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The rationalising factor of\sqrt[5]{a^{2} b^{3} c^{4}}\begin{array}{ll}{\sqrt[5]{a^{3} b^{2} c}} & {\text { (B) } \sqrt[4]{a^{3} b^{2} c}} \\ {\sqrt[3]{a^{3} b^{2} c}} & {\text { (D) } \sqrt{a^{3} b^{2} c}}\end{array} |
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Answer» A) as multiplying and dividing by root over(a³b²c²) will eliminate root over 5 in numerator |
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