Bài 1: Thực hiện phép tính a) \(\frac{3}{5}-\frac{3}{5}\left(\frac{4}{3}-\frac{1}{2}\right)^{^2}\) b) \(\sqrt{\frac{9}{100}}-\sqrt{0,25}+1,2\)
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a) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-2+\sqrt{3}\)
\(=\frac{\sqrt{3}.\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}.\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-2+\sqrt{3}\)
\(=\sqrt{3}+2+\sqrt{2}-2+\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{2}\)
b) \(\frac{-3}{2}.\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{5-4\sqrt{5}+4}+\sqrt{4^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{4^2}.\sqrt{\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\left|\sqrt{5}-2\right|+4.\left|1+\sqrt{5}\right|\)
\(=\frac{-3}{2}.\left(\sqrt{5}-2\right)+4\left(1+\sqrt{5}\right)\)
\(=\frac{-3\sqrt{5}}{2}+3+4+4\sqrt{5}\)
\(=\frac{-3\sqrt{5}}{2}+4\sqrt{5}+7\)
\(=\frac{-3\sqrt{5}}{2}+\frac{8\sqrt{5}}{2}+\frac{14}{2}\)
\(=\frac{-3\sqrt{5}+8\sqrt{5}+14}{2}=\frac{14+5\sqrt{5}}{2}\)
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\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{1}{18}\)
\(b,2^8:2^5+3^3.2-12\)
\(=2^3+9.2-12\)
\(=8+18-12\)
\(=26-12\)
\(=14\)
Câu c,d em chưa học nên không biết làm ạ, mong mọi người thông cảm!!!
Sửa lại câu b
\(=2^3+27.2-12\)
\(=8+54-12\)
\(=62-12\)
\(=50\)
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1.a) \(\left(31\frac{6}{13}+5\frac{9}{41}\right)-36\frac{6}{13}=\left(31+\frac{6}{13}+5+\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)\)
\(=\left(36+\frac{6}{13}-\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)=\left(36+\frac{6}{13}\right)-\left(36+\frac{6}{13}\right)-\frac{9}{41}=-\frac{9}{41}\)
b) \(\frac{5}{3}+\left(-\frac{2}{7}\right)-\left(-1,2\right)-\left|1.4-0,2\right|\)
\(=\frac{5}{3}-\frac{2}{7}+1,2-1,2=\frac{29}{21}\)
c) \(0,25+\frac{3}{5}-\left(\frac{1}{8}-\frac{2}{5}+1\frac{1}{4}\right)+\left|\frac{3}{5}\right|\)
\(=\frac{1}{4}+\frac{3}{5}-\frac{1}{8}+\frac{2}{5}-1-\frac{1}{4}+\frac{3}{5}\)
\(=\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{3}{5}+\frac{2}{5}-1\right)+\frac{3}{5}-\frac{1}{8}=\frac{19}{40}\)
2) \(-\frac{3}{5}-x=0,75\)
=> \(-\frac{3}{5}-x=\frac{3}{4}\)
=> \(x=-\frac{3}{5}-\frac{3}{4}=\frac{-27}{20}\)
b) \(x+\frac{1}{3}=\frac{2}{5}-\left(-\frac{1}{3}\right)\)
=> \(x+\frac{1}{3}=\frac{2}{5}+\frac{1}{3}\)
=> \(x=\frac{2}{5}\)
c) |2x - 4| + 1 = 5
=> |2x - 4| = 4
<=> \(\orbr{\begin{cases}2x-4=4\\2x-4=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)
Giúp mình với nha cả nhả :<
Cả nhà làm vài ý thui cx được ạ :<
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Bài 2 :
\(a,\frac{3}{2}.\frac{4}{9}-\frac{1}{5}+\frac{2}{15}\)
\(=\frac{2}{3}-\frac{1}{5}+\frac{2}{15}\)
\(=\frac{10}{15}-\frac{3}{15}+\frac{2}{15}\)
\(=\frac{9}{15}=\frac{3}{5}\)
\(b,\left(\frac{1}{2}\right)^2.\frac{4}{3}+\frac{8}{5}:\frac{2}{5}\)
\(=\frac{1}{4}.\frac{4}{3}+\frac{8}{5}.\frac{5}{2}\)
\(=\frac{1}{3}+4\)
\(=\frac{13}{3}\)
\(c,\sqrt{196}-\sqrt{100}+\sqrt{\frac{9}{4}}\)
\(=14-10+\frac{3}{2}\)
\(=4+\frac{3}{2}\)
\(=\frac{11}{2}\)
Học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1
\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)
\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)
\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)
\(=\frac{9}{25}+\frac{8}{9}-1\)
\(=\frac{56}{225}\)
\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)
\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)
\(=1:\frac{4}{3}=\frac{3}{4}\)
Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v
\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)
\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)
\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)
\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)
\(=-\frac{1}{2}\)
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\(\left(\sqrt{4,5}-\frac{1}{2}.\sqrt{72}+5\sqrt{\frac{1}{2}}\right).\left(42\sqrt{\frac{25}{6}}-10\sqrt{\frac{3}{2}}-12\sqrt{\frac{98}{3}}\right)\)
=\(\left(\frac{3\sqrt{2}}{2}-3\sqrt{2}+\frac{5\sqrt{2}}{2}\right).\left(35\sqrt{6}-5\sqrt{6}-28\sqrt{6}\right)\)
=\(\left(\frac{3\sqrt{2}-6\sqrt{2}+5\sqrt{2}}{2}\right).2\sqrt{6}\)
=\(2\sqrt{2}.\sqrt{6}=4\sqrt{3}\)
a) \(\frac{3}{5}-\frac{3}{5}(\frac{8}{6}-\frac{3}{6})=\frac{3}{5}-\frac{3}{5}\times\frac{5}{6}=\frac{3}{5}-\frac{1}{2}=\frac{6}{10}-\frac{5}{10}=\frac{1}{10}\)\(\frac{1}{10}\)
b)\(\sqrt{\frac{9}{100}}-\sqrt{0,25}+1,2=\frac{3}{10}-\frac{1}{2}+\frac{6}{5}=\frac{3}{10}-\frac{5}{10}+\frac{12}{10}=1\)
9/10-0,5+1,2
0,4+1,2
1,6
3/5-3/5.(5/6)^2
3/5-3/5.25/36
3/5.(1-25/36)
3/5.11/36
11/60