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16 tháng 9 2017

mình mới học lớp 6 nên không biết

16 tháng 9 2017

mày xem sẽ chưa tao xem rồi đó về mà xem

17 tháng 10 2019
https://i.imgur.com/zmqmZ1u.jpg
17 tháng 9 2017

a) \(\sqrt{\frac{2}{3}}=\sqrt{\frac{2.3}{3.3}}=\frac{\sqrt{6}}{3}\)

b) \(\sqrt{\frac{x^2}{2}}=\sqrt{\frac{2x^2}{2.2}}=\frac{x\sqrt{2}}{2}\)

c) \(\sqrt{\frac{x}{y}}=\sqrt{\frac{xy}{y.y}}=\frac{\sqrt{xy}}{y}\)

d) \(\sqrt{\frac{9x^3}{25y}}=\sqrt{\frac{9x^3.y}{25y^2}}=\frac{3x\sqrt{xy}}{5y}\)

NM
3 tháng 8 2021

a. \(\sqrt{\frac{y}{5x^3}}=\sqrt{\frac{5xy}{25x^4}}=\frac{\sqrt{5xy}}{25x^2}\)

b\(\sqrt{\frac{5}{x\left(1-\sqrt{2}\right)}}=\sqrt{\frac{5\times x\left(1+\sqrt{2}\right)}{x^2\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}}=\sqrt{\frac{-5\times x\left(1+\sqrt{2}\right)}{x^2}}=-\frac{\sqrt{-5\times x\left(1+\sqrt{2}\right)}}{x}\)

c.\(\sqrt{\frac{x-1}{2\left(\sqrt{x}-1\right)}}=\sqrt{\frac{\sqrt{x}+1}{2}}=\frac{\sqrt{2\sqrt{x}+2}}{2}\)

d.\(a\sqrt{\frac{4}{a}}=\sqrt{\frac{4a^2}{a}}=\sqrt{4a}=2\sqrt{a}\)

e.\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\left(\text{ do a< 0}\right)\)\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\)( do a <0)

f.\(\sqrt{\frac{2}{x-1}-\frac{1}{\left(x-1\right)^2}}=\sqrt{\frac{2\left(x-1\right)-1}{\left(x-1\right)^2}}=\frac{\sqrt{2x-3}}{\left|x-1\right|}\)

11 tháng 10 2020

a) \(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}=\frac{1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}=\frac{2\sqrt{x}}{x-1}\)( x > 0 ; x ≠ 1 )

b) \(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{x-4}\)

\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)( x > 0 ; x ≠ 4 )

11 tháng 10 2020

a) Với \(x>0\)và \(x\ne1\)ta có:

\(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}+1\)

\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{\sqrt{x}+1+\sqrt{x}-1+x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

b) Với \(x>0\)và \(x\ne4\)ta có: 

\(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{x-4}\)

\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(\sqrt{x}-2\right)-2\left(\sqrt{x}+2\right)+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)

a: \(=-xy\cdot\dfrac{\sqrt{xy}}{x}=-y\sqrt{yx}\)

b: \(=\sqrt{\dfrac{-105x^3}{35^2}}=\sqrt{-105x}\cdot\dfrac{x}{35}\)

c: \(=\sqrt{\dfrac{5a^3b}{49b^2}}=\sqrt{5ab}\cdot\dfrac{a}{7b}\)

d: \(=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3}\cdot\sqrt{xy}\)

19 tháng 10 2017

áp dụng bdt cauchy -schửat dạng engel ta có 

\(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{x+z}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)\(\ge\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2}=\frac{1}{2}\)

(do \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\) bn tự cm nhé)

dau = xay ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)

3 tháng 8 2020

Ta có : \(P=\frac{\frac{\left(x-y\right)^3}{\left(\sqrt{x}+\sqrt{y}\right)^3}+2x\sqrt{x}+y\sqrt{y}}{x\sqrt{x}+y\sqrt{y}}+\frac{3\left(\sqrt{xy}-y\right)}{x-y}\)

=> \(P=\frac{\frac{\left(\sqrt{x}+\sqrt{y}\right)^3\left(\sqrt{x}-\sqrt{y}\right)^3}{\left(\sqrt{x}+\sqrt{y}\right)^3}+2x\sqrt{x}+y\sqrt{y}}{\sqrt{x}^3+\sqrt{y}^3}+\frac{3\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

=> \(P=\frac{\left(\sqrt{x}-\sqrt{y}\right)^3+2x\sqrt{x}+y\sqrt{y}}{\sqrt{x}^3+\sqrt{y}^3}+\frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=> \(P=\frac{x\sqrt{x}-3x\sqrt{y}+3y\sqrt{x}-y\sqrt{y}+2x\sqrt{x}+y\sqrt{y}}{\left(x+y\right)\left(x-\sqrt{xy}+y\right)}+\frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=> \(P=\frac{3x\sqrt{x}-3x\sqrt{y}+3y\sqrt{x}}{\left(x+y\right)\left(x-\sqrt{xy}+y\right)}+\frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=> \(P=\frac{3\sqrt{x}\left(x-\sqrt{xy}+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}+\frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=> \(P=\frac{3\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=> \(P=\frac{3\sqrt{x}+3\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\frac{3\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=3\)