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a) ĐKXĐ : \(x,y\ge0;y\ne1;x+y\ne0\)
\(P=\frac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\frac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\frac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{x\left(1+\sqrt{x}\right)-y\left(1-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{\left(x-y\right)+\left(x\sqrt{x}+y\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+x-\sqrt{xy}+y-xy\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-\sqrt{y}\left(\sqrt{x}+1\right)+y\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1+\sqrt{y}\right)}\)
\(=\frac{\sqrt{x}-\sqrt{y}+x-y\sqrt{x}}{1-\sqrt{y}}=\frac{\sqrt{x}\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)-\sqrt{y}\left(1-\sqrt{y}\right)}{1-\sqrt{y}}\)
\(=\sqrt{x}+\sqrt{xy}+\sqrt{y}\)
Vậy P \(=\sqrt{x}+\sqrt{xy}+\sqrt{y}\)
b) ĐKXĐ : \(x,y\ge0;y\ne1;x+y\ne0\)
\(P=2\Leftrightarrow\) \(\sqrt{x}+\sqrt{xy}+\sqrt{y}=2\) ( * )
\(\Leftrightarrow\sqrt{x}\left(1+\sqrt{y}\right)-\left(\sqrt{y}+1\right)=1\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{y}+1\right)=1\)
Có : \(1+\sqrt{y}\ge1\Rightarrow\sqrt{x}-1\le1\Leftrightarrow0\le x\le4\Rightarrow x=0;1;2;3;4\)
Thay x = 0 ; 1 ; 2 ; 3 ;4 vào ( * )
Ta có các cặp giá trị : x =4 ; y = 0 và x = 2 ; y = 2 ( TM )
Bài 1
a, \(\left(\frac{\sqrt{y}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{y}-1\right)}{\sqrt{y}-1}\right).\sqrt{y}\left(\sqrt{x}-1\right)\)
=\(\left(\sqrt{y}+\sqrt{x}\right).\sqrt{y}\left(\sqrt{x}-1\right)\)
b,\(\sqrt{8+2.2\sqrt{2}+1}-\sqrt{8-2.2\sqrt{2}+1}\)
=\(\sqrt{\left(\sqrt{8}+1\right)^2}-\sqrt{\left(\sqrt{8}-1\right)^2}\)
=\(\sqrt{8}+1-\left(\sqrt{8}-1\right)\)
=2
Bài 2
a, ĐKXĐ : x\(\ge\)0, x\(\pm\)1
b, Q=\(\left(\frac{\sqrt{x}\left(1+\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\left(\frac{\sqrt{x}\left(1+\sqrt{x}\right)+\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\left(\frac{\sqrt{x}+x+\sqrt{x}-x}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}-\frac{3-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{2\sqrt{x}-3+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{3\sqrt{x}-3}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
=\(\frac{-3}{1+\sqrt{x}}\)
c, de Q = 2 => \(\frac{-3}{1+\sqrt{x}}\)=2 =>1+\(\sqrt{x}\)=-6 =>\(\sqrt{x}\)=-7 =>x vô nghiệm
Bài 1: \(\left(\frac{\sqrt{xy}-\sqrt{y}}{\sqrt{x}-1}+\frac{\sqrt{xy}-\sqrt{x}}{\sqrt{y}-1}\right)\cdot\left(\sqrt{xy}-\sqrt{y}\right)\)
\(=\left(\frac{\sqrt{y}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{y}-1\right)}{\sqrt{y}-1}\right)\cdot\left(\sqrt{xy}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{xy}-\sqrt{y}\right)\)
\(\sqrt{9+4\sqrt{2}}-\sqrt{9-4\sqrt{2}}=\sqrt{\left(2\sqrt{2}+1\right)^2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\\ =2\sqrt{2}+1-2\sqrt{2}+1=2\)
Bài 2:
\(Q=\left(\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}\right)+\frac{3-\sqrt{x}}{x-1}\left(ĐK:x\ge0;x\ne1\right)\)
\(=\frac{-\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-x-\sqrt{x}+x-\sqrt{x}+3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{-3}{\sqrt{x}+1}\)
Để Q=2
=> \(\frac{-3}{\sqrt{x}+1}=2\)
\(\Leftrightarrow2\left(\sqrt{x}+1\right)=-3\)
\(\Leftrightarrow2\sqrt{x}+2=-3\)
\(\Leftrightarrow2\sqrt{x}=-5\) (vô lí)
Vậy k có giá trị nào của x thỏa mãn Q=2
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
a) +) Điều kiện : x \(\ge\) 0 ; y \(\ge\) 0 ; y \(\ne\) 1 ; x; y không đồng thời bằng 0
+) \(P=\frac{x\left(\sqrt{x}+1\right)-y\left(1-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{x\sqrt{x}+x-y+y\sqrt{y}-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\left(x\sqrt{x}+y\sqrt{y}\right)+\left(x-y\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+y-\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{x+y-\sqrt{xy}+\sqrt{x}-\sqrt{y}-xy}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(x+\sqrt{x}\right)+\left(y-xy\right)-\left(\sqrt{xy}+\sqrt{y}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(1+\sqrt{x}\right)\sqrt{x}+y\left(1-x\right)-\sqrt{y}\left(\sqrt{x}+1\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\left(1+\sqrt{x}\right)\left(\sqrt{x}+y-y\sqrt{x}-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-y\sqrt{x}\right)+\left(y-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)}=\frac{\sqrt{x}\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)-\sqrt{y}\left(1-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)}\)
\(P=\sqrt{x}\left(1+\sqrt{y}\right)-\sqrt{y}=\sqrt{x}-\sqrt{y}+\sqrt{xy}\)
b) Để P = 2 <=> \(\sqrt{x}-\sqrt{y}+\sqrt{xy}=2\) <=> \(\sqrt{x}+\sqrt{xy}=\sqrt{y}+2\)
<=> \(\left(\sqrt{x}+\sqrt{xy}\right)^2=\left(\sqrt{y}+2\right)^2\)
<=> \(x+xy+2x\sqrt{y}=y+4+4\sqrt{y}\)
<=> \(x+xy-y+\left(2x-4\right)\sqrt{y}=4\)(*)
P = 2 <=> (x; y) thỏa mãn (*)