1)
A=\(\sqrt{\dfrac{9.12\left(a-2\right)^2}{27}}\)
=\(\sqrt{4\left(a-2\right)^2}\)
=\(2\left(a-2\right)\)
=2a-4
B=
B=\(\sqrt{\dfrac{ab\left(a-b\right)^2}{\left(a-b\right)^2}}\)
=\(\sqrt{ab}\)
2)
x\(^2\)=\(\left(\sqrt{\dfrac{2}{5}}+\sqrt{\dfrac{5}{2}}\right)\)=\(\dfrac{2}{5}+\dfrac{5}{2}+1\)
thay x\(^2\),x vào rồi tính. mình lười bấm quá.
1) \(A=3\cdot\sqrt{\dfrac{12\left(a-2\right)^2}{27}}\)
\(A=3\cdot\sqrt{\dfrac{4\cdot\left(a-2\right)^2}{9}}\)
\(A=3\cdot\dfrac{2\cdot\left(a-2\right)}{3}\)
\(A=\dfrac{6a-12}{3}=\dfrac{3\cdot\left(2a-4\right)}{3}=2a-4\)
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\(B=\left(a-b\right)\sqrt{\dfrac{ab}{\left(a-b\right)^2}}\)
\(B=\sqrt{\dfrac{ab\cdot\left(a-b\right)^2}{\left(a-b\right)^2}}=\sqrt{ab}\) (Đưa thừa số vào trong dấu căn)
2) \(A=x^2-x\sqrt{10}\)
Thay \(x=\sqrt{\dfrac{2}{5}}+\sqrt{\dfrac{5}{2}},x^2=\dfrac{49}{10}\) vào tính là xong