a) \(\sqrt{0,09}-\sqrt{0,64}=\frac{-1}{2}=-0,5\)

b) \(0,1\cdot\sqrt{225}-\sqrt{\frac{1}{4}}=0,1\cdot15-\frac{1}{2}=1\)

c) \(\sqrt{0,36}\cdot\sqrt{\frac{25}{16}+\frac{1}{4}}=\frac{3\sqrt{29}}{20}\)

d) đề baì có sai ko ban?

a) = \(\frac{7}{2}\)

b) = \(\frac{643}{64}\)

c) = 0

\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)

\(=\frac{-5}{9}.\frac{-1}{10}\)

\(=\frac{1}{18}\)

\(b,2^8:2^5+3^3.2-12\)

\(=2^3+9.2-12\)

\(=8+18-12\)

\(=26-12\)

\(=14\)

Câu c,d em chưa học nên không biết làm ạ, mong mọi người thông cảm!!!

Sửa lại câu b

\(=2^3+27.2-12\)

\(=8+54-12\)

\(=62-12\)

\(=50\)

= 0,6 : 5/4 + 1/4 + 2/9 : 5/9  - 1/4

 = 3/5 . 4/5  + 2/9 . 9/5

= 12/25  + 2/5

= 22/25

\(\sqrt{\frac{1}{9}+\frac{1}{16}}\)

\(=\frac{1}{3}+\frac{1}{4}\)

\(=\frac{7}{12}\)

\(\sqrt{4+36+81}\)

\(=\sqrt{121}\)

\(=\pm11\)

\(\sqrt{1^3}+2^3\)

\(=\sqrt{1}+8\)

\(=1+8\)

\(=9\)

a) \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)

\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)

\(=\frac{2}{5}-\frac{7}{5}\)

\(=-1.\)

b) \(\sqrt{36}.\sqrt{\frac{25}{16}}+\frac{1}{4}\)

\(=6.\frac{5}{4}+\frac{1}{4}\)

\(=\frac{15}{2}+\frac{1}{4}\)

\(=\frac{31}{4}.\)

c) \(1\frac{1}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)

\(=\frac{3}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)

\(=\frac{3}{2}+\left(-\frac{9}{14}\right)\)

\(=\frac{6}{7}.\)

d) \(1,17-0,4.\left(\frac{1}{2}\right)^2-\frac{1}{-5}\)

\(=\frac{117}{100}-\frac{2}{5}.\frac{1}{4}-\left(-\frac{1}{5}\right)\)

\(=\frac{117}{100}-\frac{1}{10}+\frac{1}{5}\)

\(=\frac{107}{100}+\frac{1}{5}\)

\(=\frac{127}{100}.\)

Chúc bạn học tốt!

a, \(\frac{4}{81}:\sqrt{\frac{25}{81}-1\frac{2}{5}}\)

\(\Rightarrow\frac{4}{81}:\frac{5}{9}-\frac{7}{5}\)

\(\Rightarrow\frac{4}{81}.\frac{9}{5}-\frac{7}{5}\)

\(\Rightarrow\frac{4}{9}.\frac{1}{5}-\frac{7}{5}\)

\(\Rightarrow\frac{-59}{45}\)

b,\(\sqrt{36}.\sqrt{\frac{25}{16}+\frac{1}{4}}\)

\(\Rightarrow6.\frac{5}{4}+\frac{1}{4}\)

\(\Rightarrow\frac{15}{2}+\frac{1}{4}\)

\(\Rightarrow\frac{31}{4}\)

c,\(1\frac{1}{2}+\frac{4}{7}:\frac{-8}{9}\)

\(\Rightarrow\frac{3}{2}-\frac{4}{7}.\frac{-8}{9}\)

\(\Rightarrow\frac{3}{2}-\frac{9}{14}\)

\(\Rightarrow\frac{6}{7}\)

d, \(1,17-\left(0,4.\frac{1}{2}\right)^2-\frac{1}{5}\)

\(\Rightarrow\frac{117}{100}-\left(\frac{1}{5}\right)^2-\frac{1}{5}\)

\(\Rightarrow\frac{117}{100}-\frac{1}{25}-\frac{1}{5}\)

\(\Rightarrow\frac{93}{100}\)

a, ĐK :a >= 3

\(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{\left(a-3\right)\left(a+3\right)}+6\sqrt{\left(a-3\right)\left(a+3\right)}=0\)

\(\Leftrightarrow\sqrt{a-3}\left(5-\frac{14}{3}-\sqrt{a+3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{a-3}=0\\\sqrt{a+3}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{2}{9}\left(loai\right)\end{cases}}\)

b, \(ĐK:x\ge-\frac{1}{2}\)

\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)

\(\Leftrightarrow\frac{4}{3}\sqrt{2x+1}=4\)

\(\Leftrightarrow\sqrt{2x+1}=3\)

\(\Leftrightarrow x=4\left(tm\right)\)

a) đk: \(a\ge3\)

pt \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)

\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)

\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)

\(\sqrt{\frac{25}{4}}+\left(\sqrt{\frac{1}{2}}\right)^2:\left(\frac{-\sqrt{9}}{4}\right).\sqrt{\frac{16}{81}}-4^2-\left(-2\right)^3\)

\(=\frac{5}{2}+\frac{1}{2}:\frac{-3}{4}.\frac{4}{9}-16+8\)

\(=\frac{5}{2}-\frac{8}{27}-8\)

\(=\frac{-313}{54}\)

-313/54

-9/25 nghĩa là (âm chín phần hai mươi năm ) nha bạn 

Bài 2:

a) Ta có: \(\sqrt{4}+\sqrt{25}+\sqrt{64}+...+\sqrt{2500}\)

\(=2+5+8+...+50\)

Số hạng tử là: \(\frac{50-2}{3}+1=\frac{48}{3}+1=16+1=17\)(số)

Tổng của dãy số là: \(\left(50+2\right)\cdot\frac{17}{2}=\frac{52\cdot17}{2}=26\cdot17=442\)

b) Ta có: \(\sqrt{1}+\sqrt{25}+\sqrt{81}+...+\sqrt{9409}\)

\(=1+5+9+...+97\)

Số hạng tử là:

\(\frac{97-1}{4}+1=\frac{96}{4}+1=24+1=25\)(số)

Tổng của dãy số là: \(\left(97+1\right)\cdot\frac{25}{2}=\frac{98\cdot25}{2}=49\cdot25=1225\)