Rút gọn đa thức
a) $2$ $\sqrt{a}^{2}$ $-$ $5$$a$  với a <0

b ) $\sqrt{25a}^{2}$ + 3a với a $\geq$ 0

c ) $\sqrt{9a}^{4}$ + $3a^{2}$ 

d) $5$$\sqrt{4a^{6}}$ - $3a^{3}$ với a<0
sửa 2 căn x –  Dưa Leo 15-07-16 08:12 PM
cái j kia –  Dưa Leo 15-07-16 08:12 PM
$2\sqrt{a^2}-5a =-2a-5a(vì : a<0) =-7a$
$\sqrt{25a^{2}}+3a=5a+3a(vì :a \geq 0)=8a$
$\sqrt{9a^{4}}+3a^2=3a^2+3a^2=6a^2$
$5\sqrt{4a^6}-3a^3=-5. 2a^3-3a^3(vì:a<0)=-10a^3-3a^3=-13a^3$
chi tiết v~ –  Blood 16-07-16 08:20 AM
được nès hay zạ –  Dưa Leo 15-07-16 08:56 PM
a)$ 2 \sqrt{a}^{2} -5a =-2a-5a=-7a $(vì a nhỏ hơn 0 )
b)  $\sqrt{25a}^{2}+3a=5a+3a=8a$ (vì a lớn hơn hoặc =0 )
c$\sqrt{9a}^{4} + 3a^{2} = 3a^{2}+3a^{2}=6a^{2}$
d)$5 \sqrt{4a}^{6} - 3a^{3} = -5.2a^{3} -3a^{3}=-13a^{3}$( như câu a)
ghi lộn thôi –  Dưa Leo 08-09-17 06:19 AM
E K PÍT A Ý LM Z –  Yêu Tatoo 16-07-16 09:21 PM
sao lại đánh như thế này –  AKIRA 16-07-16 09:37 AM
ặc sửa lại cái bài đi ông –  Băng Băng 15-07-16 11:38 PM

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