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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
\(=2^3+27.2-12\)
\(=8+54-12\)
\(=62-12\)
\(=50\)
e) ta có : \(E=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{1}=1\)
g) ta có : \(G=13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}=13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}\)
\(=13+30\sqrt{3+2\sqrt{2}}=13+30\sqrt{\left(\sqrt{2}+1\right)^2}=42+30\sqrt{2}\)
h) ta có : \(H=1+\sqrt{3+\sqrt{13+4\sqrt{3}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
\(=1+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{1-\sqrt{3-\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)
\(=1+\sqrt{4+2\sqrt{3}}+\sqrt{1-\sqrt{4-2\sqrt{3}}}=1+\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{1-\left(\sqrt{3}-1\right)^2}\)
\(=1+\sqrt{3}+1+\sqrt{2-\sqrt{3}}=2+\sqrt{3}+\dfrac{\sqrt{4-2\sqrt{3}}}{2}\)
\(=2+\sqrt{3}+\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}=2+\sqrt{3}+\dfrac{\sqrt{3}-1}{2}=\dfrac{3}{2}+\dfrac{3\sqrt{3}}{2}\)
cái câu mà bạn bảo kéo dài căn đến hết phải zầy o bn
\(\sqrt{3\sqrt{3\sqrt{3\sqrt{3\sqrt{...\sqrt{3}}}}}}\) nếu đúng thì bài này chỉ chứng mk giá trị của nó nhỏ hơn 3 mà thôi . bn xem lại đề nha
\(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}=\left|-13\right|\)
\(=-8+\frac{1}{2}.8-5+13\)
\(=4\)
\(\frac{1}{2}.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(-\frac{2012}{2013}\right)^0\)
\(=\frac{1}{2}.10-\frac{1}{4}+1\)
\(=5-\frac{5}{4}\)
\(=\frac{15}{4}\)
\(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+|-13|\)
\(=-12+\frac{1}{2}.8-5+13\)
\(=-12+4-5+13\)
\(=4\)
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)
\(B=\frac{300}{343}:\frac{1347}{343}\)
\(B=\frac{100}{449}\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)
\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)
\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)
\(A=\frac{-1}{2}+\frac{1710}{9}\)
\(A=\frac{-1}{2}+190\)
\(A=\frac{-1}{2}+\frac{380}{2}\)
\(A=\frac{379}{2}\)
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
\(\sqrt{10^2-6^2}-\sqrt{13^2-12^2}+\sqrt{13^2}-\sqrt{12^2}\)
\(=\sqrt{100-36}-\sqrt{169-144}+\sqrt{13^2}-\sqrt{12^2}\)
\(=\sqrt{64}-\sqrt{25}+\sqrt{13^2}-\sqrt{12^2}\)
\(=\sqrt{8^2}-\sqrt{5^2}+\sqrt{13^2}-\sqrt{12^2}\)
\(=8-5+13-12=4\)