Cho a ≥ 0. Chứng minh rằng:
a ² – √a / a+ √a +1 – a ² + √a / a – √a +1 +a+1 = ( √a -1) ²
0 bình luận về “Cho a ≥ 0. Chứng minh rằng:
a ² – √a / a+ √a +1 – a ² + √a / a – √a +1 +a+1 = ( √a -1) ²”
Đáp án:
\(\dfrac{{{a^2} – \sqrt a }}{{a + \sqrt a + 1}} – \dfrac{{{a^2} + \sqrt a }}{{a – \sqrt a + 1}} + a + 1 = {\left( {\sqrt a – 1} \right)^2}\)
Giải thích các bước giải:
\(\begin{array}{l} \dfrac{{{a^2} – \sqrt a }}{{a + \sqrt a + 1}} – \dfrac{{{a^2} + \sqrt a }}{{a – \sqrt a + 1}} + a + 1 = {\left( {\sqrt a – 1} \right)^2}\\ VT = \dfrac{{\sqrt a \left( {a\sqrt a – 1} \right)}}{{a + \sqrt a + 1}} – \dfrac{{\sqrt a \left( {a\sqrt a + 1} \right)}}{{a – \sqrt a + 1}} + a + 1\\ = \dfrac{{\sqrt a \left( {\sqrt a – 1} \right)\left( {a + \sqrt a + 1} \right)}}{{a + \sqrt a + 1}} – \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)\left( {a – \sqrt a + 1} \right)}}{{a – \sqrt a + 1}} + a + 1\\ = \sqrt a \left( {\sqrt a – 1} \right) – \sqrt a \left( {\sqrt a + 1} \right) + a + 1\\ = a – \sqrt a – a – \sqrt a + a + 1\\ = a – 2\sqrt a + 1\\ = {\left( {\sqrt a – 1} \right)^2} = VP\\ \to dpcm \end{array}\)
Đáp án:
\(\dfrac{{{a^2} – \sqrt a }}{{a + \sqrt a + 1}} – \dfrac{{{a^2} + \sqrt a }}{{a – \sqrt a + 1}} + a + 1 = {\left( {\sqrt a – 1} \right)^2}\)
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{{{a^2} – \sqrt a }}{{a + \sqrt a + 1}} – \dfrac{{{a^2} + \sqrt a }}{{a – \sqrt a + 1}} + a + 1 = {\left( {\sqrt a – 1} \right)^2}\\
VT = \dfrac{{\sqrt a \left( {a\sqrt a – 1} \right)}}{{a + \sqrt a + 1}} – \dfrac{{\sqrt a \left( {a\sqrt a + 1} \right)}}{{a – \sqrt a + 1}} + a + 1\\
= \dfrac{{\sqrt a \left( {\sqrt a – 1} \right)\left( {a + \sqrt a + 1} \right)}}{{a + \sqrt a + 1}} – \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)\left( {a – \sqrt a + 1} \right)}}{{a – \sqrt a + 1}} + a + 1\\
= \sqrt a \left( {\sqrt a – 1} \right) – \sqrt a \left( {\sqrt a + 1} \right) + a + 1\\
= a – \sqrt a – a – \sqrt a + a + 1\\
= a – 2\sqrt a + 1\\
= {\left( {\sqrt a – 1} \right)^2} = VP\\
\to dpcm
\end{array}\)
Đáp án: $\dfrac{\sqrt{a}(a\sqrt{a}-1)}{a+\sqrt{a}+1}-\dfrac{\sqrt{a}(a\sqrt{a}+1)}{a-\sqrt{a}+1}+a+1=(\sqrt{a}-1)^{2}$
Giải thích các bước giải:
$\dfrac{\sqrt{a}(a\sqrt{a}-1)}{a+\sqrt{a}+1}-\dfrac{\sqrt{a}(a\sqrt{a}+1)}{a-\sqrt{a}+1}+a+1\\=\dfrac{\sqrt{a}(\sqrt{a}-1)(a+\sqrt{a}+1)}{a+\sqrt{a}+1}-\dfrac{\sqrt{a}(\sqrt{a}+1)(a-\sqrt{a}+1)}{a-\sqrt{a}+1}+a+1\\=\sqrt{a}(\sqrt{a}-1)-\sqrt{a}(\sqrt{a}+1)+a+1\\=a-\sqrt{a}-a-\sqrt{a}+a+1\\=-2\sqrt{a}+a+1\\=(\sqrt{a}-1)^{2}\rightarrow đccm$