Question

Giải các bất phương trình sau :

a) \(\left(x+1\right)\left(2x-1\right)+x\le3+2x^2\)

b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)

c) \(x+\sqrt{x}>\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\)

d) \(\left(\sqrt{1-x}+3\right)\left(2\sqrt{1-x}-5\right)>\sqrt{1-x}-3\)

Answer 1

lời giải

a)

\(\left(x+1\right)\left(2x-1\right)+x\le2x^2+3\)

\(\Leftrightarrow2x^2+x-1+x\le2x^2+3\)

\(\Leftrightarrow2x\le4\Rightarrow x\le2\)

\(\)b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)

\(\left(x^2+3x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)

\(x^3+3x^2+3x^2+9x+2x+6-x>x^3+6x^2-5\)

\(10x+6>-5\Rightarrow x>-\dfrac{11}{10}\)

Answer 2

c)Đkxđ: x\ge0
x+\sqrt{x}>\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)
\Leftrightarrow x+\sqrt{x}>2x+\sqrt{x}-3
\Leftrightarrow x-3>0
\Leftrightarrow x>3. (tmđk).
 

Answer 3

_{d) Đkxđ: \(1-x\ge0\)\(\Leftrightarrow x\le1\)}.
\(\left(\sqrt{1-x}+3\right)\left(2\sqrt{1-x}-5\right)>\sqrt{1-x}-3\)
Đặt \(\sqrt{1-x}=t\left(t\ge0\right)\) bpt trở thành:
\(\left(t+3\right)\left(2t-5\right)>t-3\)\(\Leftrightarrow2t^2+t-15>t-3\)
\(\Leftrightarrow2t^2>12\)\(\Leftrightarrow t^2>6\)\(\Leftrightarrow t>\sqrt{6}\) ( do \(t\ge0\) ).
Trở lại phép đặt: \(\sqrt{1-x}>\sqrt{6}\)\(\Leftrightarrow1-x>6\)\(\Leftrightarrow x< -5\).