Giải phương trình
-5x=0
(x²-9)+(x+3)(3-2x)
X³+3x²=x²+3x
2x-5/x-5=3
X-3/x-2+x+2/x=2
X/x+1-2x-3/x-1=2x+3/x²-1
X-1/x+x-2/x+1=2
Giải giúp em với ạ
Giải phương trình -5x=0 (x²-9)+(x+3)(3-2x) X³+3x²=x²+3x 2x-5/x-5=3 X-3/x-2+x+2/x=2 X/x+1-2x-3/x-1=2x+3/x²-1 X-1/x+x-2/x+1=2 Giải giúp em với ạ
By Clara
Đáp án:
f) \(x = – \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a) – 5x = 0\\
\to x = 0\\
b)\left( {{x^2} – 9} \right) + \left( {x + 3} \right)\left( {3 – 2x} \right) = 0\\
\to \left( {x + 3} \right)\left( {x – 3 + 3 – 2x} \right) = 0\\
\to \left[ \begin{array}{l}
x + 3 = 0\\
– x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 3\\
x = 0
\end{array} \right.\\
c){x^2}\left( {x + 3} \right) – x\left( {x + 3} \right) = 0\\
\to \left( {x + 3} \right)\left( {{x^2} – x} \right) = 0\\
\to x\left( {x – 1} \right)\left( {x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 1\\
x = – 3
\end{array} \right.\\
d)DK:x \ne 5\\
\dfrac{{2x – 5}}{{x – 5}} = 3\\
\to 2x – 5 = 3x – 15\\
\to x = 10\\
e)DK:x \ne \left\{ {0;2} \right\}\\
\dfrac{{x – 3}}{{x – 2}} + \dfrac{{x + 2}}{x} = 2\\
\to \dfrac{{{x^2} – 3x + {x^2} – 4 – 2x\left( {x – 2} \right)}}{{x\left( {x – 2} \right)}} = 0\\
\to 2{x^2} – 3x – 4 – 2{x^2} + 4x = 0\\
\to x = 4\\
f)DK:x \ne \left\{ { – 1;0} \right\}\\
\dfrac{{x – 1}}{x} + \dfrac{{x – 2}}{{x + 1}} = 2\\
\to \dfrac{{{x^2} – 1 + {x^2} – 2x – 2x\left( {x + 1} \right)}}{{x\left( {x + 1} \right)}} = 0\\
\to 2{x^2} – 2x – 1 – 2{x^2} – 2x = 0\\
\to 4x = – 1\\
\to x = – \dfrac{1}{4}
\end{array}\)