Toán Lớp 9: Rút gọn biểu thức
a)A=4/ √3-1+ 7/3- √2 -2 √3
b) 1/2 √x-2 – 1/2 √x+2 + √x/1-x
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Toán Lớp 9: Rút gọn biểu thức
a)A=4/ √3-1+ 7/3- √2 -2 √3
b) 1/2 √x-2 – 1/2 √x+2 + √x/1-x
Comments ( 1 )
a)A = \sqrt 3 + 4\\
b) – \dfrac{1}{{\sqrt x + 1}}
\end{array}\)
a)A = \dfrac{4}{{\sqrt 3 – 1}} + \dfrac{7}{{3 – \sqrt 2 }} – 2\sqrt 3 \\
= \dfrac{{4\left( {\sqrt 3 + 1} \right)}}{{3 – 1}} + \dfrac{{7\left( {3 + \sqrt 2 } \right)}}{{9 – 2}} – 2\sqrt 3 \\
= 2\left( {\sqrt 3 + 1} \right) + \sqrt 3 + 2 – 2\sqrt 3 \\
= 2\sqrt 3 + 2 – \sqrt 3 + 2\\
= \sqrt 3 + 4\\
b)DK:x \ge 0;x \ne 1\\
\dfrac{1}{{2\sqrt x – 2}} – \dfrac{1}{{2\sqrt x + 2}} + \dfrac{{\sqrt x }}{{1 – x}}\\
= \dfrac{{\sqrt x + 1 – \sqrt x + 1 – 2\sqrt x }}{{2\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2 – 2\sqrt x }}{{2\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= – \dfrac{{2\left( {\sqrt x – 1} \right)}}{{2\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
= – \dfrac{1}{{\sqrt x + 1}}
\end{array}\)