Equazioni,disequazioni, sistemi e radicali

Esercizi sulle equazioni con i radicali

1) (\sqrt{2} + 1 )(x + 1) = 2 (2 – x)                                risultato:  \frac{ 11 -6\sqrt{2}}{7}

2) \sqrt{2}(x + \sqrt{2})+ \sqrt{5}(x - \sqrt{5})=0                       risultato:  (\sqrt{5} - \sqrt{2})

 

3) 5 \sqrt{2}(x +  \sqrt{10}) – 10  \sqrt{5} (1 +  \sqrt{5}x) – 5( \sqrt{2} – 10) = 0           risultato: 1

4)(2x - \sqrt{3})(x - \sqrt{2}) - (x - \sqrt{3})  ^{2} - (x + \sqrt{2}) ^{2} = (1 - \sqrt{3})(\sqrt{3} - \sqrt{2} 

) - \sqrt{2}(\sqrt{2}+ 3)          risultato: 1

5) \sqrt{2}\left [ 3(x + \sqrt{2})(x - \sqrt{2}) + 5\sqrt{2}(x - \sqrt{2}) \right ] = 3(1 - \sqrt{2}x)(1 - x)- (3 - 7x)                risultato: \frac{16(\sqrt{2}-1)}{3}

6)\frac{2x - 3\sqrt{2}}{\sqrt{2}} - \frac{x - 3}{3} = \frac{4x}{3\sqrt{2}}                                   risultato: 6(\sqrt{2} + 1)

Esercizi sulle equazioni fratte con i radicali

7) \frac{1}{ 3 - 3\sqrt{3}x} +\frac{2\sqrt{3}}{1 - 3x ^{2}}  = \frac{\sqrt{3}}{3 + 3\sqrt{3}x}                    risultato: \frac{6 - 7\sqrt{3}}{3}

8) \frac{5x - \sqrt{2}}{5x+\sqrt{2}} - \frac{5x + \sqrt{2}}{5x-\sqrt{2} = \frac{2\sqrt{2}(\sqrt{2}- 9x)}{25x ^{2}-2}                  risultato: -\sqrt{2}

Disequazioni con i radicali

9) x(x + 1) + 1 + \sqrt{2} > x² + \sqrt{2} (x + 1)                         risultato: x < 1 + \sqrt{2}

10) (3\sqrt{3} + 2\sqrt{2})(x + \sqrt{3}) – (x + 4\sqrt{2})(\sqrt{3} + 2\sqrt{2}) < – 7             risultato: x < \sqrt{2}

11) \frac{x - \sqrt{7}}{2\sqrt{7}} - \frac{2x - 3\sqrt{7}}{4\sqrt{7}}+ \frac{5x - \sqrt{7}}{3\sqrt{7}} > \frac{7x + 12\sqrt{7}}{12\sqrt{7}}                      risultato: x > \sqrt{7}

 

Sistemi con i radicali

13)

 sistema-con-radicali risultato: x = 2\sqrt{5};  y= -\sqrt{10}

 

sistema-con-radicalirisultato:x = 3 ;   y= 2\sqrt{3}

 

Svolgimento

1) (\sqrt{2} + 1 )(x + 1) = 2 (2 – x)

\sqrt{2} x + \sqrt{2}  + x + 1 = 4 – 2x   Portiamo tutte le x al primo membro

\sqrt{2} x +x + 2x = 4 – \sqrt{2}  – 1

3x + \sqrt{2} x = 3 – \sqrt{2}   raccogliamo la x

x(3 +  \sqrt{2} ) = 3 – \sqrt{2}

x = \frac{3 - \sqrt{2}}{3 + \sqrt{2}   Razionalizziamo il denominatore moltiplicano numeratore e denominatore per  3 - \sqrt{2}  in modo di ottenere al denominatore la somma per differenza

x = \frac{3 - \sqrt{2}}{3 + \sqrt{2} · \frac{3 - \sqrt{2}}{3 - \sqrt{2} = \frac{(3 - \sqrt{2}) ^{2}}{9 - 2} =  \frac{9 + 2 -6 \sqrt{2} }{7} =  \frac{11 -6 \sqrt{2} }{7}

x = \frac{11 -6 \sqrt{2} }{7}

2) \sqrt{2}(x + \sqrt{2})+ \sqrt{5}(x - \sqrt{5})=0       

\sqrt{2}x + 2 + \sqrt{5}x – 5 =0

\sqrt{2}x+ \sqrt{5}x = – 2 + 5

x( \sqrt{2} + \sqrt{5}) = 3

x = \frac{3}{\sqrt{2} + \sqrt{5}} ⋅ \frac{\sqrt{2} - \sqrt{5}}{\sqrt{2} - \sqrt{5}} = \frac{3\sqrt{2} - 3\sqrt{5}}{2 - 5} = -\frac{3(\sqrt{2} - \sqrt{5})}{3} = -\sqrt{2}+ \sqrt{5}

x = -\sqrt{2}+ \sqrt{5}

3) 5 \sqrt{2}(x +  \sqrt{10}) – 10 \sqrt{5} (1 +  \sqrt{5}x) – 5( \sqrt{2} – 10) = 0

5 \sqrt{2} x+ 5\sqrt{20}  – 10 \sqrt{5} – 10·5 x –  5 \sqrt{2} +50  = 0

5 \sqrt{2} x+ 10\sqrt{5} – 10 \sqrt{5} – 50 x –  5 \sqrt{2} + 50 = 0

5 \sqrt{2} x -50x  =+  5 \sqrt{2} – 50

x(5\sqrt{2} – 50) =  5 \sqrt{2} – 50

x = \frac{5 \sqrt{2} - 50}{5 \sqrt{2} - 50}  = 1

x = 1

4)(2x - \sqrt{3})(x - \sqrt{2}) - (x - \sqrt{3})  ^{2} - (x + \sqrt{2}) ^{2} = (1 - \sqrt{3})(\sqrt{3} - \sqrt{2} 

) - \sqrt{2}(\sqrt{2}+ 3) 

2x² -2\sqrt{2}x -\sqrt{3}x + \sqrt{6} – ( x² + 3 -2\sqrt{3}x) – (x² + 2 +2\sqrt{2}x) = \sqrt{3} – \sqrt{2} – 3 + \sqrt{6}  – 2 – 3\sqrt{2}

2x² -2\sqrt{2}x -\sqrt{3}x + \sqrt{6} –  x² – 3  +2\sqrt{3}x – x² – 2  -2\sqrt{2}x = \sqrt{3} – \sqrt{2} – 3 + \sqrt{6}  – 2 – 3\sqrt{2}

2x²-  x²-  x²-2\sqrt{2}x-\sqrt{3}x +2\sqrt{3}x -2\sqrt{2}x   = \sqrt{3} – \sqrt{2} – 3 + \sqrt{6}  – 2 – 3\sqrt{2} – \sqrt{6}  + 3 + 2

-4\sqrt{2}x +\sqrt{3} x = \sqrt{3} -4\sqrt{2}

x( +\sqrt{3} -4\sqrt{2}) = \sqrt{3} -4\sqrt{2}

x = \frac{\sqrt{3} -4\sqrt{2}}{\sqrt{3} -4\sqrt{2}} = 1

x = 1

5) \sqrt{2}\left [ 3(x + \sqrt{2})(x - \sqrt{2}) + 5\sqrt{2}(x - \sqrt{2}) \right ] = 3(1 - \sqrt{2}x)(1 - x)- (3 - 7x)

\sqrt{2}\left [ (3x + 3\sqrt{2})(x - \sqrt{2}) + 5\sqrt{2}x - 10 \right ] = (3 - 3\sqrt{2}x)(1 - x)- 3 + 7x

\sqrt{2}( 3x ^{2}-3\sqrt{2}x+ 3\sqrt{2}x-6 + 5\sqrt{2}x -10)= 3 - 3x - 3\sqrt{2}x+ 3\sqrt{2}x ^{2}- 3 + 7x

3\sqrt{2}x ^{2} -6x +6x-6 \sqrt{2} -10x - 10\sqrt{2}= 3 - 3x - 3\sqrt{2}x+ 3\sqrt{2}x ^{2}- 3 + 7x

3\sqrt{2}x ^{2}-3\sqrt{2}x ^{2} -6x +6x +10x  +3x+3\sqrt{2}x} -7x= 3 - 3 + 6 \sqrt{2} +10\sqrt{2}

6x +3\sqrt{2}x} =16 \sqrt{2}

x(6 +3\sqrt{2}} )=16 \sqrt{2}

x =  \frac{16 \sqrt{2} }{6 +3\sqrt{2}} } ·  \frac{6 -3\sqrt{2}}{6 -3\sqrt{2}} } = \frac{96\sqrt{2}-96}{36 - 18} = \frac{96(\sqrt{2}-1)}{18} = \frac{16(\sqrt{2}-1)}{3}

x = \frac{16(\sqrt{2}-1)}{3}

6)\frac{2x - 3\sqrt{2}}{\sqrt{2}} - \frac{x - 3}{3} = \frac{4x}{3\sqrt{2}}             

\frac{3(2x - 3\sqrt{2})-\sqrt{2}(x - 3) }{3\sqrt{2}} = \frac{4x}{3\sqrt{2}}

\frac{6x - 9\sqrt{2}-\sqrt{2}x + 3\sqrt{2} }{3\sqrt{2}} = \frac{4x}{3\sqrt{2}}

\frac{6x -\sqrt{2}x -4x  }{3\sqrt{2}} = \frac{+9\sqrt{2}- 3\sqrt{2}}{3\sqrt{2}}

\frac{2x -\sqrt{2}x  }{3\sqrt{2}} = \frac{+6\sqrt{2}}{3\sqrt{2}}

\frac{x(2 -\sqrt{2})  }{3\sqrt{2}} = \frac{2}{1}

x  = \frac{2(3\sqrt{2})}{(2 -\sqrt{2})}       = \frac{6\sqrt{2}}{(2 -\sqrt{2})}       ·\frac{(2 +\sqrt{2})}{(2 +\sqrt{2})}       = \frac{12\sqrt{2}+ 2}{4 - 2}       = \frac{2(6\sqrt{2}+ 1)}{2}

x =   6\sqrt{2}+ 1

7) \frac{1}{ 3 - 3\sqrt{3}x} +\frac{2\sqrt{3}}{1 - 3x ^{2}}  = \frac{\sqrt{3}}{3 + 3\sqrt{3}x}    

\frac{1}{ 3 (1 -\sqrt{3}x)} +\frac{2\sqrt{3}}{(1 -\sqrt{3}x)(1 +\sqrt{3}x)}  = \frac{\sqrt{3}}{3(1 + \sqrt{3}x)}

    
\frac{1 +\sqrt{3}x+3(2\sqrt{3})}{3(1 -\sqrt{3}x)(1 +\sqrt{3}x)}= \frac{\sqrt{3}(1 -\sqrt{3}x)}{3(1 + \sqrt{3}x)(1 -\sqrt{3}x)}

C.E. 1 - \sqrt{3}x ≠ 0  ⇒ x ≠\frac{1}{\sqrt{3}} ⇒ x ≠ \frac{\sqrt{3}}{3}

1 + \sqrt{3}x≠ 0⇒ x ≠ –\frac{1}{\sqrt{3}}⇒ x ≠ – \frac{\sqrt{3}}{3}

    

 1 +\sqrt{3}x+3(2\sqrt{3})= \sqrt{3}(1 -\sqrt{3}x)

    

 1 +\sqrt{3}x+6\sqrt{3}= \sqrt{3} -3x

    
\sqrt{3}x+ 3x = \sqrt{3} -1   -6\sqrt{3}

    
x(\sqrt{3}+ 3) =-1   -5\sqrt{3}

x =     

\frac{-1   -5\sqrt{3}}{\sqrt{3}+ 3} =     

\frac{-1   -5\sqrt{3}}{\sqrt{3}+ 3}•     

\frac{\sqrt{3}- 3}{\sqrt{3}- 3}

x =     

\frac{-\sqrt{3}+ 3-15+15 \sqrt{3}}{3 - 9} =     

\frac{14\sqrt{3}-12 }{-6} =     

\frac{2(7\sqrt{3}-6) }{-6}

x =     

-\frac{7\sqrt{3}-6 }{3} =     

\frac{-7\sqrt{3}+6 }{3}   questa soluzione è accettabile perchè è diversa da +\frac{\sqrt{3}}{3} e – \frac{\sqrt{3}}{3}

8) \frac{5x - \sqrt{2}}{5x+\sqrt{2}} - \frac{5x + \sqrt{2}}{5x-\sqrt{2} = \frac{2\sqrt{2}(\sqrt{2}- 9x)}{25x ^{2}-2}           

\frac{5x - \sqrt{2}}{5x + \sqrt{2}} - \frac{5x + \sqrt{2}}{5x - \sqrt{2}} = \frac{4 -18\sqrt{2}x}{(5x - \sqrt{2})(5x - \sqrt{2})}

\frac{(5x - \sqrt{2})(5x - \sqrt{2})-(5x + \sqrt{2})(5x + \sqrt{2})}{(5x + \sqrt{2})(5x - \sqrt{2})}  = \frac{4 -18\sqrt{2}x}{(5x - \sqrt{2})(5x - \sqrt{2})}

\frac{(25x ^{2}+2  - 10\sqrt{2}x)-(25x ^{2}+2  + 10\sqrt{2}x)}{(5x + \sqrt{2})(5x - \sqrt{2})}  = \frac{4 -18\sqrt{2}x}{(5x - \sqrt{2})(5x - \sqrt{2})}

C.E.

5x – \sqrt{2} ≠0 ⇒   x  ≠ \frac{\sqrt{2}}{5}

5x + \sqrt{2} ≠0 ⇒   x  ≠ –  \frac{\sqrt{2}}{5}


25x ^{2}+2  - 10\sqrt{2}x-25x ^{2}-2  - 10\sqrt{2}x = 4 -18\sqrt{2}x


25x ^{2} -25x ^{2}  - 10\sqrt{2}x - 10\sqrt{2}x +18\sqrt{2}x  = 4 -2+2

-2\sqrt{2}x= 4

x = \frac{-4}{2\sqrt{2}} = \frac{-2}{\sqrt{2}}

x =  \frac{-2}{\sqrt{2}}  • \frac{\sqrt{2}}{\sqrt{2}} = \frac{-2\sqrt{2}}{2} = -\sqrt{2}

x = -\sqrt{2}  La soluzione è accettabile perchè diversa da ± \frac{\sqrt{2}}{5}

9) x(x + 1) + 1 + \sqrt{2} > x² + \sqrt{2} (x + 1)

  +x + 1 + \sqrt{2} > \sqrt{2} x +\sqrt{2} 

x + 1 >\sqrt{2}  x

x –  \sqrt{2} x > – 1    ⇒ x(1 – \sqrt{2} )> -1

x < \frac{-1}{1-\sqrt{2}}  quindi x < \frac{-1}{1-\sqrt{2}} • \frac{1+\sqrt{2}}{1+\sqrt{2}}

x < -\frac{1+\sqrt{2}}{-1}

x < 1+\sqrt{2}

10) (3\sqrt{3} + 2\sqrt{2})(x + \sqrt{3}) – (x + 4\sqrt{2})(\sqrt{3} + 2\sqrt{2}) < – 7

3\sqrt{3}x + 9 + 2\sqrt{2}x + 2\sqrt{6 }- (\sqrt{3}x + 2 \sqrt{2}x + 4 \sqrt{6} + 16) < -7

3\sqrt{3}x + 9 + 2\sqrt{2}x + 2\sqrt{6 }- \sqrt{3}x - 2 \sqrt{2}x - 4 \sqrt{6} - 16 < -7


3\sqrt{3}x + 2\sqrt{2}x - \sqrt{3}x- 2 \sqrt{2}x < -7 -9 -2\sqrt{6}+4\sqrt{6}+16

2\sqrt{3}x < 2\sqrt{6}

x  < \frac{2\sqrt{6}}{2\sqrt{3}}  ⇒ x <\frac{2\sqrt{3}\sqrt{2}}{2\sqrt{3}}  ⇒  x <\sqrt{2}

11) \frac{x - \sqrt{7}}{2\sqrt{7}} - \frac{2x - 3\sqrt{7}}{4\sqrt{7}}+ \frac{5x - \sqrt{7}}{3\sqrt{7}} > \frac{7x + 12\sqrt{7}}{12\sqrt{7}}     

\frac{x - \sqrt{7}}{2\sqrt{7}} - \frac{2x - 3\sqrt{7}}{4\sqrt{7}}+ \frac{5x - \sqrt{7}}{3\sqrt{7}} - \frac{7x-12\sqrt{7}}{12\sqrt{7}}     > 0

\frac{6(x -\sqrt{7})- 3(2x-3\sqrt{7})+4(5x-\sqrt{7})-7x-12\sqrt{7}}{12\sqrt{7}}     > 0

6x -6\sqrt{7}- 6x+9\sqrt{7}+20x-4\sqrt{7}-7x-12\sqrt{7} > 0

6x – 6x + 20x – 7x > 
12\sqrt{7}+6\sqrt{7}-9\sqrt{7}+4\sqrt{7}

13x > 13\sqrt{7}  ⇒ x  > \frac{13\sqrt{7}}{13}

x > \sqrt{7}

13

sistema-con-radicali

sistema-con-radicali

 

x = 2\sqrt{5};  y= -\sqrt{10}

sistema-con-radicali

sistema-con-radicali

risultato:x = 3 ;   y= 2\sqrt{3}

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