Cho bieu thuc:
A = 2x/ x + 1 (x ko= 1 ; -1)
B = 3x/ x – 1 – x2/ x2 – 1
a) Tinh gia tri bieu thuc A biet |x + 1| = 2
b) Tim x de A = -3/ 5
c) Tim x de A = 3x/ x + 2
d) Tim bieu thuc P = A -B
e) Tim x thuoc Z de A thuoc Z
Đáp án:
a) A=3
Giải thích các bước giải:
\(\begin{array}{l}
a)\left| {x + 1} \right| = 2\\
\to \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = – 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = – 3
\end{array} \right.\\
Thay:x = – 3\\
\to A = \dfrac{{2.\left( { – 3} \right)}}{{ – 3 + 1}} = 3\\
b)A = – \dfrac{3}{5}\\
\to \dfrac{{2x}}{{x + 1}} = – \dfrac{3}{5}\\
\to 10x = – 3x – 3\\
\to 13x = – 3\\
\to x = – \dfrac{3}{{13}}\\
c)A = \dfrac{{3x}}{{x + 2}}\\
\to \dfrac{{2x}}{{x + 1}} = \dfrac{{3x}}{{x + 2}}\\
\to 2{x^2} + 4x = 3{x^2} + 3x\\
\to {x^2} – x = 0\\
\to x\left( {x – 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( {TM} \right)\\
x = 1\left( l \right)
\end{array} \right.\\
d)P = A – B = \dfrac{{3x}}{{x + 2}} – \left( {\dfrac{{3x}}{{x – 1}} – \dfrac{{{x^2}}}{{{x^2} – 1}}} \right)\\
= \dfrac{{3x}}{{x + 2}} – \dfrac{{3{x^2} + 3x – {x^2}}}{{{x^2} – 1}}\\
= \dfrac{{3x}}{{x + 2}} – \dfrac{{2{x^2} + 3x}}{{{x^2} – 1}}\\
= \dfrac{{3{x^3} – 3x – \left( {x + 2} \right)\left( {2{x^2} + 3x} \right)}}{{\left( {x + 2} \right)\left( {{x^2} – 1} \right)}}\\
= \dfrac{{3{x^3} – 3x – 2{x^3} – 3{x^2} – 4{x^2} – 6x}}{{\left( {x + 2} \right)\left( {{x^2} – 1} \right)}}\\
= \dfrac{{{x^3} – 7{x^2} – 9x}}{{\left( {x + 2} \right)\left( {{x^2} – 1} \right)}}\\
e)A = \dfrac{{2x}}{{x + 1}} = \dfrac{{2\left( {x + 1} \right) – 2}}{{x + 1}} = 2 – \dfrac{2}{{x + 1}}\\
A \in Z\\
\Leftrightarrow \dfrac{2}{{x + 1}} \in Z\\
\Leftrightarrow x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = – 2\\
x + 1 = 1\\
x + 1 = – 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – 3\\
x = 0\\
x = – 2
\end{array} \right.
\end{array}\)