a)(2^4.7^3+7^4.36) : (7^3.20-7^3.16)
b)(8^4.13+4^6.27):(2^13.15-2^12.6
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2 tháng 2 2020
Ta có
\(2^8.4.13=2^9.26\)
\(2^7.8.65=2^9.130\)
Nên \(2^8.4.13+2^7.8.65=2^9.156\)
Mà \(2^9:2^9=1\)
\(156:3=52\)
Do đó A = 52
MV
27 tháng 4 2017
1)
a)
\(\dfrac{-5}{11}\cdot\dfrac{4}{7}+\dfrac{-5}{11}\cdot\dfrac{3}{7}-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot1-\dfrac{8}{11}\\ =\dfrac{-5}{11}-\dfrac{8}{11}\\ =\dfrac{-5}{11}+\dfrac{-8}{11}\\ =\dfrac{-13}{11}\)
b)
\(\left(\dfrac{2}{9}:\dfrac{5}{3}+\dfrac{1}{3}:\dfrac{5}{3}\right)^2-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =\left(\dfrac{2}{9}\cdot\dfrac{3}{5}+\dfrac{1}{3}\cdot\dfrac{3}{5}\right)^2-\left(\dfrac{-7}{24}\right)\\ =\left[\dfrac{3}{5}\cdot\left(\dfrac{2}{9}+\dfrac{1}{3}\right)\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{3}{5}\cdot\dfrac{5}{9}\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{1}{3}\right]^2+\dfrac{7}{24}\\ =\dfrac{1}{9}+\dfrac{7}{24}\\ =\dfrac{29}{72}\)
c) \(14-\left|\dfrac{-3}{4}\right|-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =14-\dfrac{3}{4}-\left(\dfrac{-7}{24}\right)\\ =14+\dfrac{-3}{4}+\dfrac{7}{24}\\ =13\dfrac{13}{24}\)
T
1
3 tháng 4 2017
a)\(\frac{5}{7}:X=1-\frac{4}{7}=\frac{3}{7}\)
\(X=\frac{5}{7}:\frac{3}{7}=\frac{5}{3}\)
b)
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{19}-\frac{1}{21}\right)-X=\frac{4}{3}-\frac{221}{231}=\frac{29}{77}\)
\(\left(\frac{1}{11}-\frac{1}{21}\right)-X=\frac{29}{77}\)
\(X=\frac{10}{231}-\frac{29}{77}=-\frac{1}{3}\)
xem xong nhớ tích
25 tháng 10 2023
a: \(\sqrt{4\cdot36}=\sqrt{144}=12\)
b: \(\left(\sqrt{8}-3\sqrt{2}\right)\cdot\sqrt{2}\)
\(=\left(2\sqrt{2}-3\sqrt{2}\right)\cdot\sqrt{2}\)
\(=-\sqrt{2}\cdot\sqrt{2}=-2\)
c: \(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}=\dfrac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}=-\sqrt{7}\)
d: \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{\sqrt{5}-2}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)+2\left(\sqrt{5}+2\right)}{5-4}\)
\(=2\sqrt{5}-4+2\sqrt{5}+4=4\sqrt{5}\)
DT
2
NV
2 tháng 8 2016
\(A=\frac{a^2.a^7.b^2.c^8}{a^5.b^3.\left(-c\right)^4}=\frac{a^9.b^2.c^8}{a^5.b^3.c^4}=\frac{a^4.c^4}{b}\)
\(B=\left(\frac{3}{7}\right)^5.\frac{\left(25\right)^4.7^5}{15^4.20^2}=\frac{3^5.\left(5^2\right)^4.7^5}{7^5.\left(3.5\right)^4.\left(4.5\right)^2}=\frac{3^5.5^8.7^5}{7^5.3^4.5^4.4^2.5^2}\)
\(=\frac{3^5.5^8.7^5}{7^5.3^4.5^6.4^2}=\frac{3.5^2}{4^2}=\frac{75}{16}\)
\(91924\)\(:1372=67\)
\(163840:98304=\frac{5}{3}\)
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