Question

B1 so sánh

a, \(\sqrt{48}\) +3 và 7+ \(\sqrt{10}\)

b, \(\sqrt{65}\)+\(\sqrt{50}\) và 15

c, \(\sqrt{197}\)-\(\sqrt{15}\) và 10

B2 Tìm điều kiện của x để các căn thức sau có nghĩa

a, \(\sqrt{1}\) - 4x

b,\(\sqrt{x^{ }2-x+1}\)

c, \(\sqrt{\frac{2x+1}{2-x}}\)

d, \(\sqrt{x^{ }2-5x+4}\)

e, \(\sqrt{x+1}\) + \(\sqrt{x-2}\)

CÁC ANH CHỊ GIÚP EM VỚI

Answer 1

Bài 1:

a)

\(\sqrt{48}< \sqrt{49}\) hay \(\sqrt{48}< 7\)

\(3=\sqrt{9}< \sqrt{10}\)

\(\Rightarrow \sqrt{48}+3< 7+\sqrt{10}\)

b)

\(\sqrt{65}>\sqrt{64}=8\); \(\sqrt{50}>\sqrt{49}=7\)

\(\Rightarrow \sqrt{65}+\sqrt{50}>8+7\) hay \(\sqrt{65}+\sqrt{50}>15\)

c)

\(\sqrt{197}>\sqrt{196}=14\)

\(\Rightarrow \sqrt{197}-\sqrt{15}>14-\sqrt{15}\). Mà \(\sqrt{15}< \sqrt{16}=4\)

\(\Rightarrow \sqrt{197}-\sqrt{15}>7-\sqrt{15}>14-4\)

hay \(\sqrt{197}-\sqrt{15}>10\)

Answer 2

Bài 2:

a) ĐKXĐ: \(1-4x\geq 0\Leftrightarrow x\leq \frac{1}{4}\)

b) ĐKXĐ: \(x^2-x+1\geq 0\)

\(\Leftrightarrow (x-\frac{1}{2})^2+\frac{3}{4}\geq 0\)

\(\Leftrightarrow x\in\mathbb{R}\)

c) ĐKXĐ:

\(\left\{\begin{matrix} 2-x\neq 0\\ \frac{2x+1}{2-x}\geq 0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} 2x+1\geq 0; 2-x>0 \\ 2x+1\leq 0; 2-x< 0\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} -\frac{1}{2}\leq x< 2\\ -\frac{1}{2}\geq x> 2(\text{vô lý})\end{matrix}\right.\Rightarrow -\frac{1}{2}\leq x< 2\)

d)

ĐKXĐ: \(x^2-5x+4\geq 0\Leftrightarrow (x-1)(x-4)\geq 0\)

\(\Leftrightarrow \left[\begin{matrix} x-1\geq 0; x-4\geq 0\\ x-1\leq 0; x-4\leq 0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x\geq 4\\ x\leq 1\end{matrix}\right.\)

e)

ĐKXĐ: \(\left\{\begin{matrix} x+1\geq 0\\ x-2\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -1\\ x\geq 2\end{matrix}\right.\Rightarrow x\geq 2\)

Answer 3

Bài 2:

a) ĐKXĐ: \(1-4x\geq 0\Leftrightarrow x\leq \frac{1}{4}\)

b) ĐKXĐ: \(x^2-x+1\geq 0\)

\(\Leftrightarrow (x-\frac{1}{2})^2+\frac{3}{4}\geq 0\)

\(\Leftrightarrow x\in\mathbb{R}\)

c) ĐKXĐ:

\(\left\{\begin{matrix} 2-x\neq 0\\ \frac{2x+1}{2-x}\geq 0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} 2x+1\geq 0; 2-x>0 \\ 2x+1\leq 0; 2-x< 0\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} -\frac{1}{2}\leq x< 2\\ -\frac{1}{2}\geq x> 2(\text{vô lý})\end{matrix}\right.\Rightarrow -\frac{1}{2}\leq x< 2\)

d)

ĐKXĐ: \(x^2-5x+4\geq 0\Leftrightarrow (x-1)(x-4)\geq 0\)

\(\Leftrightarrow \left[\begin{matrix} x-1\geq 0; x-4\geq 0\\ x-1\leq 0; x-4\leq 0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x\geq 4\\ x\leq 1\end{matrix}\right.\)

e)

ĐKXĐ: \(\left\{\begin{matrix} x+1\geq 0\\ x-2\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -1\\ x\geq 2\end{matrix}\right.\Rightarrow x\geq 2\)