Tìm lim x^2 -5x +6 / √4x+1 -3 x->2 A. 3/2 B -2/3 C -3/2 D 1/2
2 câu trả lời
Đáp án:
`\bb C`
Giải thích các bước giải:
\(\lim\limits_{x \to 2} \dfrac{x^2-5x+6}{\sqrt{4x+1}-3}\)
\(=\lim\limits_{x \to 2} \dfrac{(x^2-2x-3x+6)(\sqrt{4x+1}+3)}{(\sqrt{4x+1})^2-(3)^2}\)
\(=\lim\limits_{x \to 2} \dfrac{(x-2)(x-3)(\sqrt{4x+1}+3)}{4x-8}\)
\(=\lim\limits_{x \to 2} \dfrac{(x-2)(x-3)(\sqrt{4x+1}+3)}{4(x-2)}\)
\(=\lim\limits_{x \to 2} \dfrac{(x-3)(\sqrt{4x+1}+3)}{4}\)
`=\frac{(2-3)(\sqrt{4.2+1}+3)}{4}`
`=-3/2`
Đáp án: `lim_{x->2} \frac{x²-5x+6}{\sqrt{4x+1}-3}=-3/2`
Giải thích các bước giải:
`lim_{x->2} \frac{x²-5x+6}{\sqrt{4x+1}-3}`
`= lim_{x->2} \frac{(x²-5x+6)(\sqrt{4x+1}+3)}{(\sqrt{4x+1}-3)(\sqrt{4x+1}+3)}`
`= lim_{x->2} \frac{(x-2)(x-3)(\sqrt{4x+1}+3)}{4x+1-9}`
`= lim_{x->2} \frac{(x-2)(x-3)(\sqrt{4x+1}+3)}{4(x-2)}`
`= lim_{x->2} \frac{(x-3)(\sqrt{4x+1}+3)}{4}`
`= \frac{(2-3)(\sqrt{4.2+1}+3)}{4}=-3/2`
