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1, \(\left(1,5.x-\frac{4}{5}\right).\left(\frac{1}{2019}-\frac{1}{2018}\right)\)\(=0\)
\(\Leftrightarrow\) \(1,5.x-\frac{4}{5}=0:\left(\frac{1}{2019}-\frac{1}{2018}\right)\)
\(1,5.x-\frac{4}{5}=0\)
\(1,5.x=0+\frac{4}{5}\)
\(1,5.x=\frac{4}{5}\)
\(x=\frac{4}{5}:1,5\)
\(x=\frac{4}{5}:\frac{15}{10}\)
\(x=\frac{4}{5}.\frac{10}{15}\)
\(\Rightarrow x=\frac{8}{15}\)
2, \(\frac{2x}{3}+\frac{1}{3}=\left|-\frac{2}{5}\right|\)
\(\Leftrightarrow\frac{2x+1}{3}=\frac{2}{5}\)
\(2x+1=\frac{2}{5}.3\)
\(2x+1=\frac{6}{5}\)
\(2x=\frac{6}{5}-1\)
\(2x=\frac{1}{5}\)
\(x=\frac{1}{5}:2\)
\(x=\frac{1}{5}.\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{10}\)
mk làm bài 1 thui,bài 2 chỉ qui đồng ms
3a/6 = 3b/4 => 3(a-b)/ (6-4) = 3.4,5/2= 13,5/2 =k
a = 2k=13,5
b = 4k/3 =9
Bài giải
Thay \(x=\frac{a}{m}\text{ ; }y=\frac{b}{m}\text{ ; }z=\frac{a+b}{m}\) vào \(P\) ta được :
\(P=\frac{\frac{a}{m}+\frac{b}{m}}{\frac{b}{m}+\frac{a+b}{m}}=\frac{\frac{a+m}{m}}{\frac{a+2b}{m}}=\frac{a+b}{m}\cdot\frac{m}{a+2b}=\frac{a+b}{a+2b}\)
Áp dụng :
\(\frac{\frac{1}{4}+\frac{1}{2}}{\frac{1}{2}+\frac{3}{4}}=\frac{\frac{3}{4}}{\frac{5}{4}}=\frac{3}{4}\cdot\frac{4}{5}=\frac{3}{5}\)
a, \(p=\frac{x+y}{y+z}=\frac{\frac{a}{m}+\frac{b}{m}}{\frac{b}{m}+\frac{a+b}{m}}=\frac{\frac{a+b}{m}}{\frac{a+b^2}{m}}=\frac{a+b}{a+b^2}\)
\(\frac{\frac{1}{4}+\frac{1}{2}}{\frac{1}{2}+\frac{3}{4}}=\frac{\frac{1}{4}+\frac{2}{4}}{\frac{2}{4}+\frac{1+2}{4}}=\frac{1+2}{1+2^2}=\frac{3}{5}\)
Hok tốt !!!!!!!!!
a, Ta có: \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\Rightarrow a=kb;c=kd\)
Thay:
\(\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)
=> đpcm
a) \(25^3:5^2=5^6:5^2=5^4=625\)
b) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3-1+\frac{1}{8}=\frac{17}{8}\)
A=\(\frac{1}{2}\)+\(\left(\frac{1}{2}\right)^2\)+...+\(\left(\frac{1}{2}\right)^{2020}\).
=> 2A= 1+\(\frac{1}{2}\)+\(\left(\frac{1}{2}\right)^2\)+...+\(\left(\frac{1}{2}\right)^{2019}\).
trừ 2 vế ta được A= 1 -\(\left(\frac{1}{2}\right)^{2020}\)<1 => điều phải chứng minh .