a) \(\frac{a^2}{bc^3}\cdot\sqrt{\frac{b^2\left(a-b\right)^2}{a^4}}=\frac{a^2}{bc^3}\cdot\frac{b\left(b-a\right)}{a^2}=\frac{b-a}{c^3}\)
b) \(\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
c) \(\sqrt{11-6\sqrt{2}}=\sqrt{3^2-2\cdot3\cdot\sqrt{2}+2}=\sqrt{\left(3-\sqrt{2}\right)^2}=3-\sqrt{2}\)
cho a;b;c>0 : a+b+c=1. c/m 5(a^2+b^2+c^2)-6(a^3+b^3+c^3)\(\le\)1
Với a>0, b>0 chứng minh rằng
a)\(\sqrt{a}+\sqrt{b}>\sqrt{a+b}\)
b)\(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
c)\(\dfrac{\sqrt{a^2+6}}{\sqrt{a^2+5}}>2\)
Cho a,b,c > 0 . Chứng minh : a/b+c + b/c+a + c/a+b ≥3/2
Tính
a) (3+√2) ²
b) (2-√3) ²
c) (√2+√3) ²
d) √a² (a≥0)
e) -2√a⁴ (a<0)
f) √x²-6x+9 (x>3)
g) √4(a-2) ² (a<2)
h) √9(x-5) ⁴
i) √b²(a²+2ab+b²) (b>0)
Cho a,b,c>0 thỏa a+b+c=1
Chứng minh : \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\sqrt[3]{abc}\ge\frac{10}{9\left(a^2+b^2+c^2\right)}\)
a) Với \(n\in N\). Chứng minh:
\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
b) Cho a,b,c > 0. Chứng minh:
+) Nếu \(a+b+c=\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\) thì a = b = c.
+) \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\).
P=\(\dfrac{\left(\sqrt{a^2+a\sqrt{a^2-b^2}}-\sqrt{a^2-a\sqrt{a^2-b^2}}\right)^2}{2\sqrt{a^3b}}:\left(\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}-2\right)\left(a>0,b>0\right)\)
a)rút gọn P
b)tính P biết a=7+\(4\sqrt{3}\) , b=7-\(4\sqrt{3}\)
cho a > 0, b > 0. C/m :
\(\frac{3a^2+2ab+3b^2}{a+b}\ge2\sqrt{2\left(a^2+b^2\right)}\)
a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)
b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)
c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)
d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)
e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
