(x+√(x^2-2x))/(x-√(x^2-2x)) – (x-√(x^2-2x))/(x+√(x^2-2x))
a/tìm ĐKXD
b/Rút gọn A
c/Tìm x để A<2
Mình sẽ vote 5* cho bạn nào trả lời chính xác nhất
(x+√(x^2-2x))/(x-√(x^2-2x)) – (x-√(x^2-2x))/(x+√(x^2-2x)) a/tìm ĐKXD b/Rút gọn A c/Tìm x để A<2 Mình sẽ vote 5* cho bạn nào trả lời chính xác nh
By Kinsley
Đáp án:
a)ĐKXĐ:\(\begin{cases}x^2-2x \ge 0\\x+\sqrt{x^2-2x} \ne 0\\x-\sqrt{x^2-2x} \ne 0\\\end{cases}\)
`<=>` \(\begin{cases}x(x-2) \ge 0\\x \ne 0\\x \ne 0\\\end{cases}\)
`<=>` \(\begin{cases}\left[ \begin{array}{l}x\ge2\\x \le 0\end{array} \right.\\x \ne 0\\\end{cases}\)
`<=>` \(\left[ \begin{array}{l}x\ge 2\\x<0\end{array} \right.\)
`b)A=(x+\sqrt{x^2-2x})/(x-\sqrt{x^2-2x})-(x-\sqrt{x^2-2x})/(x+\sqrt{x^2-2x})`
`=(x+\sqrt{x^2-2x})^2/(x^2-(x^2-2x))-((x-\sqrt{x^2-2x})^2)/(x^2-(x^2-2x))`(Nhân liên hợp)
`=((x+\sqrt{x^2-2x})^2-(x-\sqrt{x^2-2x})^2)/(2x)`
`=(x^2+x^2-2x-x^2-x^2+2x+2x\sqrt{x^2-2x}+2x\sqrt{x^2-2x})/(2x)`
`=(4x\sqrt{x^2-2x})/(2x)`
`=2\sqrt{x^2-2x}`
`c)A<2`
`<=>\sqrt{x^2-2x}<1`
`<=>x^2-2x<1`
`<=>x^2-2x+1<2`
`<=>(x-1)^2<2`
`<=>-\sqrt2<x-1<\sqrt2`
`<=>1-sqrt2<x<1+sqrt2`
Kết hợp đkxđ ta có:\(\left[ \begin{array}{l}2 \le x<1+\sqrt2\\1-\sqrt2<x<0\end{array} \right.\)
Đáp án:
a) \(\left[ \begin{array}{l}
x \ge 2\\
x < 0
\end{array} \right.\)
b) \(\sqrt {{x^2} – 2x} \)
c) \(\left[ \begin{array}{l}
2 \le x < 1 + \sqrt 2 \\
0 > x > 1 – \sqrt 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{{x + \sqrt {{x^2} – 2x} }}{{x – \sqrt {{x^2} – 2x} }} – \dfrac{{x – \sqrt {{x^2} – 2x} }}{{x + \sqrt {{x^2} – 2x} }}\\
a)DK:\left\{ \begin{array}{l}
x – \sqrt {{x^2} – 2x} \ne 0\\
{x^2} – 2x \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x^2} \ne {x^2} – 2x\\
\left[ \begin{array}{l}
x \ge 2\\
x \le 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x \ge 2\\
x < 0
\end{array} \right.\\
b)A = \dfrac{{x + \sqrt {{x^2} – 2x} }}{{x – \sqrt {{x^2} – 2x} }} – \dfrac{{x – \sqrt {{x^2} – 2x} }}{{x + \sqrt {{x^2} – 2x} }}\\
= \dfrac{{{x^2} + 2x\sqrt {{x^2} – 2x} + {x^2} – 2x – \left( {{x^2} – 2x\sqrt {{x^2} – 2x} + {x^2} – 2x} \right)}}{{\left( {x – \sqrt {{x^2} – 2x} } \right)\left( {x + \sqrt {{x^2} – 2x} } \right)}}\\
= \dfrac{{2{x^2} – 2x + 2x\sqrt {{x^2} – 2x} – 2{x^2} + 2x + 2x\sqrt {{x^2} – 2x} }}{{\left( {x – \sqrt {{x^2} – 2x} } \right)\left( {x + \sqrt {{x^2} – 2x} } \right)}}\\
= \dfrac{{4x\sqrt {{x^2} – 2x} }}{{{x^2} – \left( {{x^2} – 2x} \right)}}\\
= \dfrac{{4x\sqrt {{x^2} – 2x} }}{{2x}}\\
= 2\sqrt {{x^2} – 2x} \\
c)A < 2\\
\to 2\sqrt {{x^2} – 2x} < 2\\
\to \sqrt {{x^2} – 2x} < 1\\
\to {x^2} – 2x < 1\\
\to {x^2} – 2x + 1 < 2\\
\to {\left( {x – 1} \right)^2} < 2\\
\to \left[ \begin{array}{l}
x – 1 < \sqrt 2 \\
x – 1 > – \sqrt 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x < 1 + \sqrt 2 \\
x > 1 – \sqrt 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
2 \le x < 1 + \sqrt 2 \\
0 > x > 1 – \sqrt 2
\end{array} \right.
\end{array}\)