@Phan Đức Gia Linh _ Xin cảm ơn những bạn đã quan tâm tới câu hỏi của mình!
Tìm x, biết:
1) \(\dfrac{2}{x+1}\) = \(\dfrac{x+1}{8}\)
2) \(\dfrac{23+x}{40-x}\)
3) \(\dfrac{x-4}{x-7}\) = \(\left(\dfrac{-3}{5}\right)^2\)
4) \(\dfrac{3+x}{7+y}\) = \(\dfrac{3}{7}\) và x + y = 20
5) \(\dfrac{x}{x+1}\) > 0
6) \(\dfrac{2x-1}{2-x}\) \(\le\) 0
Các bạn cố gắng giải đầy đủ giúp mình!
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
Câu 2) \(\dfrac{23+x}{40-x}\) = \(\dfrac{-3}{4}\)
