3 phút 30giay x 7 +3,5x 4 -210giây
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\(a,\dfrac{x^7-2x^3}{2x^4-x^3}=\dfrac{x^3\left(x^4-2\right)}{x^3\left(2x-1\right)}=\dfrac{x^4-2}{2x-1}\)
\(b,\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{16x}=\dfrac{\left(x+2-x+2\right)\left(x+2+x-2\right)}{16x}=\dfrac{4.2x}{16x}=\dfrac{1}{2}\)
\(c,\dfrac{24,5x^2-0,5y^2}{3,5x^2-0,5y^2}=\dfrac{0,5\left(49x^2-y^2\right)}{0,5\left(7x^2-y^2\right)}=\dfrac{49x^2-y^2}{7x^2-y^2}\)
a) \(\dfrac{x^7-2x^3}{2x^4-x^3}=\dfrac{x^3\left(x^4-2\right)}{x^3\left(2x-1\right)}=\dfrac{x^4-2}{2x-1}\)
b)\(\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{16x}=\dfrac{x^2+4x+4-\left(x^2-4x+4\right)}{16x}\)=\(\dfrac{x^2+4x+4-x^2+4x-4}{16x}=\dfrac{8x}{16x}=\dfrac{1}{2}\)
a) ( 5x - 4)(4x + 6)=0
<=> \([^{5x-4=0}_{4x+6=0}< =>[^{x=\frac{4}{5}}_{x=\frac{-6}{4}}\)
Vậy S = \(\left\{\frac{4}{5};\frac{-6}{4}\right\}\)
b) ( 3,5x - 7 )( 2,1x - 6,3 ) = 0
<=> \([^{3,5x-7=0}_{2,1x-6,3=0}< =>[^{x=2}_{x=3}\)
Vậy S = \(\left\{2;3\right\}\)
c) ( 4x - 10 )( 24 + 5x ) = 0
<=> \([^{4x-10=0}_{24+5x=0}< =>[^{x=\frac{5}{2}}_{x=\frac{-24}{5}}\)
Vậy S = \(\left\{\frac{5}{2};\frac{-24}{5}\right\}\)
d) ( x - 3 )( 2x + 1 ) = 0
<=> \(\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=\frac{-1}{2}\end{matrix}\right.\)
Vậy S = \(\left\{3;\frac{-1}{2}\right\}\)
e) ( 5x - 10 )( 8 - 2x ) = 0
<=> \(\left[{}\begin{matrix}5x-10=0\\8-2x=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy S = \(\left\{2;4\right\}\)
f) ( 9 - 3x )( 15 + 3x ) = 0
<=> \(\left[{}\begin{matrix}9-3x=0\\15+3x=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy S = \(\left\{3;-5\right\}\)
Học tốt nhaaa !