Toán Lớp 10: giải bất pt : √x(x+2) + √x ≥ √(x+1) ³

Question

Toán Lớp 10:  giải bất  pt :  √x(x+2) + √x  ≥ √(x+1) ³

huyenmi76 · Answer

Điều kiện $x ge 0$   $ sqrt {x left( {x + 2}  right)}  +  sqrt x   ge  sqrt {{{ left( {x + 1}  right)}^3}}      Leftrightarrow  sqrt x  left( { sqrt {x + 2}  + 1}  right)  ge  sqrt {{{ left( {x + 1}  right)}^3}}      Leftrightarrow  sqrt x . dfrac{{x + 1}}{{ sqrt {x + 2}  - 1}}  ge  sqrt {{{ left( {x + 1}  right)}^3}}      Leftrightarrow  sqrt x . dfrac{{x + 1}}{{ sqrt {x + 2}  - 1}}  ge  left( {x + 1}  right) sqrt {x + 1}      Leftrightarrow  left( {x + 1}  right) left( { dfrac{{ sqrt x }}{{ sqrt {x + 2}  - 1}} -  sqrt {x + 1} }  right)  ge 0     Leftrightarrow  left[  begin{array}{l}  left {  begin{array}{l} x  ge  - 1    dfrac{{ sqrt x }}{{ sqrt {x + 2}  - 1}}  ge  sqrt {x + 1}   end{array}  right.    left {  begin{array}{l} x  le  - 1    dfrac{{ sqrt x }}{{ sqrt {x + 2}  - 1}}  le  sqrt {x + 1}   end{array}  right.(L)  end{array}  right.     Leftrightarrow  left {  begin{array}{l} x  ge  - 1    sqrt x   ge  sqrt { left( {x + 1}  right) left( {x + 2}  right)}  -  sqrt {x + 1}   end{array}  right.     Leftrightarrow  left {  begin{array}{l} x  ge  - 1    sqrt x  +  sqrt {x + 1}   ge  sqrt { left( {x + 1}  right) left( {x + 2}  right)}   end{array}  right.     Leftrightarrow  left {  begin{array}{l} x  ge  - 1   x + x + 1 + 2 sqrt {x left( {x + 1}  right)}   ge {x^2} + 3x + 2  end{array}  right.     Leftrightarrow  left {  begin{array}{l} x  ge  - 1   2 sqrt {x left( {x + 1}  right)}   ge {x^2} + x + 1  end{array}  right.     Leftrightarrow  left {  begin{array}{l} x  ge  - 1   2t  ge {t^2} + 1  end{array}  right. left( {t =  sqrt {x left( {x + 1}  right)} }  right)     Leftrightarrow  left {  begin{array}{l} x  ge  - 1   { left( {t - 1}  right)^2}  le 0  end{array}  right.  Leftrightarrow  left {  begin{array}{l} x  ge  - 1    sqrt {x left( {x + 1}  right)}  = 1  end{array}  right.    Leftrightarrow  left {  begin{array}{l} x  ge  - 1   x left( {x + 1}  right) = 1  end{array}  right.  Leftrightarrow  left {  begin{array}{l} x  ge  - 1   {x^2} + x - 1 = 0  end{array}  right.     Leftrightarrow  left {  begin{array}{l} x  ge  - 1   x =  dfrac{{ sqrt 5   pm 1}}{2}  end{array}  right.  Leftrightarrow x =  dfrac{{ sqrt 5  - 1}}{2} left( *  right)    left( *  right)  cap DK  Rightarrow x =  dfrac{{ sqrt 5  - 1}}{2}     Rightarrow S =  left { { dfrac{{ sqrt 5  - 1}}{2}}  right } $