\(\frac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)=\(\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}\left(\sqrt{3}+\sqrt{5}\right)}\)=\(\frac{\sqrt{2}}{\sqrt{7}}\)

Viết lại đề cho bạn nè
  \(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)

= \(\sqrt{9-6\sqrt{3}+3}-\sqrt{12-12\sqrt{3}+9}\)

= \(\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}\right)^2-2.3.\sqrt{12}+3^2}\)

= \(\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}-3\right)^2}\)

= |\(3-\sqrt{3}\)| - |\(\sqrt{12}-3\)|

= \(3-\sqrt{3}-\sqrt{12}+3\)

= \(6-\sqrt{3}-2\sqrt{3}\)

= \(6-3\sqrt{3}\)

toán lớp 9 mik mới học lớp 6 thôi

\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{9}+2\sqrt{21}\)

=\(\left(\sqrt{4}\sqrt{7}-\sqrt{7}-\sqrt{12}\right).3+2\sqrt{21}\)

=\(\left(2\sqrt{7}-\sqrt{7}-\sqrt{4}\sqrt{3}\right).3+2\sqrt{21}\)

=\(\left(\sqrt{7}-2\sqrt{3}\right).3+2\sqrt{21}\)

=\(3\sqrt{7}-6\sqrt{3}+2\sqrt{21}\)

đề có sai ko nhưng kết quả ra thế

\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{9}+2\sqrt{21}=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right).3+2\sqrt{21}=\left(\sqrt{7}-2\sqrt{3}\right).3+2\sqrt{21}=3\sqrt{7}-6\sqrt{3}+2\sqrt{21}\)

\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\frac{16a^4b^6}{128a^6b^6}}=\sqrt{\frac{1}{8a^2}}=\frac{\sqrt{1}}{\sqrt{8a^2}}=\frac{1}{\sqrt{2}\sqrt{4}\sqrt{a}}\)

=\(\frac{1}{2\sqrt{2}a}\)

tui hướng dẫn thui nha,,,,\(\sqrt{33-12\sqrt{6}}=\sqrt{24-2.2\sqrt{6}.3+9}=\sqrt{\left(2\sqrt{6}-3\right)^2}=2\sqrt{6}-3\)

ấy dễ ko???,,,bạn lm tương tự típ nhá,,,,

đk: x>=0; x khác 3

a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)

b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)

ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)

co ei giup tui vơi

-Nếu \(x\le2\) thì A=-81-25(2-x)+40=-81-50+25x+40=25x-91

-Nếu \(x\ge2\) thì A=-81-25(x-2)+40=-81-25x+50+40=9-25x

Rút gọn hử

A = -41-25.|2-x|

hết.