Question

[pham bao chi 2k5]

10 tick cho 1 tuần nếu bn nào làm đúngggg

Toán 9 khó gê có bạn nào nghĩ như v không ? cần bạn ib ns chuyện

Giải

a/  \(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}\) với a lớn hơn bằng 0

b/ \(\sqrt{13a}.\sqrt{\frac{52}{a}}\) a>0

c/\(\sqrt{5a}.\sqrt{45a}-3a\)a lớn bằng 0

d/ \(\left(3-ã\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)

Answer 1

[Ahwi]

a/   \(\sqrt{\frac{2a}{3}}\cdot\sqrt{\frac{3a}{8}}\)

\(=\sqrt{\frac{2a}{3}\cdot\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\sqrt{\frac{a^2}{2^2}}=\sqrt{\left(\frac{a}{2}\right)^2}=\left|\frac{a}{2}\right|\)

mak ta có \(a\ge0\)

\(\Rightarrow\left|\frac{a}{2}\right|=\frac{a}{2}\)\(\Rightarrow\sqrt{\frac{2a}{3}}\cdot\sqrt{\frac{3a}{8}}=\frac{a}{2}\)

b/ \(\sqrt{13a}\cdot\sqrt{\frac{52}{a}}\)

\(=\sqrt{13a\cdot\frac{52}{a}}=\sqrt{\frac{13a\cdot52}{a}}=\sqrt{13\cdot52}=\sqrt{13\cdot13\cdot4}=\sqrt{13^2\cdot2^2}=\sqrt{\left(13\cdot2\right)^2}=13\cdot2=26\)

c/ \(\sqrt{5a}\cdot\sqrt{45}-3a\)

\(=\sqrt{5a\cdot45a}-3a=\sqrt{5a\cdot5a\cdot9}-3a\)

                                        \(=\sqrt{5^2\cdot a^2\cdot3^2}-3a=\left|5\cdot a\cdot3\right|-3a\)

                                                                                      \(=15\left|a\right|-3a\)

 Có \(a\ge0\Rightarrow\left|a\right|=a\)

\(\Rightarrow15\left|a\right|-3a=15a-3a=12a\)

\(\Rightarrow\sqrt{5a}\cdot\sqrt{45}-3a=12a\)

  d/ \(\left(3-a\right)^2-\sqrt{0,2}\cdot\sqrt{180a^2}\)

\(=\left(3-a\right)^2-\sqrt{0,2\cdot180a^2}\)

\(=\left(3-a\right)^2-\sqrt{0,2\cdot9\cdot2\cdot10\cdot a^2}\)

\(=\left(3-a\right)^2-\sqrt{4\cdot9\cdot a^2}\)

\(=\left(3-a\right)^2-\sqrt{2^2\cdot3^2\cdot a^2}\)

\(=\left(3-a\right)^2-\left|2\cdot3\cdot a\right|\)

\(=\left(3-a\right)^2-6\left|a\right|=9-6a+a^2-6\left|a\right|\)

Chia làm 2 Trường Hợp:

 + TH1 : \(9-6a+a^2-6a=9-12a+a^2\left(a\ge0\right)\)

+  TH2 : \(9-6a+a^2-\left(-6a\right)=9+a^2\left(a< 0\right)\)