Question

[Đỗ Mạnh Anh Hải]

Tìm x:

a) \(\dfrac{x+1}{32}=\dfrac{2}{x+1}\)

b) \(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)

c)\(|4,5-2x|:1\dfrac{7}{4}=\dfrac{11}{14}\)

Giúp mình gấp nhoa!!! Mình sẽ ủng hộ người đó!!!!

Answer 1

[qwerty]

a) \(\dfrac{x+1}{32}=\dfrac{2}{x+1}\)

\(\Leftrightarrow\dfrac{x+1}{32}=\dfrac{2}{x+1}\left(đk:x\ne1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x+1\right)=64\)

\(\Leftrightarrow\left(x+1\right)^2-64=0\)

\(\Leftrightarrow x^2+2x+1-64=0\)

\(\Leftrightarrow x^2+6x-63=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+16}{2}\\x=\dfrac{-2-16}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\left(đk:x\ne-1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\)

Vậy \(x_1=-9;x_2=7\)

b) \(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)

\(\Leftrightarrow\dfrac{x+1}{5}=\dfrac{7}{x-1}\left(đk:x\ne1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=35\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)-35=0\)

\(\Leftrightarrow x^2-1-35=0\)

\(\Leftrightarrow x^2-36=0\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\left(đk:x\ne1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Vậy \(x_1=-6;x_2=6\)

c) \(\left|4,5-2x\right|:1\dfrac{7}{4}=\dfrac{11}{14}\)

\(\Leftrightarrow\left|4,5-2x\right|:\dfrac{11}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|4,5-2x\right|\cdot\dfrac{4}{11}=\dfrac{11}{14}\)

\(\Leftrightarrow\dfrac{4}{11}\cdot\left|4,5-2x\right|=\dfrac{11}{14}\)

\(\Leftrightarrow\left|4,5-2x\right|=\dfrac{121}{56}\)

\(\Leftrightarrow\left[{}\begin{matrix}4,5-2x=\dfrac{121}{56}\\4,5-2x=-\dfrac{121}{56}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{131}{112}\\x=\dfrac{373}{112}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{131}{112};x_2=\dfrac{373}{112}\)

Answer 2

[Nguyễn Thanh Hằng]

a) \(\dfrac{x+1}{32}=\dfrac{2}{x+1}\)

\(\Rightarrow\left(x+1\right)\left(x+1\right)=32.2\)

\(\Rightarrow\left(x+1\right)^2=64\)

\(\Rightarrow\left(x+1\right)^2=8^2\)

\(\Rightarrow x+1=8\)

\(\Rightarrow x=8-1\)

\(\Rightarrow x=7\left(TM\right)\)

Vậy \(x=7\) là giá trị cần tìm

b) \(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)

\(\Rightarrow\left(x+1\right)\left(x-1\right)=7.5\)

\(\Rightarrow\left[{}\begin{matrix}x+1=7\\x-1=5\end{matrix}\right.\) \(\Rightarrow x=6\left(TM\right)\)

Vậy \(x=6\) là giá trị cần tìm

c) \(\left|4,5-2x\right|:1\dfrac{7}{4}=\dfrac{11}{14}\)

\(\left|\dfrac{45}{10}-2x\right|:\dfrac{11}{4}=\dfrac{11}{4}\)

\(\left|\dfrac{9}{2}-2x\right|=\dfrac{11}{14}.\dfrac{11}{4}\)

\(\left|\dfrac{9}{2}-2x\right|=\dfrac{121}{56}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{9}{2}-2x=\dfrac{121}{56}\\\dfrac{9}{2}-2x=\dfrac{-121}{56}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{131}{56}\\2x=\dfrac{373}{56}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{131}{112}\\x=\dfrac{373}{112}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{131}{112};\dfrac{373}{112}\right\}\) là giá trị cần tìm