Lời giải:

a)

$5\sqrt{25a^2}-25a=5\sqrt{(5a)^2}-25a=5|5a|-25a$

Với $a\leq 0$ thì $|5a|=-5a$. Do đó:

$5\sqrt{25a^2}-25a=-25a-25a=-50a$

b)

$\sqrt{16a^4}+6a^2=\sqrt{(4a^2)^2}+6a^2=|4a^2|+6a^2=4a^2+6a^2=10a^2$

\(5\sqrt{a}-4b\sqrt{25^3}+5a\sqrt{16ab^2}-2\sqrt{9a}\)

\(=5\sqrt{a}-4b.25a\sqrt{a}+5a.4b\sqrt{a}-6\sqrt{a}\)

\(=5\sqrt{a}-20ab\sqrt{a}+20ab\sqrt{a}-6\sqrt{a}\)

\(=-\sqrt{a}\)

Mình sữa đề 1 chút nha

\(5\sqrt{a}-4b\sqrt{25a^3}+5a\sqrt{16ab^2}-2\sqrt{9a}\)

\(=5\sqrt{a}-20ab\sqrt{a}+20ab\sqrt{a}-6\sqrt{a}\)

\(=-\sqrt{a}\)

toi biet nhung ko nho

\(5\sqrt{a}-3\sqrt{25a^3}+2\sqrt{36ab^2}-2\sqrt{9a}\\ =5\sqrt{a}-3\sqrt{25a^2.a}+2\sqrt{36b^2.a}-6\sqrt{a}\\ =-\sqrt{a}-3\left|5a\right|\sqrt{a}+2\left|66\right|\sqrt{a}\\ =-\sqrt{a}-15a\sqrt{a}+12b\sqrt{a}\)

1.\(5\sqrt{a}+6\sqrt{a.\frac{1}{4}}-\sqrt{a^2.\frac{4}{a}}+\sqrt{5}=5\sqrt{a}+6.\frac{1}{2}\sqrt{a}-2\sqrt{a}\)+\(\sqrt{5}\)

bạn tự làm nốt các câu này và làm tương tự các câu kia nhé!!Nếu khó chỗ nào hãy nhắn tin cho mk!! hihi

Thanks

\(\sqrt{\left(4-\sqrt{15}\right)^2}=\left|4-\sqrt{15}\right|=4-\sqrt{15}\)

\(\Rightarrow\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}=4-\sqrt{15}+\sqrt{15}=4\)

\(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

\(\sqrt{\left(1-\sqrt{3}\right)^2}=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)

\(\Rightarrow\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}=2-\sqrt{3}+\sqrt{3}-1=1\)

a) \(=5\left|a\right|+3a=5a+3a=8a\)

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

c) \(=5.2\left|a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)

\(2\sqrt{a}-\dfrac{5}{a}\cdot\sqrt{9a^3}+a\sqrt{\dfrac{4}{a}}-\dfrac{2}{a^2}+\sqrt{25a^2}\)

\(=2\sqrt{a}-\dfrac{5}{a}\cdot3a\sqrt{a}+a\cdot\dfrac{2}{\sqrt{a}}-\dfrac{2}{a^2}\cdot5a\)

\(=2\sqrt{a}-5\cdot3\sqrt{a}+\dfrac{2a}{\sqrt{a}}-\dfrac{2}{a}\cdot5\)

\(=2\sqrt{a}-15\sqrt{a}+2\sqrt{a}-\dfrac{10}{a}\)

\(=-11\sqrt{a}-\dfrac{10}{a}\)

\(=\dfrac{-11a\sqrt{a}-10}{a}\)