1.thực hiện phép tính
a.(0,125).(-3.7).(-2)\(^3\)
b.\(\sqrt{36}.\sqrt{\frac{25}{16}}+\frac{1}{4}\)
c.\(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1^2_5\)
d.\(0,1.\sqrt{225}.\sqrt{\frac{1}{4}}\)
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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?
\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{1}{18}\)
\(b,2^8:2^5+3^3.2-12\)
\(=2^3+9.2-12\)
\(=8+18-12\)
\(=26-12\)
\(=14\)
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
\(\sqrt{\frac{1}{9}+\frac{1}{16}}\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}\)
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 \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)
\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)
\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)
\(\sqrt{\frac{25}{4}}+\left(\sqrt{\frac{1}{2}}\right)^2:\left(\frac{-\sqrt{9}}{4}\right).\sqrt{\frac{16}{81}}-4^2-\left(-2\right)^3\)
\(=\frac{5}{2}+\frac{1}{2}:\frac{-3}{4}.\frac{4}{9}-16+8\)
\(=\frac{5}{2}-\frac{8}{27}-8\)
\(=\frac{-313}{54}\)
Bài 1:
a) Ta có: \(\left(0.125\right)\cdot\left(-3\cdot7\right)\cdot\left(-2\right)^3\)
\(=\frac{1}{8}\cdot\left(-21\right)\cdot\left(-8\right)\)
\(=\frac{1}{8}\cdot168\)
\(=21\)
b) Ta có: \(\sqrt{36}\cdot\sqrt{\frac{25}{16}}+\frac{1}{4}\)
\(=\sqrt{36\cdot\frac{25}{16}}+\frac{1}{4}\)
\(=\sqrt{\frac{225}{4}}+\frac{1}{4}\)
\(=\frac{15}{2}+\frac{1}{4}\)
\(=\frac{31}{4}\)
c) Ta có: \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}=-1\)
d) Ta có: \(0,1\cdot\sqrt{225}\cdot\sqrt{\frac{1}{4}}\)
\(=0,1\cdot15\cdot\frac{1}{2}=\frac{3}{4}\)
hay quá cảm ơn bạn nhé