tìm hai số a, b \(\varepsilon Q\)
thỏa
\(\sqrt{a\sqrt{7}}-\sqrt{b\sqrt{7}}=\sqrt{11-\sqrt{28}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
7 tháng 8 2022
a: \(=6-\sqrt{15}+2\sqrt{15}=6+\sqrt{15}\)
b: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
\(=7-2\sqrt{21}+2\sqrt{21}=7\)
c: \(=10+5\sqrt{10}-5\sqrt{10}=10\)
d: \(=22-\sqrt{198}+\sqrt{198}=22\)
H9
HT.Phong (9A5)
CTVHS
12 tháng 9 2023
a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{2^2\cdot7}-\sqrt{3^2\cdot7}+\dfrac{\sqrt{7}\cdot\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)
\(=2\sqrt{7}-3\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1\)
\(=-\sqrt{7}\)
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\left[\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
\(=\dfrac{2\cdot4}{\sqrt{x}-3}\)
\(=\dfrac{8}{\sqrt{x}-3}\)
b) \(A>B\) khi
\(\dfrac{8}{\sqrt{x}-3}< -\sqrt{7}\)
\(\Leftrightarrow8< -\sqrt{7x}+3\sqrt{7}\)
\(\Leftrightarrow x< \dfrac{\left(3\sqrt{7}-8\right)^2}{7}\)
1 tháng 12 2016
- \(A=\sqrt{11-2\sqrt{10}}=\sqrt{\left(\sqrt{10}-1\right)^2}=\sqrt{10}-1\)
- \(B=\left(\sqrt{28}-2\sqrt{4}+\sqrt{7}\right).\sqrt{7}+7\sqrt{7}=\left(2\sqrt{7}-2\sqrt{4}+\sqrt{7}\right).\sqrt{7}+7\sqrt{7}\)
\(=\left(3\sqrt{7}-4\right).\sqrt{7}+7\sqrt{7}=3\sqrt{7}+3\sqrt{7}=6\sqrt{7}\)
- \(C=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\frac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
- \(D=0,2.\sqrt{10^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=2\sqrt{3}+2\left(\sqrt{3}-\sqrt{5}\right)=4\sqrt{3}-2\sqrt{5}\)