Đáp án:
$a)\left[\begin{array}{l} x=1\\ x=-\dfrac{5}{3}\end{array} \right.\\ b) \left[\begin{array}{cc} x=5\\x=3\end{array} \right.\\ c) x =0\\ d)x=\dfrac{8\sqrt{17}}{17}\\ e) x=0$
Giải thích các bước giải:
$a)y=x^3+x^2-2\\ y'=3x^2+2x\\ y'=5\\ \Leftrightarrow 3x^2+2x=5\\ \Leftrightarrow 3x^2+2x-5=0\\ \Leftrightarrow \left[\begin{array}{l} x=1\\ x=-\dfrac{5}{3}\end{array} \right.\\ b)y=\dfrac{x+3}{x-4} (x \ne 4)\\ y'=\dfrac{-7}{(x-4)^2}\\ y'=-7\\ \Leftrightarrow \dfrac{-7}{(x-4)^2}=-7\\ \Rightarrow (x-4)^2=1\\ \Leftrightarrow \left[\begin{array}{cc} x-4=1&x>4\\x-4=-1&x<4\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{cc} x=5\\x=3\end{array} \right.\\ c)y=\sqrt{1-x^2} \ \ \ \ D=[-1;1]\\ y'=\dfrac{1}{2\sqrt{1-x^2}}.(1-x^2)'\\ =-\dfrac{x}{\sqrt{1-x^2}}\\ y'=0\\ \Leftrightarrow -\dfrac{x}{\sqrt{1-x^2}}=0\\ \Leftrightarrow x =0\\ d)y=\sqrt{4-x^2} \ \ \ \ D=[-2;2]\\ y'=\dfrac{1}{2\sqrt{4-x^2}}.(4-x^2)'\\ =-\dfrac{x}{\sqrt{4-x^2}}\\ y'=-4\\ \Leftrightarrow -\dfrac{x}{\sqrt{4-x^2}}=-4(x \ne \pm 2)\\ \Leftrightarrow \dfrac{x}{\sqrt{4-x^2}}-4=0\\ \Leftrightarrow \dfrac{x-4\sqrt{4-x^2}}{\sqrt{4-x^2}}=0\\ \Leftrightarrow x-4\sqrt{4-x^2}=0\\ \Leftrightarrow x=4\sqrt{4-x^2}\\ \Leftrightarrow x^2=16(4-x^2)(x >0)\\ \Leftrightarrow 17x^2-64=0\\ \Leftrightarrow \left[\begin{array}{l} x=\dfrac{8\sqrt{17}}{17}\\ x=-\dfrac{8\sqrt{17}}{17}(L)\end{array} \right.\\ e)y=x^4+2x^2+2\\ y'=4x^3+4x\\ y'=0\\ \Leftrightarrow 4x^3+4x=0\\ \Leftrightarrow 4x(x^2+1)=0\\ \Leftrightarrow x=0$
