Các câu hỏi tương tự
1. Chứng minh:\(\left(\frac{x\sqrt{x}+27y\sqrt{y}}{3\sqrt{x}+9\sqrt{y}}-\sqrt{xy}\right).\left(\frac{3\sqrt{x}+9\sqrt{y}}{9y-x}\right)^2>\sqrt{8}\)
2. Rút gọn A= \(\frac{\sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}}{\sqrt{a+\sqrt{2a-1}}-\sqrt{a-\sqrt{2a-1}}}\)
Tính :\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+....+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
So sánh A và B :
a)
\(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\)
\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
b)
\(A=\frac{1}{\sqrt{121}}+\frac{1}{\sqrt{12321}}+\frac{1}{\sqrt{1234321}}+...+\frac{1}{\sqrt{12345678987654321}}\)
\(B=0,111111111\)
Tính \(A=\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+....+\frac{1}{100.\sqrt{99}+99.\sqrt{100}}\)
So sánh:a)Asqrt[]{21}+sqrt{42}+sqrt{63}Bsqrt{1}+sqrt{2}+sqrt{3}+sqrt{20}+sqrt{40}+sqrt{60}b)Aleft(1-frac{1}{sqrt{4}}right)left(1-frac{1}{sqrt{16}}right)left(1-frac{1}{sqrt{100}}right)Bsqrt{0,1}c) Afrac{1}{sqrt{1}}+frac{1}{sqrt{2}}+...+frac{1}{sqrt{100}}B10
Đọc tiếp
So sánh:
a)\(A=\sqrt[]{21}+\sqrt{42}+\sqrt{63}\)
\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
b)\(A=\left(1-\frac{1}{\sqrt{4}}\right)\left(1-\frac{1}{\sqrt{16}}\right)\left(1-\frac{1}{\sqrt{100}}\right)\)
\(B=\sqrt{0,1}\)
c) \(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}\)
\(B=10\)
Cho A=\(\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{2+\sqrt{x}}{x-5\sqrt{x}+6}+\frac{3+\sqrt{x}}{\sqrt{x}-2}-\frac{2+\sqrt{x}}{\sqrt{x}-3}\right)\)
Tìm tập xác định và rút gọn A
TÍnh giá trị biểu thức sau:
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+........+\frac{1}{1000\sqrt{999}+999\sqrt{1000}}\)
Chứng minh rằng a,\(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}< 24\)
b,\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
Rút gọn : A=\(\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{2+\sqrt{x}}{x-5\sqrt{x}+6}+\frac{3+\sqrt{x}}{\sqrt{x}-2}-\frac{2+\sqrt{x}}{\sqrt{x}-3}\right)\)