\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{8\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}-x-3}{x-1}-\frac{1}{\sqrt{x}-1}\right)\)
a) Tính A với \(x=6-2\sqrt{5}\)
b) C/m: A \(\le1\)
Gpt: a) \(\sqrt[4]{3\left(x+5\right)}-\sqrt[4]{11-x}=\sqrt[4]{13+x}-\sqrt[4]{3\left(3-x\right)}\)
b) \(\frac{1+2\sqrt{x}-x\sqrt{x}}{3-x-\sqrt{2-x}}=2\left(\frac{1+x\sqrt{x}}{1+x}\right)\) c) \(\sqrt{x+1}+\frac{4\left(\sqrt{x+1}+\sqrt{x-2}\right)}{3\left(\sqrt{x-2}+1\right)^2}=3\)
d) \(\sqrt{\frac{x-2}{x+1}}+\frac{x+2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}=1\) e) \(2x+1+x\sqrt{x^2+2}+\left(x+1\right)\sqrt{x^2+2x+2}=0\)
f) \(\sqrt{2x+3}\cdot\sqrt[3]{x+5}=x^2+x-6\)
1/ Tính:
a) \(\frac{\sqrt{6+\sqrt{11}}-\sqrt{7-\sqrt{33}}}{\sqrt{6}+\sqrt{2}}\)
b) \(\frac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\frac{2}{4+\sqrt{15}}-\frac{5\sqrt{5}+3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
2/ Rút Gọn: với a ≥ 0, a ≠ 1
B=\(\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\left(\frac{1+\sqrt{a}}{a-1}\right)^2\)
3/ Cho biểu thức: A = \(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{3-3\sqrt{x}}{x-5\sqrt{x}+6}\)
a) Tìm điều kiện xác định của A
b) Rút gọn A
c) Tìm x để A < -1
Rút gọn các biểu thức sau:
a) A=\(\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-1}+\frac{2\sqrt{2}-\sqrt{6}}{\sqrt{2}}\)
b)B=\(\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\left(x\ge0;x\ne1\right)\)
Rút gọn A = \(\left(\frac{x+2\sqrt{x}+4}{x\sqrt{x}-8}+\frac{x+2\sqrt{x}+1}{x-1}\right) :\left(3+\frac{1}{\sqrt{x}-2}+\frac{2}{\sqrt{x}+1}\right)\)
a, Rút gọn A b , Tìm x thỏa mãn A > 1 c,Tính A với \(x=\frac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\)\(A=\frac{\sqrt{x}+1}{3\left(\sqrt{x}-1\right)}\)
P = \(\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}+1}{x+2\sqrt{x}+1}\right)\)
c) Tìm x để A = \(\frac{1}{P}\) đạt GTNN
Tìm GTLN:
\(A=\frac{\sqrt{10x-49}}{2020}\\ B=\frac{\sqrt{2x^2-25}}{2020x^2}\\ C=\frac{7x^8+256}{x^7}\left(x>0\right)\\ D=\frac{\sqrt{x}+6\sqrt{x}+34}{\sqrt{x}+3}\\ E=x+\frac{1}{x-1}\left(x>1\right)\)
Giải hệ:
\(\left\{{}\begin{matrix}x^2+y^2+xy=5\\27x^3+6y^2x=2+y^3+30x^2y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y^2+\frac{8xy}{x+y}=16\\\frac{x^2}{8y}+\frac{2x}{3}=\sqrt{\frac{x^3}{3y}+\frac{x^2}{4}}-\frac{y}{2}\end{matrix}\right.\), \(\left\{{}\begin{matrix}\frac{1}{3x}+\frac{2x}{3y}=\frac{x+\sqrt{y}}{2x^2+y}\\2\left(2x+\sqrt{y}\right)=\sqrt{2x+6}-y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2y-3x-1=3x\sqrt{y}\left(\sqrt{1-x}-1\right)^3\\\sqrt{8x^2-3xy+4y^2}+\sqrt{xy}=4y\end{matrix}\right.\)
Cho các số a,b,c là các số k âm sao cho tổng hai số bất kì đều dương.CMR \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}+\frac{16\sqrt{ab+bc+ac}}{a+b+c}\ge8\)
Cho biểu thức A=\(\left(\frac{x+2\sqrt{x}+4}{x\sqrt{x}-8}+\frac{x+2\sqrt{x}+1}{x-1}\right):\left(3+\frac{1}{\sqrt{x}-2}+\frac{2}{\sqrt{x}+1}\right)\)
Rút gọn A?
b, Tính A biết x=\(\frac{\sqrt{7+\sqrt{5}}+\sqrt{7-\sqrt{5}}}{\sqrt{7+2\sqrt{11}}}+\sqrt{83-18\sqrt{2}}\)