Cho a,b,c > 0. Tìm GTNN:
a, \(A=\dfrac{a^2}{2b+5c}+\dfrac{b^2}{2c+5a}+\dfrac{c^2}{2a+5b}\) với abc = 8
b, \(B=\dfrac{b+c}{a^2}+\dfrac{c+a}{b^2}+\dfrac{a+b}{c^2}\) với abc = 1
c, \(C=\dfrac{a+bc}{b+c}+\dfrac{b+ca}{c+a}+\dfrac{c+ab}{a+b}\) với a + b + c = 1
d, \(D=\dfrac{a^3}{2b+3c}+\dfrac{b^3}{2c+3a}+\dfrac{c^3}{2a+3b}\) với \(a^2+b^2+c^2\ge3\)
Rút gọn và tính giá trị các biểu thức :
a, \(\sqrt{\dfrac{3+\sqrt{5}}{2x^2}}-\sqrt{\dfrac{3-\sqrt{5}}{2}}\left(x>0\right)T\text{ại}:x=1\)
\(b,\dfrac{\sqrt{a^3+4a^2+4a}}{\sqrt{a\left(a^2-2ab+b^2\right)}}-\dfrac{\sqrt{b^3-4b^2+4b}}{\sqrt{b\left(a^2-2ab+b^2\right)}}+ab\) ( a > b > 2 ) tại a = 4 ; b = 3
c, \(ab^2.\sqrt{\dfrac{4}{a^2.b^4}}+ab\left(a;b\ne0;a>0\right)\) Tại a = 1 ; b = - 2
d,\(\dfrac{a+b}{b^2}.\sqrt{\dfrac{a^2b^2}{a^2+2ab+b^2}}\left(a;b>0\right)\) Tại a = 1 ; b = 2
Cho a,b,c >0.Chứng minh:
\(P=\dfrac{a^2b}{ab^2+1}+\dfrac{b^2c}{bc^2+1}+\dfrac{c^2a}{ca^2+1}\ge\dfrac{3abc}{1+abc}\)
Cho a,b,c >0 thỏa a+b+c \(\ge9\)
Tìm Min:
\(P=2\sqrt{a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}}+\sqrt{\dfrac{1}{a}+\dfrac{9}{b}+\dfrac{25}{c}}\)
cho a,b,c>0
CMR:
1) \(a+b+\dfrac{1}{4}\ge\sqrt{a+b}\)
2) \(\left(a+b+\dfrac{1}{2}\right)^2+\left(b+c+\dfrac{1}{2}\right)^2+\left(c+a+\dfrac{1}{2}\right)^2\ge4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\right)\)
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
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,\(E=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a)cho a>b>0 chứng minh rằng : \(\dfrac{1}{a+b}\le\dfrac{1}{2\sqrt{ab}}\)
b) Chứng minh \(\dfrac{\sqrt{2}-\sqrt{1}}{3}+\dfrac{\sqrt{3}-\sqrt{2}}{5}+\dfrac{\sqrt{4}-\sqrt{3}}{7}+...+\dfrac{\sqrt{2011}-\sqrt{2010}}{4021}< \dfrac{1}{2}\)
giúp mk vs
Cho \(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\) . Chứng minh \(\dfrac{\sqrt{a}}{1+a}+\dfrac{\sqrt{b}}{1+b}+\dfrac{\sqrt{c}}{1+c}=\dfrac{2}{\sqrt{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
