Ta có: \(\left(x^2+2x-3\right)>0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x< -3\\x>1\end{matrix}\right.\)
Ta có: x2+2x−3>0
⇔(x+3)(x−1)>0
⇔\(\left[{}\begin{matrix}x>-3\\x>1\end{matrix}\right.\)
Ta có: \(\left(x^2+2x-3\right)>0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x< -3\\x>1\end{matrix}\right.\)
Ta có: x2+2x−3>0
⇔(x+3)(x−1)>0
⇔\(\left[{}\begin{matrix}x>-3\\x>1\end{matrix}\right.\)
,giải pt
a,\(\sqrt{x-2}-3\sqrt{x^2-4}=0\)
b,\(\sqrt{x^2-1}-x^2+1=0\)
giải phương trình a)x-2√x +1=0 b)x-2√x -3=0
đưa thừa số ra ngoài dấu căn :
a) a2\(\sqrt{\dfrac{2}{3a}}\)( a > 0 )
b) \(\dfrac{x-3}{x}\)\(\sqrt{\dfrac{x^3}{9-x^2}}\)(0<x<3)
Giải phương trình :
a) x^2 + ( √3 + √2)x + √6 =0
b) 2 - x^2 = √2 - x
c) √(x^2+ 4x +4) = √(x+2)
d) 20 - √(3-2x) = |2x - 3|
e) √(x +1) + √(x^2- 2x +1) =0
bài 1: rút gọn
A= \(\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\) X>=0; x#0
B= \(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\left(\dfrac{2}{x^2-2x+1}\right)\) x>=0; x\(\ne\)1
bài 2:
\(\dfrac{1}{2x-3\sqrt{x}+2}\) x>=0
Tìm GTNN của A
Giải các phương trình sau:
a) \(x\sqrt{x-1}+\left(2x+1\right)\sqrt{x+2}+x^3-4x^2+x-6=0\)
b) \(\left(2x+3\right)\sqrt{2x-1}+x\sqrt{x+3}+x^2-5x-3=0\)
c) \(x\sqrt{2x+3}+\left(x+1\right)\sqrt{4x-1}+2\left(x^2-x-1\right)=0\)
Tìm x
\(a.\sqrt{2+\sqrt{3+\sqrt{x}}=3}\)
\(b.\sqrt{x^2-4}+\sqrt{x+2}=0\)
\(c.\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)
giải các phương trình sau. x2-11=0. x2-2√13x+13=0. x2-9√x+14=0. √x-6=13. √x+9=3. √x2-2x+4=x-1
Gải pt:
a) \(\sqrt{x+5}+\sqrt{3-x}-2\left(\sqrt{15-2x-x^2}+1\right)=0\)
b) \(3\sqrt{3}\left(x^2+4x+2\right)-\sqrt{x+8}=0\)
c) \(x^2-x-2\sqrt{1+16x}=2\)
Chứng minh BĐT:
a) x2 + x + 1 > 0 ∀ x
b) x - \(\sqrt{x}\) + 1 > 0 ∀ x
c) x2 - xy + y2 > 0 ∀ xy , x; y ≠0
d) x2 + x\(\sqrt{2}\) + 1 > 0 ∀ x
e) ( x + y + z )2 ≤ 3( x2 + y2 + z2) ∀ xyz