Câu hỏi của Anna Truong

Question

Rút gọn:a/$\sqrt{-9a}-\sqrt{9+12a+4a^2}$     b/$1+\frac{3m}{m-2}\sqrt{m^2-4m+4}$c/$4x-\sqrt{9x^2+6x+1}$

Huy Hoang · Accepted Answer

$a)\sqrt{-9a}-\sqrt{9+12a+4a^2}$$==\sqrt{3^2.\left(-a\right)}-\sqrt{3^2-2.3.2a+\left(2a\right)^2}$$=3\sqrt{-a}-\sqrt{\left(3+2a\right)^2}$$=3\sqrt{a}-\left|3+2a\right|$$b)1+\frac{3m}{m-2}\sqrt{m^2-4m+4}$$=1+\frac{3m}{m-2}\sqrt{\left(m\right)^2-2.2m+2^2}$$=1+\frac{3m}{m-2}\sqrt{\left(m-2\right)^2}$$=1+\frac{3m}{m-2}|m-2|$$c)4x-\sqrt{9x^2+6x+1}$$=4x-\sqrt{\left(3x\right)^2+2.3x+1}$$=4x-\sqrt{\left(3x+1\right)^2}$$=4x-|3x+1|$Câu trả lời: $a)\sqrt{-9a}-\sqrt{9+12a+4a^2}$$==\sqrt{3^2.\left(-a\right)}-\sqrt{3^2-2.3.2a+\left(2a\right)^2}$$=3\sqrt{-a}-\sqrt{\left(3+2a\right)^2}$$=3\sqrt{a}-\left|3+2a\right|$$b)1+\frac{3m}{m-2}\sqrt{m^2-4m+4}$$=1+\frac{3m}{m-2}\sqrt{\left(m\right)^2-2.2m+2^2}$$=1+\frac{3m}{m-2}\sqrt{\left(m-2\right)^2}$$=1+\frac{3m}{m-2}|m-2|$$c)4x-\sqrt{9x^2+6x+1}$$=4x-\sqrt{\left(3x\right)^2+2.3x+1}$$=4x-\sqrt{\left(3x+1\right)^2}$$=4x-|3x+1|$

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