a) 2/x+7=x+7/32
<=> (x+7)^2=64
=> x+7=8 hoặc x+7=-8
=> x=-1 hoặc x=-15
b) - (x+5)^2= (x-2).(x+8)
<=> -(x+5)^2=x^2+8x-2x-16
<=> - (x+5)^2 =(x-4)^2
+> Không có giá trị x thỏa mãn
a. ĐK: x\(\ne\)-7
2.32=(x+7)2
<=> 64=x2+ 14x+ 49
<=>x2+ 14x- 15=0
<=>x2+ 15x- x- 15=0
<=>(x-1)(x+15)=0
\(\Leftrightarrow\hept{\begin{cases}x=1\\x=-15\end{cases}}\)
b, ĐK: x\(\ne\)-5;-8
(x-2)(x+8)=(x-5)(x+5)
<=>x2+ 6x- 16=x2- 25
<=>6x+9=0
\(\Leftrightarrow x=-\frac{3}{2}\)
\(\frac{2}{x+7}=\frac{x+7}{32}\)
\(\Leftrightarrow\left(x+7\right)\left(x+7\right)=64\)
\(\Leftrightarrow\left(x+7\right)^2=8^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+7=8\\x+7=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-15\end{cases}}\)
vậy x=1 hoặc x=-15
b)\(\frac{x-2}{x+5}=\frac{x-5}{x+8}\)ĐKXĐ : \(x\ne-5\) và \(x\ne8\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)=\left(x-2\right)\left(x+8\right)\)
\(\Leftrightarrow x^2-25=x^2+6x-16\)
\(\Leftrightarrow6x+9=0\)
\(\Leftrightarrow6x=-9\)
\(\Leftrightarrow x=-\frac{9}{6}=-\frac{3}{2}\)
a. \(\frac{2}{x+7}=\frac{x+7}{32}\)
\(\Leftrightarrow\left(x+7\right)^2=64\)
\(\Leftrightarrow\left(x+7\right)^2=8^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+7=8\\x+7=-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-15\end{cases}}\)
b. \(\frac{x-2}{x+5}=\frac{x-5}{x+8}\)
\(\Leftrightarrow\left(x-2\right)\left(x+8\right)=\left(x-5\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+8x-2x-16=x^2+5x-5x-25\)
\(\Leftrightarrow x^2+6x-16=x^2-25\)
\(\Leftrightarrow6x-16+25=0\)
\(\Leftrightarrow6x+9=0\)
\(\Leftrightarrow6x=-9\)
\(\Leftrightarrow x=-\frac{3}{2}\)
a,\(\frac{2}{x+7}=\frac{x+7}{32}\left(ĐK:x\ne-7\right)\)
\(< =>2.32=\left(x+7\right)^2\)
\(< =>64=x^2+14x+49\)
\(< =>x^2+14x-15=0\)
\(< =>x^2-x+15x-15=0\)
\(< =>x\left(x-1\right)+15\left(x-1\right)=0\)
\(< =>\left(x+15\right)\left(x-1\right)=0\)
\(< =>\orbr{\begin{cases}x+15=0\\x-1=0\end{cases}< =>\orbr{\begin{cases}x=-15\\x=1\end{cases}}\left(tmđk\right)}\)
\(\frac{x-2}{x+5}=\frac{x-5}{x+8}\left(ĐK:x\ne-5;-8\right)\)
\(< =>\left(x-2\right)\left(x+8\right)=\left(x+5\right)\left(x-5\right)\)
\(< =>x^2+8x-2x-16=x^2-25\)
\(< =>x^2+6x-16-x^2+25=0\)
\(< =>6x+9=0< =>6x=-9< =>x=-\frac{3}{2}\left(tmđk\right)\)
