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Bài làm:
Ta có: \(P=\frac{2x-1}{x-1}=\frac{\left(2x-2\right)+1}{x-1}=2+\frac{1}{x-1}\)
Để P đạt GTLN
=> \(\frac{1}{x-1}\) đạt GTLN => \(x-1\) đạt giá trị dương nhỏ nhất
Mà x nguyên => x - 1 nguyên
=> \(x-1=1\Rightarrow x=2\)
Vậy Max(P) = 3 khi x = 2
\(P=\frac{2x-1}{x-1}=\frac{2\left(x-1\right)+1}{x-1}=2+\frac{1}{x-1}\)( ĐKXĐ : x khác 1 )
Để P đạt GTLN => \(\frac{1}{x-1}\)đạt GTNN
=> x - 1 là số dương nhỏ nhất
=> x - 1 = 1
=> x = 2 ( tmđk )
Vậy PMax = \(2+\frac{1}{2-1}=2+1=3\), đạt được khi x = 2
Mình không chắc nha -.-
\(\Rightarrow A=4.\left[\frac{6}{2.\left(2.4\right)}+\frac{5}{\left(2.4\right).13}+\frac{3}{13.\left(4.4\right)}+\frac{2}{\left(4.4\right).18}+\frac{10}{18.\left(7.4\right)}\right]\)
\(=4.\left(\frac{6}{2.8}+\frac{5}{8.13}+\frac{3}{13.16}+\frac{2}{16.18}+\frac{10}{18.28}\right)=4.\left(\frac{1}{2}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{18}+\frac{1}{18}-\frac{1}{28}\right)\)
\(=4.\left(\frac{1}{2}-\frac{1}{28}\right)=4.\frac{13}{28}=\frac{13}{7}\)
\(\frac{x-2}{-\frac{2}{9}}=\frac{-2}{x-2}\)
=> (x - 2)2 = \(\frac{-2}{9}.\left(-2\right)\)
=> (x - 2)2 = 9
=> \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(\frac{x-2}{\frac{-2}{9}}=\frac{-2}{x-2}\)
\(\Rightarrow\left(x-2\right).\left(x-2\right)=\frac{-2}{9}.\left(-2\right)\)
\(\Rightarrow\left(x-2\right)^2=\frac{4}{9}\)
\(\Rightarrow\left(x-2\right)^2=\left(\frac{2}{3}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-2=\frac{2}{3}\\x-2=-\frac{2}{3}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}+2\\x=-\frac{2}{3}+2\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=\frac{4}{3}\end{cases}}\)
Vậy \(x=\frac{8}{3}\) hoặc \(x=\frac{4}{3}\)
Học tốt
a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)
=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
=> x + 1 = 0
=> x = -1
b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)
=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)
=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
=> x - 2021 = 0
=> x = 2021
c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
=> \(-\frac{1}{12}x+6=7\)
=> \(-\frac{1}{12}x=1\)
=> x = -12
Ta có :\(\frac{6^8.2^4-4^5.18^4}{27^3.8^4-3^9.2^{13}}=\frac{\left(2.3\right)^8.2^4-\left(2^2\right)^5.\left(3^2.2\right)^4}{\left(3^3\right)^3.\left(2^3\right)^4-3^9.2^{13}}=\frac{2^{12}.3^8-2^{14}.3^8}{3^9.2^{12}-3^9.2^{13}}=\frac{3^8.2^{12}.\left(2^2-1\right)}{3^9.2^{12}.\left(1-2\right)}\)
\(=\frac{3^9.2^{12}}{-3^9.2^{12}}=-1\)
\(\frac{6^8\cdot2^2-4^5\cdot18^4}{27^3\cdot8^4-3^9\cdot2^{13}}\)
\(=\frac{\left(2.3\right)^8.2^4-\left(2^2\right)^5.\left(3^2.2\right)^4}{\left(3^3\right)^3.\left(2^3\right)^4-3^9.2^{13}}\)
\(=\frac{2^{12}.3^8-2^{14}.3^8}{3^9.2^{12}-3^9.2^{14}}\)
\(=\frac{3^8.2^{12}.\left(2^2-1\right)}{3^9.2^{12}.\left(1-2\right)}\)
\(=\frac{3^9.2^{12}}{-3^9.2^{12}}=-1\)