a, A= √(x-3) – √1/4-x
b, B= 1/(√x-1) + 2/(√ x^2-4x +4)
giúp mik với mik sẽ vote 5 sao
Cảm ơn ạ
0 bình luận về “a, A= √(x-3) – √1/4-x
b, B= 1/(√x-1) + 2/(√ x^2-4x +4)
giúp mik với mik sẽ vote 5 sao
Cảm ơn ạ”
Đáp án:
b. \(\left[ \begin{array}{l} B = \dfrac{1}{{\sqrt x – 1}} + \dfrac{2}{{x – 2}}\\ B = \dfrac{1}{{\sqrt x – 1}} + \dfrac{2}{{ – x + 2}} \end{array} \right.\)
Đáp án:
b. \(\left[ \begin{array}{l}
B = \dfrac{1}{{\sqrt x – 1}} + \dfrac{2}{{x – 2}}\\
B = \dfrac{1}{{\sqrt x – 1}} + \dfrac{2}{{ – x + 2}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:3 \le x < 4\\
A = \sqrt {x – 3} – \sqrt {\dfrac{1}{{4 – x}}} \\
= \dfrac{{\sqrt {\left( {x – 3} \right)\left( {4 – x} \right)} – 1}}{{\sqrt {4 – x} }}\\
= \dfrac{{\sqrt { – {x^2} + 7x – 12} – 1}}{{\sqrt {4 – x} }}\\
b.DK:x \ge 0;x \ne \left\{ {1;2} \right\}\\
B = \dfrac{1}{{\sqrt x – 1}} + \dfrac{2}{{\sqrt {{x^2} – 4x + 4} }}\\
= \dfrac{1}{{\sqrt x – 1}} + \dfrac{2}{{\sqrt {{{\left( {x – 2} \right)}^2}} }}\\
= \dfrac{1}{{\sqrt x – 1}} + \dfrac{2}{{\left| {x – 2} \right|}}\\
\to \left[ \begin{array}{l}
B = \dfrac{1}{{\sqrt x – 1}} + \dfrac{2}{{x – 2}}\left( {DK:x > 2} \right)\\
B = \dfrac{1}{{\sqrt x – 1}} + \dfrac{2}{{ – x + 2}}\left( {DK:0 \le x < 2;x \ne 1} \right)
\end{array} \right.
\end{array}\)