bn ơi thiếu đề hả?

a: ĐKXĐ: \(\left\{{}\begin{matrix}y\ge0\\y\ne1\end{matrix}\right.\)

Ta có: \(P=\left(\dfrac{1}{1-\sqrt{y}}+\dfrac{1}{1+\sqrt{y}}\right):\left(\dfrac{1}{1-\sqrt{y}}-\dfrac{1}{1+\sqrt{y}}\right)+\dfrac{1}{1-\sqrt{y}}\)

\(=\dfrac{1+\sqrt{y}+1-\sqrt{y}}{\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)}:\dfrac{1+\sqrt{y}-1+\sqrt{y}}{\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)}+\dfrac{1}{1-\sqrt{y}}\)

\(=\dfrac{2}{2\sqrt{y}}-\dfrac{1}{\sqrt{y}-1}\)

\(=\dfrac{\sqrt{y}-1-\sqrt{y}}{\sqrt{y}\left(\sqrt{y}-1\right)}\)

\(=\dfrac{-1}{\sqrt{y}\left(\sqrt{y}-1\right)}\)

a: \(Q=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)

b: Khi x=4+2căn 3 thì \(Q=\dfrac{\sqrt{3}+1-2}{\sqrt{3}+1+2}=\dfrac{-3+2\sqrt{3}}{3}\)

c: Q=3

=>3căn x+6=căn x-2

=>2căn x=-8(loại)

d: Q>1/2

=>Q-1/2>0

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{1}{2}>0\)

=>2căn x-4-căn x-2>0

=>căn x>6

=>x>36

d: Q nguyên

=>căn x+2-4 chia hết cho căn x+2

=>căn x+2 thuộc Ư(-4)

=>căn x+2 thuộc {2;4}

=>x=0 hoặc x=4(nhận)

Có :

\(x=\dfrac{1}{\sqrt{5}-2}\Rightarrow x^2=\dfrac{1}{\left(\sqrt{5}-2\right)^2}=\dfrac{1}{5-4\sqrt{5}+4}\\ =\dfrac{1}{9-4\sqrt{5}}\\ y=\dfrac{1}{5+4\sqrt{5}}=\dfrac{1}{5+4\sqrt{5}+2}=\dfrac{1}{\left(\sqrt{5}+2\right)^2}\\ \Rightarrow\sqrt{y}=\sqrt{\dfrac{1}{\left(\sqrt{5}+2\right)^2}}=\dfrac{1}{\sqrt{5}+2}\) 

\(\Rightarrow A=\dfrac{1}{9-4\sqrt{5}}-3.\dfrac{1}{\sqrt{5}-2}.\dfrac{1}{\sqrt{5}+2}+\dfrac{2}{9+4\sqrt{5}}\\ =\dfrac{1}{9-4\sqrt{5}}-\dfrac{3}{5-4}+\dfrac{2}{9+4\sqrt{5}}\\ =\dfrac{9+\sqrt{5}+2\left(9-4\sqrt{5}\right)}{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}-3=\dfrac{27-4\sqrt{5}}{81-80-3}\\ =27-4\sqrt{5}-3=24-4\sqrt{5}\)

a) Ta có:

\(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}=-\sqrt{n}+\sqrt{n+1}\)

\(\Rightarrow A=...=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{48}+\sqrt{49}=-1+7=6\)

Ta có :

\(A=\sqrt{y}-\sqrt{x}=\sqrt{4+2\sqrt{3}}-\sqrt{3}=\sqrt{3+2\sqrt{3}+1}-\sqrt{3}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{3}=\sqrt{3}+1-\sqrt{3}=1\)

a: \(Q=\dfrac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\)

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

b: \(Q=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}+\sqrt{2}-1+1}=\dfrac{2\sqrt{2}-1}{7}\)

Lời giải:

a) ĐK: \(x>0; y> 0\)

\(P=\frac{(\sqrt{x}-\sqrt{y})^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)

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

\(=\frac{x+2\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}-(\sqrt{x}-\sqrt{y})\)

\(=\frac{(\sqrt{x}+\sqrt{y})^2}{\sqrt{x}+\sqrt{y}}-(\sqrt{x}-\sqrt{y})=(\sqrt{x}+\sqrt{y})-(\sqrt{x}-\sqrt{y})=2\sqrt{y}\)

b)

Khi \(y=4-2\sqrt{3}=3+1-2\sqrt{3.1}=(\sqrt{3}-1)^2\)

\(\Rightarrow \sqrt{y}=\sqrt{3}-1\)

\(\Rightarrow P=2\sqrt{y}=2(\sqrt{3}-1)\)

ĐK: y>0,y khác 1

a, \(M=\frac{y\sqrt{y}}{\sqrt{y}\left(\sqrt{y}-1\right)}-\frac{2y-\sqrt{y}}{\sqrt{y}\left(\sqrt{y}-1\right)}=\frac{y\sqrt{y}-2y+\sqrt{y}}{\sqrt{y}\left(\sqrt{y}-1\right)}\)

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

b, Thay y vào M ta dc: \(M=\sqrt{3+\sqrt{8}}-1=\sqrt{2+2\sqrt{2}.1+1}-1\)

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

\(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{2-\sqrt{3}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(3+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{9-3}\)

\(\dfrac{x}{\sqrt{2}}=\dfrac{3+\sqrt{3}+3-\sqrt{3}}{6}=\dfrac{6}{6}=1\)

\(x=\sqrt{2}\)

\(y=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

\(y\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)

\(y\sqrt{2}=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)

\(y\sqrt{2}=\sqrt{7}+1-\sqrt{7}+1\)

\(y\sqrt{2}=2\)

\(y=\dfrac{2}{\sqrt{2}}\)

Thay \(x=\sqrt{2},y=\dfrac{2}{\sqrt{2}}\) vào A ta có:

\(A=\dfrac{\sqrt{2}.\dfrac{2}{\sqrt{2}}-1}{\sqrt{2}+\dfrac{2}{\sqrt{2}}}-\dfrac{1-\sqrt{2}.\dfrac{2}{\sqrt{2}}}{2\sqrt{2}-\dfrac{2}{\sqrt{2}}}\)

\(=\dfrac{2-1}{2\sqrt{2}}-\dfrac{1-2}{\sqrt{2}}\)

\(=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{\sqrt{2}}\)

\(=\dfrac{3\sqrt{2}}{4}\)

Tự kết luận nha