rút gọn
a. \(\sqrt{\left(-3\right)^2\cdot5}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
b. \(\sqrt{\left(a-b\right)^2}-\sqrt{\left(b-c\right)^2}-\sqrt{\left(c-d\right)^2}\) với a <b <c <d
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 +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
chứng tỏ
\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)=1
cho a,b,c,x,y,z>0
CMR \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)⇔\(\sqrt{ax}+\sqrt{by}+\sqrt{cz}=\sqrt{\left(a+b+c\right)\left(x+y+z\right)}\)
giúp mình với mình cần gấp lắm =))))))
Cho a,b > 0, a\(\ne\)b
C/m : \(\frac{a+b}{2}>\frac{\left(a-b\right)^2}{4\left(\sqrt{a}-\sqrt{b}\right)}>\sqrt{ab}\)
a) chứng minh: \(\sqrt{a^2}+\sqrt{b^2}>\sqrt{\left(a+b\right)^2}\)
b) Tìm min của A=\(\sqrt{\left(2021-x\right)^2}+\sqrt{\left(2022-x\right)^2}\)
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.
+) \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\).
Cho các số thực dương a,b,c. Chứng minh rằng:
\(\sqrt{\left(a^2b+b^2c+c^2a\right)\left(ab^2+bc^2+ca^2\right)}\ge abc+\sqrt[3]{\left(a^3+abc\right)\left(b^3+abc\right)\left(c^3+abc\right)}\)
Cho a,b,c là các số thực thỏa mãn \(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\)
CM \(\dfrac{\sqrt{a}}{1+a}+\dfrac{\sqrt{b}}{1+b}+\dfrac{\sqrt{c}}{1+c}=\dfrac{2}{\sqrt{\left(1+a\right).\left(1+b\right)\left(1+c\right)}}\)
