrow\dfrac{x^2+25x-2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(5-x\right)\left(x-5\right)}{2x\left(x+5\right)\left(x-5\right)}\\ \Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=-\left(x^2-10x+25\right)\\ \Leftrightarrow x^2+25x-2x^2-20x-50=-x^2+10x-25\\ \Leftrightarrow-5x=25\\ \Leftrightarrow x=-5\)

Tick nha

Ta có: \(\left(x^2+4x+3\right)\left(x^2+6x+8\right)=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\left(x+2\right)\left(x+4\right)=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)(1)

Ta có: \(1\cdot2\cdot3\cdot4=24\)(2)

Từ (1) và (2) suy ra \(\left\{{}\begin{matrix}x+1=1\\x+2=2\\x+3=3\\x+4=4\end{matrix}\right.\Leftrightarrow x=0\)

Vậy: x=0

Sai từ chỗ (1)

(1)\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)(2)

Đặt \(y=x^2+5x+4=\left(x+1\right)\left(x+4\right)\)

\(\left(1\right)\Leftrightarrow y^2+2y-24=0\)

\(\Leftrightarrow\left(y-4\right)\left(y+6\right)=0\Rightarrow\left[{}\begin{matrix}y=4\\y=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+5x=0\\x^2+5x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\left(x+5\right)=0\\\left(x+\frac{5}{2}\right)^2+\frac{15}{4}>0\end{matrix}\right.\)

Vậy x=0 hoặc x=-5

Ta cos (x²+4x+3)(x²+6x+8)=24

<=>(x+1)(x+2)(x+3)(x+4)-24=0

<=>(x+1)(x+4)(x+2)(x+3)-24=0

<=>(x²+5x+4)(x²+5x+6)-24=0

Đặt a=x²+5x+5 =>(a-1)(a+1)-24=0

<=>a²-1-24=0

<=>a²-25=0

<=>(a-5)(a+5)=0

<=>a=5

hoặc a=-5

<=>x²+5x+5=5

Hoặc x²+5x+5=-5

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

Hoặc x²+5x+10=0(vô lí)

Vậy tập nghiệm của pt là\(\left\{0,5\right\}\)