\(\text{Theo đề bài: }=\dfrac{3\sqrt{2}+6\sqrt{3}+2\sqrt{5}-\sqrt{6}}{2}\)

a: ĐKXĐ: b>=0; b<>1

\(B=\dfrac{1-\sqrt{b}+1+\sqrt{b}}{2\left(1-b\right)}-\dfrac{b^2+1}{1-b^2}\)

\(=\dfrac{1}{1-b}+\dfrac{b^2+1}{b^2-1}\)

\(=\dfrac{-b-1+b^2+1}{b^2-1}=\dfrac{b\left(b-1\right)}{\left(b-1\right)\left(b+1\right)}=\dfrac{b}{b+1}\)

b: B>1/3

=>B-1/3>0

=>b/b+1-1/3>0

=>(3b-b-1)/(3b+3)>0

=>2b-1>0

=>b>1/2

\(=\dfrac{\sqrt{a}+2+\sqrt{a}-2}{a-4}:\dfrac{\sqrt{a}+2-2}{\sqrt{a}+2}\)

\(=\dfrac{2\sqrt{a}}{a-4}\cdot\dfrac{\sqrt{a}+2}{\sqrt{a}}=\dfrac{2}{\sqrt{a}-2}\)

rút gọn à bạn?

\(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^3}+\sqrt{b^3}}{a-b}\)

\(=\dfrac{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{a^3}-\sqrt[]{b^3}}{\left(\sqrt{x}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{a\sqrt{a}-b\sqrt{a}+a\sqrt{b}-b\sqrt{b}-a\sqrt{a}-b\sqrt[]{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{-b\sqrt{a}+\left(a-2b\right)\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

Bo thi:>

+ đk x > 0 , x khác 1

a: \(P=-5\sqrt{\dfrac{160}{90}}=-5\cdot\dfrac{4}{3}=-\dfrac{20}{3}\)

b: \(Q=\sqrt{a}-\sqrt{b}+2\sqrt{b}=\sqrt{a}+\sqrt{b}\)

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)

\(P=\sqrt{\dfrac{yz}{x^2+1}}+\sqrt{\dfrac{zx}{y^2+1}}+\sqrt{\dfrac{xy}{z^2+1}}\)

\(P=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}+\sqrt{\dfrac{zx}{y^2+xy+yz+zx}}+\sqrt{\dfrac{xy}{z^2+xy+yz+zx}}\)

\(P=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{zx}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)

\(P\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right)+\dfrac{1}{2}\left(\dfrac{z}{y+z}+\dfrac{x}{x+y}\right)+\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{3}{2}\)

\(P_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(a=b=c=\sqrt{3}\)

ĐKXĐ: \(a>0;a\ne1\)
\(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\dfrac{a-1}{\sqrt{a}+1}\)
\(=\left[\dfrac{\sqrt{a}.\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right].\dfrac{a-1}{\sqrt{a}+1}\)
\(=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right].\dfrac{a-1}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}-2}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(=\sqrt{a}-2\)