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)

risultato: x = $2\sqrt{5}$ ; y= $-\sqrt{10}$

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

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

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

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

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

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

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

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

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

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