\(\frac{x+2,4\sqrt{x}}{3}\)
\(\Leftrightarrow\frac{x+\frac{12}{5}.\sqrt{x}}{3}\)
\(\Leftrightarrow\frac{\frac{5x+12\sqrt{x}}{5}}{3}\)
\(\Leftrightarrow\frac{5x+12\sqrt{x}}{15}\)
\(\frac{x+2,4\sqrt{x}}{3}\)
\(\Leftrightarrow\frac{x+\frac{12}{5}.\sqrt{x}}{3}\)
\(\Leftrightarrow\frac{\frac{5x+12\sqrt{x}}{5}}{3}\)
\(\Leftrightarrow\frac{5x+12\sqrt{x}}{15}\)
Tính : \(S=\frac{7}{\sqrt{2.2}+\sqrt{2.3}}+\frac{7}{\sqrt{2.3}+\sqrt{2.4}}+.....+\frac{7}{\sqrt{2.2018}+\sqrt{2.2019}}\)
Tính
3) \(\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{2x-\sqrt{x}-1}{x-\sqrt{x}+1}-\frac{3x\sqrt{x}-2x+\sqrt{x}-3}{x\sqrt{x}+1}\)
4) \(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
5)\(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-5\sqrt{x}+6}\)
Help !!! Mk đang cần gấp ,thank các ben
\(\frac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\frac{2\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
Bài 1: Tính :
\(C=\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
\(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(D=\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
Bài 2 : Cho \(P=\left(\frac{1}{\sqrt{x}-1}+\frac{x-\sqrt{x}+6}{x+\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{x-\sqrt{x}-2}{x+\sqrt{x}+2}\right)\)
a, Rút gọn P
b, Tìm GTNN
c, Tìm x để \(P.\frac{x-1}{x^2+8x}< -2\)
\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{\sqrt{x}+1}{1-\sqrt{x}}+\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(\left(\frac{1}{\sqrt{x}+3}-\frac{2\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-1}{x-3\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
a) Rút gọn biểu thức sau A=\(\sqrt{3+2\sqrt{2}}-\frac{1}{1+\sqrt{2}}\)
b)Chứng minh rằng:\(\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{3}{\sqrt{x}-3}\right).\frac{\sqrt{x}+3}{x+9}=\frac{1}{\sqrt{x}-3}\)với x≥0 và x ≠ 9