Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Khi \(x=1,44\): \(A=\frac{1,44+7}{\sqrt{1,44}}=\frac{8,44}{1,2}=\frac{211}{30}\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-1}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\)(ĐK: \(x\ge0,x\ne9\))
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2x-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+5\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(S=\frac{1}{B}+A=\frac{\sqrt{x}-3}{\sqrt{x}}+\frac{x+7}{\sqrt{x}}=\frac{x+\sqrt{x}+4}{\sqrt{x}}=\sqrt{x}+\frac{4}{\sqrt{x}}+1\)
\(\ge2\sqrt{\sqrt{x}.\frac{4}{\sqrt{x}}}+1=5\)
Dấu \(=\)khi \(\sqrt{x}=\frac{4}{\sqrt{x}}\Leftrightarrow x=4\)(thỏa mãn)
a/ Với x = \(23-12\sqrt{3}\) ta có:
\(x-11=23-12\sqrt{3}-11=12-12\sqrt{3}=12\left(1-\sqrt{3}\right)\)
\(\sqrt{x-2}-3=\sqrt{23-12\sqrt{3}-2}-3=\sqrt{21-12\sqrt{3}}-3=\sqrt{3^2-2.3.2\sqrt{3}+\left(2\sqrt{3}\right)^2}-3=\sqrt{\left(3-2\sqrt{3}\right)^2}-3=2\sqrt{3}-6\) \(=2\sqrt{3}\left(1-\sqrt{3}\right)\)
=>\(\frac{x-11}{\sqrt{x-2}-3}=\frac{12\left(1-\sqrt{3}\right)}{2\sqrt{3}\left(1-\sqrt{3}\right)}=\frac{12}{2\sqrt{3}}=\frac{2\sqrt{3}.2\sqrt{3}}{2\sqrt{3}}=2\sqrt{3}\)
b/ \(\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}=\frac{1-\sqrt{a}}{2\left(1-a\right)}+\frac{1+\sqrt{a}}{2\left(1-a\right)}-\frac{a^2+2}{\left(1-a\right)\left(1-a+a^2\right)}\)
=\(\frac{2}{2\left(1-a\right)}-\frac{a^2+2}{\left(1-a\right)\left(1-a+a^2\right)}=\frac{1-a+a^2-a^2-2}{\left(1-a\right)\left(1-a+a^2\right)}=\frac{-a-1}{1-a^3}\)
Thay : \(a=\sqrt{2}tacó:\)
\(\frac{-\sqrt{2}-1}{1-\sqrt{2}^3}=\frac{-\left(1+\sqrt{2}\right)}{1-2\sqrt{2}}\)
a, Ta có: \(x=4-2\sqrt{3}\)\(=3-2\sqrt{3}+1\)\(=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}\)\(=\sqrt{3}-1\)
Thay \(\sqrt{x}=\sqrt{3}-1\) vào biểu thức P ta có:
\(P=\frac{\sqrt{3}-1+1}{\sqrt{3}-1-4}\)\(=\frac{\sqrt{3}}{\sqrt{3}-5}\)\(=\frac{\sqrt{3}.\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right).\left(\sqrt{3}+5\right)}\)\(=\frac{3-5\sqrt{3}}{3-25}\)\(=\frac{5\sqrt{3}-3}{22}\)
Vậy \(P=\frac{5\sqrt{3}-3}{22}\)khi \(x=4-2\sqrt{3}\)
b, \(E=\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\)\(=\frac{\sqrt{3}+1}{\left(\sqrt{3}-1\right).\left(\sqrt{3}+1\right)}\)\(-\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)}\)
\(=\frac{\sqrt{3}+1-\sqrt{3}+1}{3-1}\) \(=\frac{2}{2}=1\)
a, Ta có : \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
Thay vào P ta được : \(P=\frac{\sqrt{3}-1+1}{\sqrt{3}-1-4}=\frac{\sqrt{3}}{\sqrt{3}-5}=\frac{\sqrt{3}\left(\sqrt{3}+5\right)}{-22}=-\frac{3+5\sqrt{3}}{22}\)
b, \(E=\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}+1-\sqrt{3}+1}{2}=1\)
F = [3/(√1 + a) + (√1 - a)] : [3/(√1 - a^2) + 1] .
=[3+√1-a^2)/√1+a]:[(3+√1-a^2/√1-a^2]
=(3+√1-a^2/√1+a].[√1-a^2/(3+√1-a^2]
=√1-a.√1+a/√1+a=√1-a
thay a=√3/(2+√3) vào F ta được
√[1-(√3/(2+√3)]=√2/(2+√3)
=√2/(2+√3)=(√4-2√3)/4-3=√(√3-1)^2=|√3-1}
=√3-1