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\(2\dfrac{3}{x}=\dfrac{13}{x}\)(đk \(x\ne0\))
\(\dfrac{2x+3}{x}=\dfrac{13}{x}\)
\(\Rightarrow\left(2x+3\right).x=13.x\)
\(\Rightarrow2x+3=13\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\left(tmđk\right)\)
Vậy x=5
(5 - \(x\))(9\(x^2\) - 4) =0
\(\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=-\dfrac{2}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x\) \(\in\) { - \(\dfrac{2}{3}\); \(\dfrac{2}{3}\); \(5\)}
72\(x\) + 72\(x\) + 3 = 344
72\(x\) \(\times\) ( 1 + 73) = 344
72\(x\) \(\times\) (1 + 343) = 344
72\(x\) \(\times\) 344 = 344
72\(x\) = 344 : 344
72\(x\) = 1
72\(x\) = 70
\(2x\) = 0
\(x\) = 0
Kết luận: \(x\) = 0
Bạn tham khảo tại đây:
https://hoc24.vn/cau-hoi/giup-minh-voiiiii-minh-cam-on-tim-xy-biet-dfracx4-dfrac2y13-dfracx-2y-1y-voi-y-0.4107067269450
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\)
\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)
\(\Leftrightarrow\left(x-3\right).\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right).\left(x-4\right)\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=3\end{matrix}\right.\left(\text{thỏa}\right)\)
\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\left(x\ne\left\{2;4\right\}\right)\\ =>\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1\\ =>x^2-3x-4x+12+x^2-4x+4=-\left(x-2\right)\left(x-4\right)\\ =>2x^2-11x+16=-x^2+6x-8\\ =>3x^2-17x+24=0\\ =>\left(x-3\right)\left(3x-8\right)=0\\ =>\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\left(TMDK\right)\)
Câu 11:
=>4,6x=6,21
=>x=1,35
12: \(A=-\left(1.4-x\right)^2-1.4< =-1.4\)
=>x=-1,4
Câu 9:
\(\Leftrightarrow\dfrac{10a+b}{100c+90+d}=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{92}-\dfrac{1}{97}=\dfrac{1}{2}-\dfrac{1}{97}=\dfrac{95}{194}\)
=>a=9; b=5; c=1; d=4
=>a+b+c+d=9+5+1+4=19
a) Ta có: \(\left(x-\dfrac{3}{4}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{3}{4}=0\)
hay \(x=\dfrac{3}{4}\)
b) Ta có: \(\left(x+\dfrac{4}{9}\right)^2=\dfrac{49}{144}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{9}=\dfrac{7}{12}\\x+\dfrac{4}{9}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{36}\\x=\dfrac{-37}{36}\end{matrix}\right.\)
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b, \(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2014}+1\right)+\left(\dfrac{x+3}{2015}+1\right)=\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+1}{2017}+1\right)\)\(\Leftrightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\Leftrightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}-\dfrac{x+2018}{2016}-\dfrac{x+2018}{2017}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow xx+2018=0\Leftrightarrow x=-2018\)
Vậy x = -2018
Nguyễn Huy Tú, cho mk hỏi sao câu a bt đó lại bằng 0 vậy ? Mk ko hiểu lắm
a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
Mong bạn xem lại đề bài.
\(2\dfrac{3}{x}\) là hỗn số hay là \(2\times\dfrac{3}{x}\) vậy bạn .
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