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a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)
mà 112<117
nên \(4\sqrt{7}< 3\sqrt{13}\)
b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)
\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)
mà \(\dfrac{21}{4}>\dfrac{36}{7}\)
nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)
d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)
\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)
mà 108>64
nên \(3\sqrt{12}>2\sqrt{16}\)
b: \(\sqrt{3}-1=\sqrt{4-2\sqrt{3}}\)
mà \(4-3\sqrt{3}< 4-2\sqrt{3}\)
nên \(\sqrt{4-3\sqrt{3}}< \sqrt{3}-1\)
Đề này sai rồi bạn vì \(4-3\sqrt{3}< 0\)
\(A=\sqrt{24+8\sqrt{5}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{5+2.4\sqrt{5}+16}+\sqrt{4-2.2\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{5}+4\right)}^2+\sqrt{\left(2-\sqrt{3}\right)}^2\)
\(=|\sqrt{5}+4|+|2-\sqrt{3}|\)
\(=\sqrt{5}+4+4-\sqrt{3}\)
\(=\sqrt{5}-\sqrt{3}+8\)
Ko biết đề sai ko?
c.
(\sqrt{5}-\sqrt{3})-(\sqrt{10}-\sqrt{7})=(\sqrt{5}+\sqrt{7})-(\sqrt{3}+\sqrt{10})
Mà:
\((\sqrt{5}+\sqrt{7})^2=12+\sqrt{35}< 12+\sqrt{36}=18\)
\((\sqrt{3}+\sqrt{10})^2=13+\sqrt{30}>13+\sqrt{25}=18\)
\(\Rightarrow \sqrt{3}+\sqrt{10}> \sqrt{5}+\sqrt{7}\Rightarrow \sqrt{5}-\sqrt{3}< \sqrt{10}-\sqrt{7}\)
Lời giải:
a.
$5+\sqrt{2}>5+\sqrt{1}=6$
$4+\sqrt{3}< 4+\sqrt{4}=6$
$\Rightarrow 5+\sqrt{2}>4+\sqrt{3}$
b.
$\sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}$
$\sqrt{5}-\sqrt{3}=\frac{5-3}{\sqrt{5}+\sqrt{3}}=\frac{2}{\sqrt{5}+\sqrt{3}}< \frac{2}{\sqrt{2}}=\sqrt{2}$
Vậy $\sqrt{8}-\sqrt{2}>\sqrt{5}-\sqrt{2}$
1) \(\sqrt[3]{x+1}=5\)
\(\Rightarrow x+1=125\)
\(\Rightarrow x=124\)
2) \(\sqrt[3]{1-3x^3}=-2\)
\(\Rightarrow1-3x^3=-8\)
\(\Rightarrow3x^3=9\)
\(\Rightarrow x=\sqrt[3]{3}\)
\(a)\)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\)\(\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}\)
\(=\)\(\sqrt{\left(\sqrt{6}+3\right)}+\sqrt{\left(\sqrt{24}+3\right)}\)
\(=\)\(\left|\sqrt{6}+3\right|+\left|\sqrt{24}+3\right|\)
\(=\)\(\sqrt{6}+3+\sqrt{24}+3\)
\(=\)\(\sqrt{6}\left(1+\sqrt{4}\right)+9\)
\(=\)\(3\sqrt{6}+9\)
Chúc bạn học tốt ~
\(b)\)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=\)\(\left|2-\sqrt{3}\right|+\sqrt{3-2\sqrt{3}+1}\)
\(=\)\(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\) ( vì \(2=\sqrt{4}>\sqrt{3}\) )
\(=\)\(2-\sqrt{3}+\left|\sqrt{3}-1\right|\)
\(=\)\(2-\sqrt{3}+\sqrt{3}-1\) ( vì \(\sqrt{3}>\sqrt{1}=1\) )
\(=\)\(1\)
Chúc bạn học tốt ~
PS : mới lớp 8 sai thì thông cảm >.<
a, \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{17-2.3.2\sqrt{2}}-\sqrt{17+2.3.2\sqrt{2}}\)
\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}=\left|3-2\sqrt{2}\right|-\left|3+2\sqrt{2}\right|\)
\(=3-2\sqrt{2}-3-2\sqrt{2}=-4\sqrt{2}\)
b, \(\sqrt{31-12\sqrt{3}}-\sqrt{31+12\sqrt{3}}\)
\(=\sqrt{31-2.2.3\sqrt{3}}-\sqrt{31+2.2.3\sqrt{3}}\)
\(=\sqrt{\left(3\sqrt{3}-2\right)^2}-\sqrt{\left(3\sqrt{3}+2\right)^2}=\left|3\sqrt{3}-2\right|-\left|3\sqrt{3}+2\right|\)
\(=3\sqrt{3}-2-3\sqrt{3}-2=-4\)
\(\sqrt{7}+\sqrt{5}>\sqrt{12}\)
\(\sqrt{8}+3>6+\sqrt{2}\)
Ta có:
\(a.\)Ta có:
\(7>4\) nên \(\sqrt{7}>\sqrt{4}\)
\(\Rightarrow\) \(\sqrt{7}>2\) \(\left(1\right)\)
và \(5>4\) nên \(\sqrt{5}>\sqrt{4}\)
\(\Rightarrow\) \(\sqrt{5}>2\) \(\left(2\right)\)
Mặt khác, ta lại có: \(\sqrt{12}< \sqrt{16}=4\) \(\left(i\right)\)
Do đó, từ hai bđt \(\left(1\right)\) và \(\left(2\right)\) , kết hợp với chú ý \(\left(i\right)\) ta suy ra được:
\(\sqrt{7}+\sqrt{5}>\sqrt{12}\)