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Lời giải:
ĐKXĐ: $x\geq 5$
$2x^2-8x-6=2\sqrt{x-5}\leq (x-5)+1$ theo BĐT Cô-si
$\Leftrightarrow 2x^2-9x-2\leq 0$
$\Leftrightarrow 2x(x-5)+(x-2)\leq 0$
Điều này vô lý do $2x(x-5)\geq 0; x-2\geq 3>0$ với mọi $x\geq 5$
Vậy pt vô nghiệm nên không có đáp án nào đúng.
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
+) ta có : \(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
+) ta có : \(B=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1=1\)
+) điều kiện : \(x\ge1\)
ta có : \(C=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}C=2\sqrt{x-1}\left(x\ge2\right)\\C=2\left(1\le x< 2\right)\end{matrix}\right.\)
câu c này mk sữa đề nhát
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
Bài 2:
Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
12) \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
13) \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=3\sqrt{5}-1-2\sqrt{5}+3\)
\(=\sqrt{5}+2\)
a)
\(3\sqrt{5}=\sqrt{9.5}=\sqrt{45}\)
\(2\sqrt{6}=\sqrt{4.6}=\sqrt{24}\)
\(4\sqrt{2}=\sqrt{16.2}=\sqrt{32}\)
Do 24 < 29 < 32 < 45 => \(\sqrt{24}< \sqrt{29}< \sqrt{32}< \sqrt{45}\)
=> \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)
b)
\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\\ 3\sqrt{8}=\sqrt{9.8}=\sqrt{72}\\ 2\sqrt{15}=\sqrt{4.15}=\sqrt{60}\)
Do 39 < 50 < 60 < 72 nên \(\sqrt{39}< \sqrt{50}< \sqrt{60}< \sqrt{72}\)
=> \(\sqrt{39}< 5\sqrt{2}< 2\sqrt{15}< 3\sqrt{8}\)
a: 3căn5=căn 45
2căn 6=căn 24
căn 29=căn 29
4căn2=căn 32
=>2căn6<căn29<4căn2<3căn5
b: 5căn 2=căn 50
căn 39=căn 39
3căn 8=căn 72
2căn 15=căn60
=>căn 39<5căn2<2căn15<3căn8
Xét :\(\dfrac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Do đó :
\(S< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{5}\right)=\dfrac{2}{5}\)
Đáp án A