Toán Lớp 9: 1. Giải bất phương trình
a) 3(x-1)-2x ( x-1) =0
b) 5( x+2)-4x (2-x)=0
c) x(2x-4) = (3x-1)(2x-4)
d) x^2 -x+1=x^2 -x-1
e) x^2 -7x+12 =0
f) (x+2)/(2x-3)=(1-x )/(3-2x)+1
g) (x+2)/(x-2)-(x-2)/(x+2)= -1/4-x^2
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Toán Lớp 9: 1. Giải bất phương trình
a) 3(x-1)-2x ( x-1) =0
b) 5( x+2)-4x (2-x)=0
c) x(2x-4) = (3x-1)(2x-4)
d) x^2 -x+1=x^2 -x-1
e) x^2 -7x+12 =0
f) (x+2)/(2x-3)=(1-x )/(3-2x)+1
g) (x+2)/(x-2)-(x-2)/(x+2)= -1/4-x^2
Comments ( 1 )
a)3\left( {x – 1} \right) – 2x\left( {x – 1} \right) = 0\\
\Leftrightarrow \left( {x – 1} \right)\left( {3 – 2x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 1 = 0\\
3 – 2x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{3}{2}
\end{array} \right.\\
\text{Vậy}\,x = 1;x = \dfrac{3}{2}\\
b)5\left( {x + 2} \right) – 4x\left( {2 – x} \right) = 0\\
\Leftrightarrow 5x + 10 – 8x + 4{x^2} = 0\\
\Leftrightarrow 4{x^2} – 3x + 10 = 0\left( {vn} \right)\\
\text{Vậy pt vô nghiệm}\\
c)x\left( {2x – 4} \right) = \left( {3x – 1} \right)\left( {2x – 4} \right)\\
\Leftrightarrow \left( {2x – 4} \right)\left( {x – 3x + 1} \right) = 0\\
\Leftrightarrow 2.\left( {x – 2} \right).\left( {1 – 2x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 2 = 0\\
1 – 2x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \dfrac{1}{2}
\end{array} \right.\\
\text{Vậy}\,x = \dfrac{1}{2};x = 2\\
d){x^2} – x + 1 = {x^2} – x – 1\\
\Leftrightarrow 1 = – 1\left( {ktm} \right)\\
\text{Vậy pt vô nghiệm}\\
e){x^2} – 7x + 12 = 0\\
\Leftrightarrow {x^2} – 4x – 3x + 12 = 0\\
\Leftrightarrow \left( {x – 4} \right)\left( {x – 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 3
\end{array} \right.\\
\text{Vậy}\,x = 3;x = 4\\
f)Dkxd:x\# \dfrac{3}{2}\\
\dfrac{{\left( {x + 2} \right)}}{{2x – 3}} = \dfrac{{1 – x}}{{3 – 2x}} + 1\\
\Leftrightarrow \dfrac{{x + 2}}{{2x – 3}} – \dfrac{{x – 1}}{{2x – 3}} = 1\\
\Leftrightarrow \dfrac{{x + 2 – x + 1}}{{2x – 3}} = 1\\
\Leftrightarrow \dfrac{3}{{2x – 3}} = 1\\
\Leftrightarrow 2x – 3 = 3\\
\Leftrightarrow x = 3\left( {tmdk} \right)\\
\text{Vậy}\,x = 3\\
g)Dkxd:x\# 2;x\# – 2\\
\dfrac{{x + 2}}{{x – 2}} – \dfrac{{x – 2}}{{x + 2}} = \dfrac{{ – 1}}{{4 – {x^2}}}\\
\Leftrightarrow \dfrac{{{{\left( {x + 2} \right)}^2} – {{\left( {x – 2} \right)}^2}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{1}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
\Leftrightarrow {x^2} + 4x + 4 – {x^2} + 4x – 4 = 1\\
\Leftrightarrow 8x = 1\\
\Leftrightarrow x = \dfrac{1}{8}\left( {tmdk} \right)\\
\text{Vậy}\,x = \dfrac{1}{8}
\end{array}$