Tìm đkxđ 1, y= 4sin 3x/2x+1 2, y=-1/√x^2 – 3x + 2 3,y= 5cot(3x-π/3) 02/08/2021 Bởi Eloise Tìm đkxđ 1, y= 4sin 3x/2x+1 2, y=-1/√x^2 – 3x + 2 3,y= 5cot(3x-π/3)
Đáp án: $1) x\neq \dfrac{-1}{2}\\2) {\left[\begin{aligned}x\neq 1\\x\neq 2\end{aligned}\right.}\\3) x\neq \dfrac{\pi}{9}+\dfrac{k\pi}{3},(k\in\mathbb{Z}) \\$ Giải thích các bước giải: $1) y=4\sin\dfrac{3x}{2x+1}\\ĐK: 2x+1\neq 0\Leftrightarrow 2x\neq -1\Leftrightarrow x\neq \dfrac{-1}{2}\\2) y=\dfrac{-1}{\sqrt{x^2-3x+2}}\\ĐK: x^2-3x+2\neq 0\Leftrightarrow {\left[\begin{aligned}x\neq 1\\x\neq 2\end{aligned}\right.}\\3) y=5\cot(3x-\dfrac{\pi}{3})\\ĐK: 3x-\dfrac{\pi}{3}\neq k\pi\\\Leftrightarrow 3x\neq \dfrac{\pi}{3}+k\pi\\\Leftrightarrow x\neq \dfrac{\pi}{9}+\dfrac{k\pi}{3},(k\in\mathbb{Z}) \\$ Bình luận
$#lam$
Đáp án:
$1) x\neq \dfrac{-1}{2}\\
2) {\left[\begin{aligned}x\neq 1\\x\neq 2\end{aligned}\right.}\\
3) x\neq \dfrac{\pi}{9}+\dfrac{k\pi}{3},(k\in\mathbb{Z}) \\$
Giải thích các bước giải:
$1) y=4\sin\dfrac{3x}{2x+1}\\
ĐK: 2x+1\neq 0\Leftrightarrow 2x\neq -1\Leftrightarrow x\neq \dfrac{-1}{2}\\
2) y=\dfrac{-1}{\sqrt{x^2-3x+2}}\\
ĐK: x^2-3x+2\neq 0\Leftrightarrow {\left[\begin{aligned}x\neq 1\\x\neq 2\end{aligned}\right.}\\
3) y=5\cot(3x-\dfrac{\pi}{3})\\
ĐK: 3x-\dfrac{\pi}{3}\neq k\pi\\
\Leftrightarrow 3x\neq \dfrac{\pi}{3}+k\pi\\
\Leftrightarrow x\neq \dfrac{\pi}{9}+\dfrac{k\pi}{3},(k\in\mathbb{Z}) \\$