( Giải phương trình )
a.2x(x-3)+5(x-3)=0
b.(x^2-4)-(x-2)(3-2x)=0
c.(2x+5)^2 = (x+2)^2
d.x^2 – 5x+6=0
e.2x^3 + 6x^2 = x^2 + 3x
( Giải phương trình )
a.2x(x-3)+5(x-3)=0
b.(x^2-4)-(x-2)(3-2x)=0
c.(2x+5)^2 = (x+2)^2
d.x^2 – 5x+6=0
e.2x^3 + 6x^2 = x^2 + 3x
a,2x(x-3)+5(x-3)=0
<=>(x-3)(2x+5)=0
<=>x-3=0 hoặc 2x+5=0
<=>x=3 hoặc x=$\frac{-5}{2}$
b,(x^2-4)-(x-2)(3-2x)=0
<=>(x-2)(x+2)-(x-2)(3-2x)=0
<=>(x-2)(x+2-3+2x)=0
<=>x-2=0 hoặc 3x-1=0
<=>x=2 hoặc x=$\frac{1}{3}$
c,(2x+5)^2 = (x+2)^2
<=>(2x+5)^2-(x+2)^2=0
<=>(2x+5-x-2)(2x+5+x+2)=0
<=>x-3=0 hoặc 3x+7=0
<=>x=3 hoặc x=$\frac{-7}{3}$
d,.x^2-5x+6=0
<=>(x^2-6x+9)+(x-3)=0
<=>(x-3)^2+(x-3)=0
<=>(x-3)(x-3+1)=0
<=>x=3 hoặc x=2
e,2x^3 + 6x^2 = x^2 + 3x
<=>2x^3 + 6x^2-x^2-3x=0
<=>(2x^3 + 6x^2)-(x^2+3x)=0
<=>2x^2(x+3)-x(x+3)=0
<=>(x+3)(2x^2-x)=0
<=>x+3=0 hoặc 2x^2-x=0
<=>x=-3 hoặc x=$\frac{1}{2}$
Đáp án:
e. \(\left[ \begin{array}{l}
x = 0\\
x = \frac{1}{2}\\
x = – 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left( {x – 3} \right)\left( {2x + 5} \right) = 0\\
\to \left[ \begin{array}{l}
x – 3 = 0\\
2x + 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = – \frac{5}{2}
\end{array} \right.\\
b.\left( {x – 2} \right)\left( {x + 2} \right) – \left( {x – 2} \right)\left( {3 – 2x} \right) = 0\\
\to \left( {x – 2} \right)\left( {x + 2 – 3 + 2x} \right) = 0\\
\to \left[ \begin{array}{l}
x – 2 = 0\\
3x – 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = \frac{1}{3}
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
c.4{x^2} + 20x + 25 = {x^2} + 4x + 4\\
\to 3{x^2} + 16x + 21 = 0\\
\to 3{x^2} + 9x + 7x + 21 = 0\\
\to 3x\left( {x + 3} \right) + 7\left( {x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x + 3 = 0\\
3x + 7 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 3\\
x = – \frac{7}{3}
\end{array} \right.\\
d.{x^2} – 3x – 2x + 6 = 0\\
\to x\left( {x – 3} \right) – 2\left( {x – 3} \right) = 0\\
\to \left[ \begin{array}{l}
x – 3 = 0\\
x – 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\\
e.2{x^3} + 5{x^2} – 3x = 0\\
\to x\left( {2{x^2} + 5x – 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
2{x^2} – x + 6x – 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x\left( {2x – 1} \right) + 3\left( {2x – 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
2x – 1 = 0\\
x + 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = \frac{1}{2}\\
x = – 3
\end{array} \right.
\end{array}\)