Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,=\(\dfrac{\left(2-\dfrac{1}{3}+\dfrac{1}{4}\right).12}{\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right).12}\)+\(\dfrac{\left(\dfrac{3}{5}-\dfrac{1}{4}+\dfrac{1}{2}\right).20}{\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{2}{5}\right).20}\)
=\(\dfrac{24-4+3}{24+2-3}\) +\(\dfrac{12-5+10}{10+15-8}\)(nhân từng số hạng với 12;20)
=\(\dfrac{23}{23}\)+\(\dfrac{17}{17}\) =1+1=2
b,=\(\dfrac{5.\left(\dfrac{1}{79}\right)+5.\left(\dfrac{1}{83}\right)+\dfrac{1}{17}}{17.\left(\dfrac{1}{79}\right)+17.\left(\dfrac{1}{83}\right)+\dfrac{1}{5}}\)=\(\dfrac{5.\left(\dfrac{1}{79}+\dfrac{1}{83}\right)+\dfrac{1}{17}}{17.\left(\dfrac{1}{79}+\dfrac{1}{83}\right)+\dfrac{1}{5}}\)
= 
= 
= 
= 
. 
a) \(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{6}{14}+\dfrac{7}{17}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{13^2}{12^2}=\dfrac{169}{144}\)
b)\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2=\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2=\left(\dfrac{-1}{12}\right)^2=\dfrac{\left(-1\right)^2}{12^2}=\dfrac{1}{144}\)
c)\(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.5^4.2^8}{5^{10}.2^{10}}=\dfrac{5^8.2^8}{5^8.5^2.2^8.2^2}=\dfrac{1}{5^2.2^2}=\dfrac{1}{25.4}=\dfrac{1}{100}\)
d)\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4=\dfrac{\left(-10\right)^5.\left(-6\right)^4}{3^5.5^4}=\dfrac{\left(-2\right)^5.5^5.2^4.3^4}{3^4.3.5^4}=\dfrac{\left(-2\right)^5.5.5^42^4}{3.5^4}=\dfrac{\left(-2\right)^5.5.2^4}{3}=\dfrac{-2560}{3}=-853\dfrac{1}{3}\)
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
a: \(\Leftrightarrow\dfrac{-13}{3}\cdot\dfrac{1}{3}< =x< =\dfrac{-2}{5}\cdot\dfrac{-11}{12}\)
=>-13/9<=x<=22/60
mà x là số nguyên
nên \(x\in\left\{-1;0\right\}\)
b: \(\Leftrightarrow\dfrac{-22}{5}\cdot\dfrac{10}{3}< =x< =\dfrac{-13}{5}:\dfrac{7}{5}\)
=>-44/3<=x<=-13/7
mà x là số nguyên
nên \(x\in\left\{-14;-13;...;-2\right\}\)
Lời giải:
Ta có:
\(A=\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\) \(=\frac{3\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}=\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)
\(=\dfrac{-1}{8}\)
\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{71}{35}\)
\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)
\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)
\(=\dfrac{-433}{70}\)
a) Ta có: \(A=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+\dfrac{4}{7\cdot10}+...+\dfrac{4}{31\cdot34}\)
\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{31\cdot34}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{34}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{34}=\dfrac{22}{17}\)