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23 tháng 12 2021

\(=2:\dfrac{3}{3}-\dfrac{1}{3}\)

\(=2\cdot\dfrac{2}{3}-\dfrac{1}{3}=1\)

23 tháng 12 2021

\(=2\cdot\dfrac{3}{2}-\dfrac{1}{3}=3-\dfrac{1}{3}=\dfrac{8}{3}\)

15 tháng 12 2021

\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)

\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)

15 tháng 12 2021

thanks nhìu

19 tháng 3 2021

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}:\sqrt{\dfrac{25}{9}}=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}:\dfrac{5}{3}\)

\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\dfrac{5}{3}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1-5\right)}.\dfrac{5}{3}=\dfrac{1-3}{1-5}.\dfrac{5}{3}=\dfrac{1}{2}.\dfrac{5}{3}=\dfrac{5}{6}\)

19 tháng 3 2021

\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\div\sqrt{\dfrac{25}{9}}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\div\dfrac{5}{4}\)

=\(\dfrac{2^{10}\cdot3^8\left(1-2\cdot3\right)}{2^{10}\cdot3^8\left(1+5\right)}\div\dfrac{5}{4}\)

=\(\dfrac{1-6}{1+5}\cdot\dfrac{4}{5}\)

=\(-\dfrac{5}{6}\cdot\dfrac{4}{5}\)

=\(-\dfrac{2}{3}\)

11 tháng 2 2018

\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)

\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)

\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)

\(B=\frac{300}{343}:\frac{1347}{343}\)

\(B=\frac{100}{449}\)

11 tháng 2 2018

\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)

\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)

\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)

\(A=\frac{-1}{2}+\frac{1710}{9}\)

\(A=\frac{-1}{2}+190\)

\(A=\frac{-1}{2}+\frac{380}{2}\)

\(A=\frac{379}{2}\)

14 tháng 9 2021

\(a.\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=-\dfrac{25}{6}\)

\(b.-\dfrac{1}{12}-\dfrac{5}{4}+\dfrac{1}{3}=-\dfrac{1}{12}-\dfrac{15}{12}+\dfrac{4}{12}=-1\)

14 tháng 9 2021

a) \(\dfrac{-7}{2}+\dfrac{3}{4}-\dfrac{17}{12}=\dfrac{-42}{12}+\dfrac{9}{12}-\dfrac{17}{12}=\dfrac{-42+9-17}{12}=-\dfrac{50}{12}=-\dfrac{25}{6}\)

b)\(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)=-\dfrac{2}{24}-\left(\dfrac{63}{24}-\dfrac{8}{24}\right)\)

\(=-\dfrac{2}{24}-\dfrac{63}{24}+\dfrac{8}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)

23 tháng 8 2023

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(\Rightarrow A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(\Rightarrow A=\left(7-6-5\right)-\dfrac{3}{4}-\dfrac{5}{4}+\dfrac{7}{4}+\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\)

\(\Rightarrow A=-4-\dfrac{1}{4}+0\)

\(\Rightarrow A=-\dfrac{17}{4}\)

Cách 2 tính trong dấu ngoặc đơn của A, bạn tự tính nhé.

24 tháng 8 2023

A = -17/4

24 tháng 12 2023

a: Sửa đề: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)

Để A là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3+4⋮\sqrt{x}-3\)

=>\(4⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)

=>\(\sqrt{x}\in\left\{4;2;5;1;7\right\}\)

=>\(x\in\left\{16;4;25;1;49\right\}\)

b: loading...

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1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25