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 \(x_1=\dfrac{131}{112};x_2=\dfrac{373}{112}\)
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
b) x +1/5 = 7/x - 1
Suy ra: ( x+1).(x - 1) = 7.5
Nếu x +1= 7 Nếu x -1 = 5
X = 6 x = 6
Vậy x cần tìm là 6
C) |4,5-2x|:1(7/4)= 11/4
9/2-2x : 11/4 = 11/4
2x : 11/4 = 9/2 - 11/4
2x : 11/4 = 7/4
2x=7/4.11/4
2x= 77/16
x= 77/16:2
x=77/32
\(\dfrac{x+1}{32}=\dfrac{2}{x+1}\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)=32.2\)
\(\Leftrightarrow x\left(x+1\right)+1\left(x+1\right)=64\)
\(\Leftrightarrow x^2+x+x+1=64\)
\(\Leftrightarrow x^2+2x+1=64\)
\(\Leftrightarrow x^2+2x=63\)
\(x\left(x+2\right)=63\)
\(\Leftrightarrow x;x+2\inƯ\left(63\right)\)
Tìm các ước của 63 rồi tính ra nha!
\(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=5.7=35\)
\(x\left(x-1\right)+1\left(x-1\right)=35\)
\(x^2-x+x-1=35\)
\(\Leftrightarrow x^2=36\Leftrightarrow x=\left\{\pm6\right\}\)
c) \(\left|4,5-2x\right|:1\dfrac{7}{4}=\dfrac{11}{14}\)
\(\left|4,5-2x\right|:\dfrac{11}{4}=\dfrac{11}{14}\Leftrightarrow\left|4,5-2x\right|=\dfrac{121}{56}\)
\(\Leftrightarrow4,5-2x=\dfrac{121}{56}\Leftrightarrow2x=\dfrac{131}{56}\Leftrightarrow x=\dfrac{131}{112}\)
\(\Leftrightarrow4,5-2x=-\dfrac{121}{56}\Leftrightarrow2x=\dfrac{373}{56}\Leftrightarrow x=\dfrac{373}{112}\)
Câu c tính hoài vẫn sai, chỉ có b là đúng.
