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x : \(\dfrac{2}{7}\)= 0,8
= > x = 0,8 x\(\dfrac{2}{7}\)
=> x = \(\dfrac{8}{35}\)
Vậy x = \(\dfrac{8}{35}\)
x : \(\dfrac{2}{7}\) = 0,8
x : \(\dfrac{2}{7}\) = \(\dfrac{8}{10}\)
x = \(\dfrac{4}{5}.\dfrac{2}{7}\)
x = \(\dfrac{8}{35}\)
1) \(\dfrac{2}{x+1}=\dfrac{x+1}{8}\Leftrightarrow\left(x+1\right)\left(x+1\right)=2.8\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\) vậy \(x=3;x=-5\)
2) thiếu quế phải nha
3) \(\dfrac{x-4}{x-7}=\left(\dfrac{-3}{5}\right)^2\Leftrightarrow\dfrac{x-4}{x-7}=\dfrac{9}{25}\Leftrightarrow9.\left(x-7\right)=25.\left(x-4\right)\)
\(\Leftrightarrow9x-63=25x-100\Leftrightarrow25x-9x=-63+100\)
\(\Leftrightarrow16x=37\Leftrightarrow x=\dfrac{37}{16}\) vậy \(x=\dfrac{37}{16}\)
4) ta có : \(x+y=20\Leftrightarrow y=20-x\)
\(\dfrac{3+x}{7+y}=\dfrac{3}{7}\Leftrightarrow7\left(3+x\right)=3\left(7+y\right)\Leftrightarrow21+7x=21+3y\)
\(\Leftrightarrow7x=3y\Leftrightarrow7x-3y=0\Leftrightarrow7x-3\left(20-x\right)=0\)
\(\Leftrightarrow7x-60+3x=0\Leftrightarrow10x=60\Leftrightarrow x=6\)
\(\Rightarrow6+y=20\Leftrightarrow y=14\) vậy \(x=6;y=14\)
a.\(\dfrac{2}{x+1}=\dfrac{x+1}{8}\)
\(\Rightarrow\left(x+1\right)\left(x+1\right)=2.8\)
\(\Rightarrow\left(x+1\right)^2=16\)
\(\Rightarrow\left(x+1\right)^2=(\pm4)^2\)
\(\rightarrow x+1=\pm4\)
\(+)x+1=-4\)
\(x=\left(-4\right)-1\)
\(x=-5\)
\(+)x+1=4\)
\(x=4-1\)
\(x=3\)
Vậy \(x\in\left\{-5;3\right\}\)
Câu còn lại tương tự nhé. Trừ câu 2 bạn xem lại nhé
Ta có :
\(\dfrac{1}{A}=\dfrac{2^{20}-3}{2^{18}-3}=\dfrac{2^2.\left(2^{18}-3\right)+9}{2^{18}-3}=4+\dfrac{9}{2^{18}-3}\left(1\right)\)
\(\dfrac{1}{B}=\dfrac{2^{22}-3}{2^{20}-3}=\dfrac{2^2.\left(2^{20}-3\right)+9}{2^{20}-3}=4+\dfrac{9}{2^{20}-3}\left(2\right)\)
Từ (1) và (2) ta có \(\dfrac{1}{A}>\dfrac{1}{B}\Leftrightarrow A< B\)
\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{\left(3x+1\right).\left(3x+4\right)}\)=\(\dfrac{1344}{2017}\)
\(A=\dfrac{2}{3}(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{3x+1}-\dfrac{1}{3x+4}\))=\(\dfrac{1344}{2017}\)
\(A=\dfrac{2}{3}(1-\dfrac{1}{3x+4})\)=\(\dfrac{1344}{2017}\)
\(A=1-\dfrac{1}{3x+4}=\dfrac{1344}{2017}:\dfrac{2}{3}\)
\(A=1-\dfrac{1}{3x+4}=\dfrac{2016}{2017}\)
\(A=\dfrac{1}{3x+4}=1-\dfrac{2016}{2017}\)
\(A=\dfrac{1}{3x+4}=\dfrac{1}{2017}\)
\(\Rightarrow\)\(3x+4=2017\)
\(3x=2017-4\)
\(3x=2013\)
\(x=671\)
\(\Leftrightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)
\(\rightarrowđpcm\)
a) \(\dfrac{1}{3}+\dfrac{3}{8}-\dfrac{7}{12}\)
\(=\dfrac{17}{24}-\dfrac{7}{12}\)
\(=\dfrac{1}{8}\)
b) \(\dfrac{-3}{14}+\dfrac{5}{8}-\dfrac{1}{2}\)
\(=\dfrac{23}{56}-\dfrac{1}{2}\)
\(=\dfrac{-5}{56}\)
c) \(\dfrac{1}{4}-\dfrac{2}{3}-\dfrac{11}{18}\)
\(=\dfrac{-5}{12}-\dfrac{11}{18}\)
\(=\dfrac{-37}{36}\)
d) \(\dfrac{1}{4}+\dfrac{5}{12}-\dfrac{1}{13}-\dfrac{7}{8}\)
\(=\dfrac{2}{3}-\dfrac{1}{13}-\dfrac{7}{8}\)
\(=\dfrac{23}{39}-\dfrac{7}{8}\)
\(=\dfrac{-89}{312}\)
\(-2\dfrac{3}{4}\)+\(\dfrac{7}{8}\)
= \(\dfrac{-11}{4}\)+\(\dfrac{7}{8}\)
= \(\dfrac{-15}{8}\)
Vậy tổng trên bằng \(\dfrac{-15}{8}\)