cunei mă poate ajuta?,?

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Bună,
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[tex]a)x \sqrt{2} - \sqrt{8} = 2 - x \\ x \sqrt{2} - 2 \sqrt{2} = 2 - x \\ x \sqrt{2} + x = 2 + 2 \sqrt{2} \\ x( \sqrt{2 } + 1 ) = 2(1 + \sqrt{2} ) \\ x( \sqrt{2} + 1) = 2( \sqrt{2} + 1) \\ \blue{x = 2} [/tex]
[tex]b) \sqrt{2} (x - 2 \sqrt{2} ) - x( \sqrt{2} - 1) = 2x - 5 \\ x \sqrt{2} - 2 \sqrt{2} \times \sqrt{2} - x \sqrt{2} + x = 2x - 5 \\ x \sqrt{2} - 4 - x \sqrt{2} + x = 2x - 5 \\ - 4 + x = 2x - 5 \\ x - 2x = - 5 + 4 \\ - x = - 1 | \times ( - 1) \\ \green{x = 1} [/tex]
[tex]c)1 - \frac{x - 2}{ \sqrt{2} } = \sqrt{2} \\ 1 - \frac{ \sqrt{2}(x - 2) }{ \sqrt{2} \times \sqrt{2} } = \sqrt{2} \\ 1 - \frac{x \sqrt{2} - 2 \sqrt{2} }{2} = \sqrt{2} | \times 2 \\ 2 - (x \sqrt{2} - 2 \sqrt{2}) = 2 \sqrt{2} \\ 2 - x \sqrt{2} + 2 \sqrt{2} = 2\sqrt{2} \\ 2 - x \sqrt{2} = 0 \\ - x \sqrt{2} = - 2 \\ x = \frac{2}{ \sqrt{2} } \\ x = \frac{2 \sqrt{2} }{2} \\ \purple{x = \sqrt{2} } [/tex]
[tex]d) \frac{x}{ \sqrt{2} } - \frac{x}{ \sqrt{3} } = \sqrt{3} - \sqrt{2} \\ \frac{x \sqrt{2} }{2} - \frac{x \sqrt{3} }{3} = \sqrt{3} - \sqrt{2} \: | \times 6 \\ 3x \sqrt{2} - 2x \sqrt{3} = 6 \sqrt{3} - 6 \sqrt{2} \\ (3 \sqrt{2} - 2 \sqrt{3} )x = 6 \sqrt{3} - 6 \sqrt{2} \\ x = \frac{6 \sqrt{3} - 6 \sqrt{2} }{3 \sqrt{2} - 2 \sqrt{3} } \\ x = \frac{(6 \sqrt{3} - 6 \sqrt{2} )(3 \sqrt{2} - 2 \sqrt{3}) }{9 \times 2 - 4 \times 3} \\ x = \frac{6( \sqrt{3} - \sqrt{2}) (3 \sqrt{2 } + 2 \sqrt{3} ) }{6} \\ x = ( \sqrt{3} - \sqrt{2} )(3 \sqrt{2} + 2 \sqrt{3} ) \\ x = \sqrt{3} (3 \sqrt{2} + 2 \sqrt{3} ) - \sqrt{2} (3 \sqrt{2} + 2 \sqrt{3} ) \\ x = 3 \sqrt{6} + 2 \times 3 - 3 \times 2 - 2 \sqrt{6} \\ x = 3 \sqrt{6} + 6 \times 6 - 2 \sqrt{6} \\ x = 3 \sqrt{6} - 2 \sqrt{6} \\ \orange{x = \sqrt{6} } [/tex]
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Sper că te-am ajutat.