\(\sqrt{10}A=\sqrt{10}\left(\sqrt{0,1}+\sqrt{0,9}+\sqrt{6,4}+\sqrt{0,4}+\sqrt{44,1}\right)\)

\(=\sqrt{1}+\sqrt{9}+\sqrt{64}+\sqrt{4}+\sqrt{441}\)

\(=1+3+8+2+21=35\)

\(\Rightarrow A=\frac{35}{\sqrt{10}}\)

\(=4-\dfrac{12}{5}\sqrt{5}-6+\dfrac{18}{5}\sqrt{5}+2\sqrt{5}-6\)

\(=-8+\dfrac{16}{5}\sqrt{5}\)

a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)

b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)

\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)

( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))

c, Với \(a\ge0;a\ne1\)

\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)

\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)

Lời giải:

\(\sqrt[-0,4]{(-0,4)^2}=\sqrt[-0,4]{\frac{4}{25}}=(\frac{4}{25})^{\frac{1}{-0,4}}=(\frac{4}{25})^{\frac{-5}{2}}=\frac{1}{(\frac{4}{25})^{\frac{5}{2}}}=\frac{1}{\sqrt{(\frac{4}{25})^5}}=\frac{1}{(\frac{2}{5})^5}=\frac{3125}{32}\)

\(\sqrt[-0.4]{\left(-0.4\right)^2}=\sqrt[-0.4]{\dfrac{4}{25}}=\left(\dfrac{4}{25}\right)^{\dfrac{1}{-0.4}}=\dfrac{3125}{32}\)

\(\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0.4}\right)\)

\(=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}-3\sqrt{0.4}\right)\)

\(=2\sqrt{5}-6-2+\frac{6\sqrt{5}}{5}\)

\(=\frac{16\sqrt{5}-40}{5}\)

\(=\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0.4}\right)\)

\(=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}-3\sqrt{0.4}\right)\)

\(=2\sqrt{5}-6-2+\frac{6\sqrt{5}}{5}=\frac{16\sqrt{5}-40}{5}\)

2.5=10/4

0.4*2.5=0.4*10/4=4/4=1

3,04*2,5*0,4 = 3,04 nha bạn

úm ba la xin tích

là 3.04