( √a/2-1/2 √a) ∧2 × (√a-1/ √a+1- √a+1/ √a-1) 26/10/2021 Bởi Vivian ( √a/2-1/2 √a) ∧2 × (√a-1/ √a+1- √a+1/ √a-1)
ĐK: $a>0, a\ne 1$ $\Big( \dfrac{ \sqrt{a} }{2}-\dfrac{1}{2\sqrt{a}} \Big)^2.\Big( \dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\Big)$ $=\Big(\dfrac{ a-1}{2\sqrt{a}}\Big)^2.\Big( \dfrac{ (\sqrt{a}-1)^2-(\sqrt{a}+1)^2 }{a-1}\Big)$ $=\dfrac{(a-1)^2}{4a}.\dfrac{ a-2\sqrt{a}+1-a-2\sqrt{a}-1 }{(a-1)}$ $=\dfrac{a-1}{4a}.(-4\sqrt{a})$ $=-\dfrac{a-1}{\sqrt{a}}$ Bình luận
Đáp án: $\frac{1 – a}{√a}$ Giải thích các bước giải: $ĐK : a > 0 ; a \neq 1$ $(\frac{√a}{2}-\frac{1}{2√a})² . (\frac{√a-1}{√a + 1} – \frac{√a + 1}{√a – 1})$ $=(\frac{a – 1}{2√a})². \frac{(√a – 1)² – (√a + )²}{(√a – 1)(√a + 1)}$ $=\frac{(a – 1)²}{4a}.\frac{a – 2√a + 1 – a – 2√a – 1}{a – 1}$ $= \frac{(a – 1)²}{4a}.\frac{-4√a}{a – 1}$ $= \frac{-(a – 1)}{√a}$ $= \frac{1 – a}{√a}$ Bình luận
ĐK: $a>0, a\ne 1$
$\Big( \dfrac{ \sqrt{a} }{2}-\dfrac{1}{2\sqrt{a}} \Big)^2.\Big( \dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\Big)$
$=\Big(\dfrac{ a-1}{2\sqrt{a}}\Big)^2.\Big( \dfrac{ (\sqrt{a}-1)^2-(\sqrt{a}+1)^2 }{a-1}\Big)$
$=\dfrac{(a-1)^2}{4a}.\dfrac{ a-2\sqrt{a}+1-a-2\sqrt{a}-1 }{(a-1)}$
$=\dfrac{a-1}{4a}.(-4\sqrt{a})$
$=-\dfrac{a-1}{\sqrt{a}}$
Đáp án:
$\frac{1 – a}{√a}$
Giải thích các bước giải:
$ĐK : a > 0 ; a \neq 1$
$(\frac{√a}{2}-\frac{1}{2√a})² . (\frac{√a-1}{√a + 1} – \frac{√a + 1}{√a – 1})$
$=(\frac{a – 1}{2√a})². \frac{(√a – 1)² – (√a + )²}{(√a – 1)(√a + 1)}$
$=\frac{(a – 1)²}{4a}.\frac{a – 2√a + 1 – a – 2√a – 1}{a – 1}$
$= \frac{(a – 1)²}{4a}.\frac{-4√a}{a – 1}$
$= \frac{-(a – 1)}{√a}$
$= \frac{1 – a}{√a}$