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a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
2.1
\(A=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5.1}+1}-\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}=|\sqrt{5}+1|-|\sqrt{5}-1|=2\)
2.2
\(B\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{3+2\sqrt{3.5}+5}+\sqrt{3-2\sqrt{3.5}+5}-2\sqrt{5-2\sqrt{5.1}+1}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5})^2}+\sqrt{(\sqrt{3}-\sqrt{5})^2}-2\sqrt{(\sqrt{5}-1)^2}\)
\(=|\sqrt{3}+\sqrt{5}|+|\sqrt{3}-\sqrt{5}|-2|\sqrt{5}-1|=2\)
$\Rightarrow B=\sqrt{2}$
Bài 1:
1. ĐKXĐ: \(\left\{\begin{matrix} 2x-1\geq 0\\ x-3\geq 0\\ 5-x>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ x\geq 3\\ x< 5\end{matrix}\right.\Leftrightarrow 3\leq x< 5\)
2.
ĐKXĐ: \(\left\{\begin{matrix} x-1\geq 0\\ 2-x\geq 0\\ x+1>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x\leq 2\\ x>-1\end{matrix}\right.\Leftrightarrow 1\leq x\leq 2\)
a)
\(4\sqrt{7}=\sqrt{4^2.7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2.13}=\sqrt{117}\)
\(\sqrt{112}< \sqrt{117}\Rightarrow 4\sqrt{7}< 3\sqrt{13}\)
b) \(3\sqrt{12}=\sqrt{3^2.12}=\sqrt{9.2^2.3}=2\sqrt{27}>2\sqrt{16}\)
c)
\(\frac{1}{4}\sqrt{82}=\sqrt{\frac{82}{16}}=\sqrt{\frac{41}{8}}=\sqrt{5+\frac{1}{8}}\)
\(6\sqrt{\frac{1}{7}}=\sqrt{\frac{36}{7}}=\sqrt{5+\frac{1}{7}}\)
\(\sqrt{5+\frac{1}{8}}< \sqrt{5+\frac{1}{7}}\Rightarrow \frac{1}{4}\sqrt{82}< 6\sqrt{\frac{1}{7}}\)
d)
\(\frac{1}{2}\sqrt{\frac{17}{2}}=\sqrt{\frac{17}{8}}=\sqrt{2+\frac{1}{8}}\)
\(\frac{1}{3}\sqrt{19}=\sqrt{\frac{19}{9}}=\sqrt{2+\frac{1}{9}}\)
\(\sqrt{2+\frac{1}{8}}>\sqrt{2+\frac{1}{9}}\Rightarrow \frac{1}{2}\sqrt{\frac{17}{2}}> \frac{1}{3}\sqrt{19}\)
e)
\(3\sqrt{3}-2\sqrt{2}=\sqrt{27}-\sqrt{8}\)
Mà \(\sqrt{27}>\sqrt{25}; \sqrt{8}< \sqrt{9}\Rightarrow \sqrt{27}-\sqrt{8}> \sqrt{25}-\sqrt{9}=5-3=2\)
Vậy \(3\sqrt{3}-2\sqrt{2}>2\)
f)
\(\sqrt{7}+\sqrt{5}< \sqrt{9}+\sqrt{9}=6\)
\(\sqrt{49}=7\)
\(\Rightarrow \sqrt{7}+\sqrt{5}< 6< 7=\sqrt{49}\)
g)
\(\sqrt{2}< \sqrt{3}; \sqrt{11}< \sqrt{25}=5\)
\(\Rightarrow \sqrt{2}+\sqrt{11}< \sqrt{3}+5\)
h) Lặp lại câu d
i)
\(\sqrt{21}>\sqrt{20}\); \(\sqrt{5}< \sqrt{6}\)
\(\Rightarrow \sqrt{21}-\sqrt{5}> \sqrt{20}-\sqrt{6}\)
1) \(2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\)\(6+2\sqrt{2}< 6+3=9\)
2) \(4\sqrt{5}=\sqrt{80}>\sqrt{49}=7\)
\(\Rightarrow\)\(9+4\sqrt{5}>9+7=16\)
3) \(2=\sqrt{4}>\sqrt{3}\)
\(\Rightarrow\)\(2-1>\sqrt{3}-1\)
hay \(1>\sqrt{3}-1\)
4) \(9-4\sqrt{5}< 16\)
5) \(\sqrt{2}>\sqrt{1}=1\)
\(\Rightarrow\)\(\sqrt{2}+1>2\)
MÌNH KHÔNG VIẾT LẠI ĐỀ NHA
\(A=\left[\left(\sqrt{2}+1\right).\left(\sqrt{3}+1\right)\right].\left[\left(\sqrt{6}+1\right).\left(5-2\sqrt{2}-\sqrt{3}\right)\right]\)
\(A=\left(\sqrt{6}+\sqrt{2}+\sqrt{3}+1\right).\left(5\sqrt{6}-2\sqrt{12}-\sqrt{18}+5-2\sqrt{2}-\sqrt{3}\right)\)
\(A=\left(\sqrt{6}+\sqrt{2}+\sqrt{3}+1\right).\left(5\sqrt{6}-4\sqrt{3}-3\sqrt{2}+5-2\sqrt{2}-\sqrt{3}\right)\)
\(A=\left(\sqrt{6}+\sqrt{2}+\sqrt{3}+1\right).\left(5\sqrt{6}-5\sqrt{3}-5\sqrt{2}+5\right)\)
\(A=30-15\sqrt{2}-10\sqrt{3}+5\sqrt{6}+10\sqrt{3}-5\sqrt{6}-10+5\sqrt{2}+15\sqrt{2}\) \(-15-5\sqrt{6}+5\sqrt{3}+5\sqrt{6}-5\sqrt{3}-5\sqrt{2}+5\)
\(A=10+0\sqrt{2}+0\sqrt{3}+0\sqrt{6}\) ( hàng này bạn có thể không ghi cũng được )
\(A=10+0+0+0\)
\(A=10\)
vậy A = 10
chúc bạn học tốt !!!