Thay \(a=b=c=0,25\)thì ta có:
\(\dfrac{1}{\sqrt{0,25}}+\dfrac{1}{\sqrt{0,25}}+\dfrac{2\sqrt{2}}{\sqrt{0,25}}\approx9,657\)
\(\dfrac{8}{0,25+0,25+0,25}\approx10,667\)
Vậy đề sai
a, \(A=\left(\sqrt{2}+1\right)[\left(\sqrt{2}\right)^2+1][(\sqrt{2})^4+1][\left(\sqrt{2}\right)^8+1][1\left(\sqrt{2}\right)^{16}+1]\)
b, \(B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+1\sqrt{2020}}\)
c,\(C=^3\sqrt[]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\)
chứng minh các bát đẳng thức sau
a)Cho a>0 chứng minh rằng \(a+\dfrac{1}{a}\)≥2
b)\(\dfrac{a^2+a+2}{\sqrt{a^2+a+1}}\)≥2
c)\(\sqrt{a+1}-\sqrt{a}< \dfrac{1}{2\sqrt{a}}\)
1. Cho biểu thức: A = \(1+\left(\dfrac{2a+\sqrt{a}-1}{1-a}-\dfrac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right).\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}.\)
a) Rút gọn A.
b) Tìm a để A = \(\dfrac{\sqrt{6}}{1+\sqrt{6}}\).
c) CMR: A \(\ge\dfrac{2}{3}\).
câu 1 rút gọn
A=\(\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}\)
B=\(\dfrac{2}{\sqrt{3}-\sqrt{5}}+\dfrac{3-2\sqrt{3}}{\sqrt{3}-2}\)
C = \(\dfrac{\sqrt{2}+1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8}+2\sqrt{15}}\)
Câu 2 cho pt
B= \(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
a, tìm ĐKXĐ và rút gọn
b, tính B khi x =\(3+2\sqrt{2}\)
c, tìm x để B nguyên
Cho a,b,c>0 và thỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\dfrac{7-abc}{\sqrt{2}}\)
Chứng minh rằng \(M=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{3}{2}\)
cho a,b,c>0 thỏa mãn abc\(\ge1\)
chứng minh rằng
\(\dfrac{a}{\sqrt{b+\sqrt{ac}}}+\dfrac{b}{\sqrt{c+\sqrt{ab}}}+\dfrac{c}{\sqrt{a+\sqrt{bc}}}\ge\dfrac{3}{\sqrt{2}}\)
Cho \(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\) . Chứng minh \(\dfrac{\sqrt{a}}{1+a}+\dfrac{\sqrt{b}}{1+b}+\dfrac{\sqrt{c}}{1+c}=\dfrac{2}{\sqrt{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Rút gọn các biểu thức sau
a,\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
b,\(B=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{2x\sqrt{x}-2x+2\sqrt{x}-3}{x\sqrt{x}+1}\)
c,\(C=\left(1-\dfrac{x+3\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\right)\)
d,\(D=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
e,\(E=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Câu 1: Rút gọn biểu thức
a) \(N=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
b) \(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
Câu 2:
a) Cho a > 0. Chứng minh: \(a+\dfrac{1}{a}\ge2\)
b) Cho \(a\ge0\) , \(b\ge0\) . Chứng minh: \(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\)
c) Cho a, b > 0. Chứng minh: \(\sqrt{a}+\sqrt{b}\le\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\)
d) Chứng minh: \(\dfrac{a^2+2}{\sqrt{a^2+1}}\ge2\) với mọi a
