1, tìm x
a, (3x-1)^2-25=0
b, 8x^3-50x=0
c,(x-2)(x^2+3x+7)+2(x^2-4)-5(x-2)=0
2, tìm x
a, 3x(x-1)+x-1=0
b,2(x+3)-x^2-3x=0
c, 4x^2-25-(2x-5)(2x+7)=0
d, x^3+27+(x+3)(x-9)=0
1, tìm x a, (3x-1)^2-25=0 b, 8x^3-50x=0 c,(x-2)(x^2+3x+7)+2(x^2-4)-5(x-2)=0 2, tìm x a, 3x(x-1)+x-1=0 b,2(x+3)-x^2-3x=0 c, 4x^2-25-(2x-5)(2x+7)=0 d, x
By Vivian
ĐKXĐ: x khác + 5, x khác 3
`a,A=[3(x+5)/(x-5)(x+5)-x(x-5)/(x-5)(x+5)+(x2+25)/(x-5)(x+5)](1-2/x-3)`
`=(3x+15-x2+5x+x2+25)/(x-5)(x+5).(x-3-2)/x-3`
`=(8x+40)/(x-5)(x+5).(x-5)/x-3=8/x-3`
b, thay x=1/3 vào ta được `A= 8/1/3-3=-3`
`c, A=-3/5 <-> 8/x-3=-3/5 <-> x-3=-40/3 <-> x=-31/3` (t/m)
d, A lớn hơn `0<-> 8/x-3` lớn hơn 0 <-> x-3 lớn hơn 0 <-> x lớn hơn 3
đối chiếu ĐK ta có x lớn hơn 3 và x khác 5
e, A nguyên <-> 8/x-3 nguyên `<-> x-3` thuộc ước của 8 <-> x-3 thuộc
`{-8;-4;-2;-1;1;2;4;8}<-> x thuộc {-5;-1;1;2;4;5;7;11}`
đối chiếu đk ta được x thuộc `{-1;1;2;4;7;11}`
Giải thích các bước giải:
$\begin{array}{l}
B1:\\
a,{(3x – 1)^2} – 25 = 0\\
\Leftrightarrow {\left( {3x – 1} \right)^2} = 25\\
\Leftrightarrow \left[ \begin{array}{l}
3x – 1 = 5\\
3x – 1 = – 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 6\\
3x = – 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \dfrac{{ – 4}}{3}
\end{array} \right.\\
b,8{x^3} – 50x = 0\\
\Leftrightarrow 2x\left( {4{x^2} – 25} \right) = 0\\
\Leftrightarrow 2x\left( {2x – 5} \right)\left( {2x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 0\\
2x – 5 = 0\\
2x + 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{5}{2}\\
x = \dfrac{{ – 5}}{2}
\end{array} \right.\\
c,(x – 2)({x^2} + 3x + 7) + 2({x^2} – 4) – 5(x – 2) = 0\\
\Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + 3x + 7} \right) + 2\left( {x – 2} \right)\left( {x + 2} \right) – 5\left( {x – 2} \right) = 0\\
\Leftrightarrow \left( {x – 2} \right)\left[ {\left( {{x^2} + 3x + 7} \right) + 2\left( {x + 2} \right) – 5} \right] = 0\\
\Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + 5x + 6} \right) = 0\\
\Leftrightarrow \left( {x – 2} \right)\left( {x + 2} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 2 = 0\\
x + 2 = 0\\
x + 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = – 2\\
x = – 3
\end{array} \right.\\
B2:\\
a,{\rm{ }}3x\left( {x – 1} \right) + x – 1 = 0\\
\Leftrightarrow \left( {x – 1} \right)\left( {3x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 1 = 0\\
3x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{{ – 1}}{3}
\end{array} \right.\\
b,2(x + 3) – {x^2} – 3x = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {2 – x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
2 – x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 3\\
x = 2
\end{array} \right.\\
c,4{x^2} – 25 – (2x – 5)(2x + 7) = 0\\
\Leftrightarrow \left( {2x – 5} \right)\left( {2x + 5} \right) – \left( {2x – 5} \right)\left( {2x + 7} \right) = 0\\
\Leftrightarrow \left( {2x – 5} \right)\left( {2x + 5 – \left( {2x + 7} \right)} \right) = 0\\
\Leftrightarrow – 2\left( {2x – 5} \right) = 0\\
\Leftrightarrow 2x – 5 = 0\\
\Leftrightarrow x = \dfrac{5}{2}\\
d,{x^3} + 27 + (x + 3)(x – 9) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {{x^2} – 3x + 9} \right) + \left( {x + 3} \right)\left( {x – 9} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {{x^2} – 3x + 9 + x – 9} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {{x^2} – 2x} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)x\left( {x – 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x = 0\\
x – 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 3\\
x = 0\\
x = 2
\end{array} \right.
\end{array}$