cho bt A= $\frac{2x^{2}+2x}{1-x^{2}}$ va B=$\frac{1-2x}{x^{2}-3x+2}$ =$\frac{x+1}{x-2}$
a, rút gọn a
b, với P =A.B, tìm x để P=9/2
c, tìm x để p<1
d, tìm x để A.B là số nguyên
cho bt A= $\frac{2x^{2}+2x}{1-x^{2}}$ va B=$\frac{1-2x}{x^{2}-3x+2}$ =$\frac{x+1}{x-2}$ a, rút gọn a b, với P =A.B, tìm x để P=9/2 c, tìm x để p<1 d
By Camila
Đáp án:
a. \(A = \dfrac{{2x}}{{1 – x}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \pm 1\\
A = \dfrac{{2x\left( {x + 1} \right)}}{{\left( {1 – x} \right)\left( {1 + x} \right)}} = \dfrac{{2x}}{{1 – x}}\\
b.DK:x \ne \left\{ {1;2} \right\}\\
B = \dfrac{{1 – 2x}}{{\left( {x – 1} \right)\left( {x – 2} \right)}} + \dfrac{{x + 1}}{{x – 2}}\\
= \dfrac{{1 – 2x + {x^2} – 1}}{{\left( {x – 1} \right)\left( {x – 2} \right)}}\\
= \dfrac{{x\left( {x – 2} \right)}}{{\left( {x – 1} \right)\left( {x – 2} \right)}} = \dfrac{x}{{x – 1}}\\
P = A.B = \dfrac{{2x}}{{1 – x}}.\dfrac{x}{{x – 1}} = \dfrac{{ – 2{x^2}}}{{{x^2} – 2x + 1}}\\
P = \dfrac{9}{2}\\
\to \dfrac{{ – 2{x^2}}}{{{x^2} – 2x + 1}} = \dfrac{9}{2}\\
\to 9{x^2} – 18x + 9 = – 4{x^2}\\
\to 13{x^2} – 18x + 9 = 0\\
\to {\left( {x\sqrt {13} } \right)^2} – 2x\sqrt {13} .\dfrac{9}{{\sqrt {13} }} + {\left( {\dfrac{9}{{\sqrt {13} }}} \right)^2} + \dfrac{{36}}{{13}} = 0\\
\to {\left( {x\sqrt {13} – \dfrac{9}{{\sqrt {13} }}} \right)^2} + \dfrac{{36}}{{13}} = 0\left( {voly} \right)\\
Do:{\left( {x\sqrt {13} – \dfrac{9}{{\sqrt {13} }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt {13} – \dfrac{9}{{\sqrt {13} }}} \right)^2} + \dfrac{{36}}{{13}} > 0\\
\to x \in \emptyset \\
c.P < 1\\
\to \dfrac{{ – 2{x^2}}}{{{x^2} – 2x + 1}} < 1\\
\to \dfrac{{ – 2{x^2} – {x^2} + 2x – 1}}{{{x^2} – 2x + 1}} < 0\\
\to \dfrac{{ – 3{x^2} + 2x – 1}}{{{{\left( {x – 1} \right)}^2}}} < 0\\
\to – 3{x^2} + 2x – 1 < 0\left( {do:{{\left( {x – 1} \right)}^2} > 0\forall x \ne 1} \right)\\
Mà: – 3{x^2} + 2x – 1 < 0\left( {ld} \right)\forall x \ne 1\\
KL:P < 1\forall x \ne 1
\end{array}\)