Tìm x ( giải phương trình )
a) (x-2)(3+x)-(x+1)62=2-4x
b) x^2-3x / x-3 + 2x =7
c) x+2 / 2 = 2x-1 / 3 + 1
d) x^2+(x / x+1 )^2=5/4
Tìm x ( giải phương trình )
a) (x-2)(3+x)-(x+1)62=2-4x
b) x^2-3x / x-3 + 2x =7
c) x+2 / 2 = 2x-1 / 3 + 1
d) x^2+(x / x+1 )^2=5/4
Đáp án:
$\begin{array}{l}
a)\left( {x – 2} \right)\left( {3 + x} \right) – {\left( {x + 1} \right)^2} = 2 – 4x\\
\Leftrightarrow 3x + {x^2} – 6 – 2x\\
– \left( {{x^2} + 2x + 1} \right) – 2 + 4x = 0\\
\Leftrightarrow {x^2} + x – 6 – {x^2} – 2x – 1 – 2 + 4x = 0\\
\Leftrightarrow 3x – 9 = 0\\
\Leftrightarrow x = 3\\
Vậy\,x = 3\\
b)Dkxd:x\# 3\\
\dfrac{{{x^2} – 3x}}{{x – 3}} + 2x = 7\\
\Leftrightarrow \dfrac{{x\left( {x – 3} \right)}}{{x – 3}} + 2x = 7\\
\Leftrightarrow x + 2x = 7\\
\Leftrightarrow 3x = 7\\
\Leftrightarrow x = \dfrac{7}{3}\\
Vậy\,x = \dfrac{7}{3}\\
c)\dfrac{{x + 2}}{2} = \dfrac{{2x – 1}}{3} + 1\\
\Leftrightarrow \dfrac{{3\left( {x + 2} \right) – 2\left( {2x – 1} \right) – 6}}{6} = 0\\
\Leftrightarrow 3x + 6 – 4x + 2 – 6 = 0\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
d){x^2} + {\left( {\dfrac{x}{{x + 1}}} \right)^2} = \dfrac{5}{4}\\
\Leftrightarrow \dfrac{{{x^2}{{\left( {x + 1} \right)}^2} + {x^2}}}{{{{\left( {x + 1} \right)}^2}}} = \dfrac{5}{4}\\
\Leftrightarrow 5\left( {{x^2} + 2x + 1} \right) = 4\left( {{x^4} + 2{x^3} + {x^2} + {x^2}} \right)\\
\Leftrightarrow 4{x^4} + 8{x^3} + 3{x^2} – 10x – 5 = 0
\end{array}$
(Em kiểm tra lại đề bài câu cuối)